Mixing
In how many ways $1000000$ can be expressed as a product of $3$ integers?
$begingroup$
In how many ways $1000000$ can be expressed as a product of $3$ integers where each integer is greater than 1? Here $a times b times c$, $b times a times c...$ are considered to be same.
I've tried many ways. But couldn't solve anyway.
please help me...
number-theory
asked Jan 13 at 1:48
T. A.T. A.
154
$endgroup$
|
show 3 more comments
$begingroup$
In how many ways $1000000$ can be expressed as a product of $3$ integers where each integer is greater than 1? Here $a times b times c$, $b times a times c...$ are considered to be same.
I've tried many ways. But couldn't solve anyway.
please help me...
number-theory
asked Jan 13 at 1:48
T. A.T. A.
154
$endgroup$
5
$begingroup$
Show us the ways you've tried. Then we'll be able to give you relevant advice.
$endgroup$
– saulspatz
Jan 13 at 1:49
2
$begingroup$
$3121$ is not a factor of $1000$. It should be $1000=2^35^3$ and $1000000=2^65^6$
$endgroup$
– Ross Millikan
Jan 13 at 2:03
$begingroup$
Sorry..I have made a mistake. It should be $3125$ and $3125 = 5 times 5 times 5 times 5 times 5$
$endgroup$
– T. A.
Jan 13 at 2:05
$begingroup$
Where does $3121$ come from? That's obvious not a factor of $1000000=10^=2^6cdot5^6$
$endgroup$
– saulspatz
Jan 13 at 2:06
$begingroup$
$1000000 = 2 times 2 times 2 times 2 times 2 times 2 times 5 times 5 times 5 times 5 times 5 times 5$. I tried to find out in how many ways we can create three groups using 2,2,2,2,2,2,5,5,5,5,5,5. But failed to do it.
$endgroup$
– T. A.
Jan 13 at 2:11
|
show 3 more comments
$begingroup$
In how many ways $1000000$ can be expressed as a product of $3$ integers where each integer is greater than 1? Here $a times b times c$, $b times a times c...$ are considered to be same.
I've tried many ways. But couldn't solve anyway.
please help me...
number-theory
asked Jan 13 at 1:48
T. A.T. A.
154
$endgroup$
In how many ways $1000000$ can be expressed as a product of $3$ integers where each integer is greater than 1? Here $a times b times c$, $b times a times c...$ are considered to be same.
I've tried many ways. But couldn't solve anyway.
please help me...
number-theory
number-theory
asked Jan 13 at 1:48
T. A.T. A.
154
asked Jan 13 at 1:48
T. A.T. A.
154
asked Jan 13 at 1:48
T. A.T. A.
154
asked Jan 13 at 1:48
T. A.T. A.
154
asked Jan 13 at 1:48
T. A.T. A.
154
154
5
$begingroup$
Show us the ways you've tried. Then we'll be able to give you relevant advice.
$endgroup$
– saulspatz
Jan 13 at 1:49
2
$begingroup$
$3121$ is not a factor of $1000$. It should be $1000=2^35^3$ and $1000000=2^65^6$
$endgroup$
– Ross Millikan
Jan 13 at 2:03
$begingroup$
Sorry..I have made a mistake. It should be $3125$ and $3125 = 5 times 5 times 5 times 5 times 5$
$endgroup$
– T. A.
Jan 13 at 2:05
$begingroup$
Where does $3121$ come from? That's obvious not a factor of $1000000=10^=2^6cdot5^6$
$endgroup$
– saulspatz
Jan 13 at 2:06
$begingroup$
$1000000 = 2 times 2 times 2 times 2 times 2 times 2 times 5 times 5 times 5 times 5 times 5 times 5$. I tried to find out in how many ways we can create three groups using 2,2,2,2,2,2,5,5,5,5,5,5. But failed to do it.
$endgroup$
– T. A.
Jan 13 at 2:11
|
show 3 more comments
5
$begingroup$
Show us the ways you've tried. Then we'll be able to give you relevant advice.
$endgroup$
– saulspatz
Jan 13 at 1:49
2
$begingroup$
$3121$ is not a factor of $1000$. It should be $1000=2^35^3$ and $1000000=2^65^6$
$endgroup$
– Ross Millikan
Jan 13 at 2:03
$begingroup$
Sorry..I have made a mistake. It should be $3125$ and $3125 = 5 times 5 times 5 times 5 times 5$
$endgroup$
– T. A.
Jan 13 at 2:05
$begingroup$
Where does $3121$ come from? That's obvious not a factor of $1000000=10^=2^6cdot5^6$
$endgroup$
– saulspatz
Jan 13 at 2:06
$begingroup$
$1000000 = 2 times 2 times 2 times 2 times 2 times 2 times 5 times 5 times 5 times 5 times 5 times 5$. I tried to find out in how many ways we can create three groups using 2,2,2,2,2,2,5,5,5,5,5,5. But failed to do it.
$endgroup$
– T. A.
Jan 13 at 2:11
5
5
$begingroup$
Show us the ways you've tried. Then we'll be able to give you relevant advice.
$endgroup$
– saulspatz
Jan 13 at 1:49
$begingroup$
Show us the ways you've tried. Then we'll be able to give you relevant advice.
$endgroup$
– saulspatz
Jan 13 at 1:49
2
2
$begingroup$
$3121$ is not a factor of $1000$. It should be $1000=2^35^3$ and $1000000=2^65^6$
$endgroup$
– Ross Millikan
Jan 13 at 2:03
$begingroup$
$3121$ is not a factor of $1000$. It should be $1000=2^35^3$ and $1000000=2^65^6$
$endgroup$
– Ross Millikan
Jan 13 at 2:03
$begingroup$
Sorry..I have made a mistake. It should be $3125$ and $3125 = 5 times 5 times 5 times 5 times 5$
$endgroup$
– T. A.
Jan 13 at 2:05
$begingroup$
Sorry..I have made a mistake. It should be $3125$ and $3125 = 5 times 5 times 5 times 5 times 5$
$endgroup$
– T. A.
Jan 13 at 2:05
$begingroup$
Where does $3121$ come from? That's obvious not a factor of $1000000=10^=2^6cdot5^6$
$endgroup$
– saulspatz
Jan 13 at 2:06
$begingroup$
Where does $3121$ come from? That's obvious not a factor of $1000000=10^=2^6cdot5^6$
$endgroup$
– saulspatz
Jan 13 at 2:06
$begingroup$
$1000000 = 2 times 2 times 2 times 2 times 2 times 2 times 5 times 5 times 5 times 5 times 5 times 5$. I tried to find out in how many ways we can create three groups using 2,2,2,2,2,2,5,5,5,5,5,5. But failed to do it.
$endgroup$
– T. A.
Jan 13 at 2:11
$begingroup$
$1000000 = 2 times 2 times 2 times 2 times 2 times 2 times 5 times 5 times 5 times 5 times 5 times 5$. I tried to find out in how many ways we can create three groups using 2,2,2,2,2,2,5,5,5,5,5,5. But failed to do it.
$endgroup$
– T. A.
Jan 13 at 2:11
|
show 3 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Given number is $2^6times 5^6$. Any factor of it of the form $2^xtimes 5^y$ with both $x,y$ not exceeding 6, possibly zero.
We want $$2^6times 5^6= 2^{x_1}times 5^{y_1}times 2^{x_2}times 5^{y_2}times 2^{x_3}times 5^{y_3}= 2^{x_1+x_2+x_3}times 5^{y_1+y_2+y_3}$$
Now it is clear that we have to find $0le x_i,y_ile 6$ such that $x_1+x_2+x_3=6 $ and $y_1+y_2+y_3=6$. The number of solutions is the answer. This is not difficult as the problem is only about addition and that too with numbers upto 6.
answered Jan 13 at 2:19
P VanchinathanP Vanchinathan
15.1k12136
$endgroup$
$begingroup$
$x_1 + x_2 + x_3 = 6$ can be formed in $7!$ ways. $y_1 + y_2 + y_3 = 6$ can also be formed in $7!$ ways. So the answer is $(7!)^2$. Isn't it?
$endgroup$
– T. A.
Jan 13 at 2:38
2
$begingroup$
@T.A. $x_1+x_2+x_3=6$ will not have $7!$ solutions. It will have $binom{8}{2}=28$ solutions. Hint: stars and bars.
$endgroup$
– Anurag A
Jan 13 at 2:53
$begingroup$
We also need to discard solutions where $x_1 = y_1 = 0$ (and similarly for the two other pairs of exponents) since the problem statement requires all factors to be greater than $1.$
$endgroup$
– David K
Jan 13 at 4:04
$begingroup$
Also, stars and bars gives you factors $atimes btimes c$ and $btimes atimes c$ as two different results if $aneq b.$ The problem statement says these should only count once.
$endgroup$
– David K
Jan 13 at 4:06
$begingroup$
This answer gives a good starting point, however. Next you can figure out how many times unallowed factorizations were counted and how many times allowed factorizations were overcounted, and subtract all of those.
$endgroup$
– David K
Jan 13 at 4:14
add a comment |
$begingroup$
There are seven ascending partitions $(006)$–$(222)$ of $6$ into three $geq0$ parts. We have to choose such a partition for the three exponents $x_1leq x_2leq x_3$ of $2$ and independently a partition for the three exponents $y_j$ of $5$. For each of the $49$ possible combinations we have to find out how many different factor triples we can form by permuting the entries of the chosen $y_j$-partition. If I have done this correctly I obtain the following (symmetric) table, whereby the rows are corresponding to the $(x_1x_2x_3)$ and the columns to the $(y_1y_2y_3)$:
$$matrix{
&006&015&024&033&114&123&222cr
cr
006:quad &0&1&1&1&2&3&1cr
015:quad&1&4&4&2&3&6&1cr
024:quad&1&4&4&2&3&6&1cr
033:quad&1&2&2&1&2&3&1cr
114:quad&2&3&3&2&2&3&1cr
123:quad&3&6&6&3&3&6&1cr
222:quad&1&1&1&1&1&1&1cr}$$
Adding all entries in this table gives $114$.
answered Jan 13 at 17:49
Christian BlatterChristian Blatter
173k7113326
$endgroup$
$begingroup$
OP said the order of factors does not matter
$endgroup$
– Ross Millikan
Jan 14 at 3:26
$begingroup$
I see, I was counting a factor of $1$. So far so good, then.
$endgroup$
– Chris Custer
Jan 14 at 6:57
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@RossMillikan: In my counting the order does not matter. The pairing $(114, 015)$ means that we have $2^1 5^{y_1}cdot 2^1 5^{y_2}cdot2^4 5^{y_3}$ with ${y_1,y_2,y_3}={0,1,5}$.
$endgroup$
– Christian Blatter
Jan 14 at 8:40
add a comment |
$begingroup$
I get $114$. Here's how:
Exponents on the three powers of $2$ can be (in ascending order):
all different
- $0,1,5$
- $0,2,4$
- $1,2,3$
two the same
- $0,0,6$
- $0,3,3$
- $1,1,4$
all three same
- $2,2,2$
and likewise for the three powers of $5$.
So for the "all different powers on the $2$s and all different powers on the $5$s" we have:
Example: pairing $0,1,5$ for $2$s with $0,1,5$ for $5$s:
- $(2^0 5^5) (2^1 5^0) (2^5 5^1)$
- $(2^0 5^0) (2^1 5^5) (2^4 5^1)$
- $(2^0 5^1) (2^1 5^0) (2^4 5^5)$
- etc...
But... as pointed out by @farruhota, the calculation includes cases where there is a factor of $1$ (i.e., $2^0 5^0)$ so we do not include those.
Go through all such cases and find $114$.
Here's a list:
{{2, 2, 250000}, {2, 4, 125000}, {2, 5, 100000}, {2, 8, 62500}, {2,
10, 50000}, {2, 16, 31250}, {2, 20, 25000}, {2, 25, 20000}, {2, 32,
15625}, {2, 40, 12500}, {2, 50, 10000}, {2, 80, 6250}, {2, 100,
5000}, {2, 125, 4000}, {2, 160, 3125}, {2, 200, 2500}, {2, 250,
2000}, {2, 400, 1250}, {2, 500, 1000}, {2, 625, 800}, {4, 4,
62500}, {4, 5, 50000}, {4, 8, 31250}, {4, 10, 25000}, {4, 16,
15625}, {4, 20, 12500}, {4, 25, 10000}, {4, 40, 6250}, {4, 50,
5000}, {4, 80, 3125}, {4, 100, 2500}, {4, 125, 2000}, {4, 200,
1250}, {4, 250, 1000}, {4, 400, 625}, {4, 500, 500}, {5, 5,
40000}, {5, 8, 25000}, {5, 10, 20000}, {5, 16, 12500}, {5, 20,
10000}, {5, 25, 8000}, {5, 32, 6250}, {5, 40, 5000}, {5, 50,
4000}, {5, 64, 3125}, {5, 80, 2500}, {5, 100, 2000}, {5, 125,
1600}, {5, 160, 1250}, {5, 200, 1000}, {5, 250, 800}, {5, 320,
625}, {5, 400, 500}, {8, 8, 15625}, {8, 10, 12500}, {8, 20,
6250}, {8, 25, 5000}, {8, 40, 3125}, {8, 50, 2500}, {8, 100,
1250}, {8, 125, 1000}, {8, 200, 625}, {8, 250, 500}, {10, 10,
10000}, {10, 16, 6250}, {10, 20, 5000}, {10, 25, 4000}, {10, 32,
3125}, {10, 40, 2500}, {10, 50, 2000}, {10, 80, 1250}, {10, 100,
1000}, {10, 125, 800}, {10, 160, 625}, {10, 200, 500}, {10, 250,
400}, {16, 20, 3125}, {16, 25, 2500}, {16, 50, 1250}, {16, 100,
625}, {16, 125, 500}, {16, 250, 250}, {20, 20, 2500}, {20, 25,
2000}, {20, 40, 1250}, {20, 50, 1000}, {20, 80, 625}, {20, 100,
500}, {20, 125, 400}, {20, 200, 250}, {25, 25, 1600}, {25, 32,
1250}, {25, 40, 1000}, {25, 50, 800}, {25, 64, 625}, {25, 80,
500}, {25, 100, 400}, {25, 125, 320}, {25, 160, 250}, {25, 200,
200}, {32, 50, 625}, {32, 125, 250}, {40, 40, 625}, {40, 50,
500}, {40, 100, 250}, {40, 125, 200}, {50, 50, 400}, {50, 80,
250}, {50, 100, 200}, {50, 125, 160}, {64, 125, 125}, {80, 100,
125}, {100, 100, 100}}
answered Jan 13 at 5:41
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
But the choices with $(2^05^0)$ must be ignored, because by requirement each integer must be greater than $1$.
$endgroup$
– farruhota
Jan 13 at 6:55
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Given number is $2^6times 5^6$. Any factor of it of the form $2^xtimes 5^y$ with both $x,y$ not exceeding 6, possibly zero.
We want $$2^6times 5^6= 2^{x_1}times 5^{y_1}times 2^{x_2}times 5^{y_2}times 2^{x_3}times 5^{y_3}= 2^{x_1+x_2+x_3}times 5^{y_1+y_2+y_3}$$
Now it is clear that we have to find $0le x_i,y_ile 6$ such that $x_1+x_2+x_3=6 $ and $y_1+y_2+y_3=6$. The number of solutions is the answer. This is not difficult as the problem is only about addition and that too with numbers upto 6.
answered Jan 13 at 2:19
P VanchinathanP Vanchinathan
15.1k12136
$endgroup$
$begingroup$
$x_1 + x_2 + x_3 = 6$ can be formed in $7!$ ways. $y_1 + y_2 + y_3 = 6$ can also be formed in $7!$ ways. So the answer is $(7!)^2$. Isn't it?
$endgroup$
– T. A.
Jan 13 at 2:38
2
$begingroup$
@T.A. $x_1+x_2+x_3=6$ will not have $7!$ solutions. It will have $binom{8}{2}=28$ solutions. Hint: stars and bars.
$endgroup$
– Anurag A
Jan 13 at 2:53
$begingroup$
We also need to discard solutions where $x_1 = y_1 = 0$ (and similarly for the two other pairs of exponents) since the problem statement requires all factors to be greater than $1.$
$endgroup$
– David K
Jan 13 at 4:04
$begingroup$
Also, stars and bars gives you factors $atimes btimes c$ and $btimes atimes c$ as two different results if $aneq b.$ The problem statement says these should only count once.
$endgroup$
– David K
Jan 13 at 4:06
$begingroup$
This answer gives a good starting point, however. Next you can figure out how many times unallowed factorizations were counted and how many times allowed factorizations were overcounted, and subtract all of those.
$endgroup$
– David K
Jan 13 at 4:14
add a comment |
$begingroup$
Given number is $2^6times 5^6$. Any factor of it of the form $2^xtimes 5^y$ with both $x,y$ not exceeding 6, possibly zero.
We want $$2^6times 5^6= 2^{x_1}times 5^{y_1}times 2^{x_2}times 5^{y_2}times 2^{x_3}times 5^{y_3}= 2^{x_1+x_2+x_3}times 5^{y_1+y_2+y_3}$$
Now it is clear that we have to find $0le x_i,y_ile 6$ such that $x_1+x_2+x_3=6 $ and $y_1+y_2+y_3=6$. The number of solutions is the answer. This is not difficult as the problem is only about addition and that too with numbers upto 6.
answered Jan 13 at 2:19
P VanchinathanP Vanchinathan
15.1k12136
$endgroup$
$begingroup$
$x_1 + x_2 + x_3 = 6$ can be formed in $7!$ ways. $y_1 + y_2 + y_3 = 6$ can also be formed in $7!$ ways. So the answer is $(7!)^2$. Isn't it?
$endgroup$
– T. A.
Jan 13 at 2:38
2
$begingroup$
@T.A. $x_1+x_2+x_3=6$ will not have $7!$ solutions. It will have $binom{8}{2}=28$ solutions. Hint: stars and bars.
$endgroup$
– Anurag A
Jan 13 at 2:53
$begingroup$
We also need to discard solutions where $x_1 = y_1 = 0$ (and similarly for the two other pairs of exponents) since the problem statement requires all factors to be greater than $1.$
$endgroup$
– David K
Jan 13 at 4:04
$begingroup$
Also, stars and bars gives you factors $atimes btimes c$ and $btimes atimes c$ as two different results if $aneq b.$ The problem statement says these should only count once.
$endgroup$
– David K
Jan 13 at 4:06
$begingroup$
This answer gives a good starting point, however. Next you can figure out how many times unallowed factorizations were counted and how many times allowed factorizations were overcounted, and subtract all of those.
$endgroup$
– David K
Jan 13 at 4:14
add a comment |
$begingroup$
Given number is $2^6times 5^6$. Any factor of it of the form $2^xtimes 5^y$ with both $x,y$ not exceeding 6, possibly zero.
We want $$2^6times 5^6= 2^{x_1}times 5^{y_1}times 2^{x_2}times 5^{y_2}times 2^{x_3}times 5^{y_3}= 2^{x_1+x_2+x_3}times 5^{y_1+y_2+y_3}$$
Now it is clear that we have to find $0le x_i,y_ile 6$ such that $x_1+x_2+x_3=6 $ and $y_1+y_2+y_3=6$. The number of solutions is the answer. This is not difficult as the problem is only about addition and that too with numbers upto 6.
answered Jan 13 at 2:19
P VanchinathanP Vanchinathan
15.1k12136
$endgroup$
Given number is $2^6times 5^6$. Any factor of it of the form $2^xtimes 5^y$ with both $x,y$ not exceeding 6, possibly zero.
We want $$2^6times 5^6= 2^{x_1}times 5^{y_1}times 2^{x_2}times 5^{y_2}times 2^{x_3}times 5^{y_3}= 2^{x_1+x_2+x_3}times 5^{y_1+y_2+y_3}$$
Now it is clear that we have to find $0le x_i,y_ile 6$ such that $x_1+x_2+x_3=6 $ and $y_1+y_2+y_3=6$. The number of solutions is the answer. This is not difficult as the problem is only about addition and that too with numbers upto 6.
answered Jan 13 at 2:19
P VanchinathanP Vanchinathan
15.1k12136
answered Jan 13 at 2:19
P VanchinathanP Vanchinathan
15.1k12136
answered Jan 13 at 2:19
P VanchinathanP Vanchinathan
15.1k12136
answered Jan 13 at 2:19
P VanchinathanP Vanchinathan
15.1k12136
15.1k12136
$begingroup$
$x_1 + x_2 + x_3 = 6$ can be formed in $7!$ ways. $y_1 + y_2 + y_3 = 6$ can also be formed in $7!$ ways. So the answer is $(7!)^2$. Isn't it?
$endgroup$
– T. A.
Jan 13 at 2:38
2
$begingroup$
@T.A. $x_1+x_2+x_3=6$ will not have $7!$ solutions. It will have $binom{8}{2}=28$ solutions. Hint: stars and bars.
$endgroup$
– Anurag A
Jan 13 at 2:53
$begingroup$
We also need to discard solutions where $x_1 = y_1 = 0$ (and similarly for the two other pairs of exponents) since the problem statement requires all factors to be greater than $1.$
$endgroup$
– David K
Jan 13 at 4:04
$begingroup$
Also, stars and bars gives you factors $atimes btimes c$ and $btimes atimes c$ as two different results if $aneq b.$ The problem statement says these should only count once.
$endgroup$
– David K
Jan 13 at 4:06
$begingroup$
This answer gives a good starting point, however. Next you can figure out how many times unallowed factorizations were counted and how many times allowed factorizations were overcounted, and subtract all of those.
$endgroup$
– David K
Jan 13 at 4:14
add a comment |
$begingroup$
$x_1 + x_2 + x_3 = 6$ can be formed in $7!$ ways. $y_1 + y_2 + y_3 = 6$ can also be formed in $7!$ ways. So the answer is $(7!)^2$. Isn't it?
$endgroup$
– T. A.
Jan 13 at 2:38
2
$begingroup$
@T.A. $x_1+x_2+x_3=6$ will not have $7!$ solutions. It will have $binom{8}{2}=28$ solutions. Hint: stars and bars.
$endgroup$
– Anurag A
Jan 13 at 2:53
$begingroup$
We also need to discard solutions where $x_1 = y_1 = 0$ (and similarly for the two other pairs of exponents) since the problem statement requires all factors to be greater than $1.$
$endgroup$
– David K
Jan 13 at 4:04
$begingroup$
Also, stars and bars gives you factors $atimes btimes c$ and $btimes atimes c$ as two different results if $aneq b.$ The problem statement says these should only count once.
$endgroup$
– David K
Jan 13 at 4:06
$begingroup$
This answer gives a good starting point, however. Next you can figure out how many times unallowed factorizations were counted and how many times allowed factorizations were overcounted, and subtract all of those.
$endgroup$
– David K
Jan 13 at 4:14
$begingroup$
$x_1 + x_2 + x_3 = 6$ can be formed in $7!$ ways. $y_1 + y_2 + y_3 = 6$ can also be formed in $7!$ ways. So the answer is $(7!)^2$. Isn't it?
$endgroup$
– T. A.
Jan 13 at 2:38
$begingroup$
$x_1 + x_2 + x_3 = 6$ can be formed in $7!$ ways. $y_1 + y_2 + y_3 = 6$ can also be formed in $7!$ ways. So the answer is $(7!)^2$. Isn't it?
$endgroup$
– T. A.
Jan 13 at 2:38
2
2
$begingroup$
@T.A. $x_1+x_2+x_3=6$ will not have $7!$ solutions. It will have $binom{8}{2}=28$ solutions. Hint: stars and bars.
$endgroup$
– Anurag A
Jan 13 at 2:53
$begingroup$
@T.A. $x_1+x_2+x_3=6$ will not have $7!$ solutions. It will have $binom{8}{2}=28$ solutions. Hint: stars and bars.
$endgroup$
– Anurag A
Jan 13 at 2:53
$begingroup$
We also need to discard solutions where $x_1 = y_1 = 0$ (and similarly for the two other pairs of exponents) since the problem statement requires all factors to be greater than $1.$
$endgroup$
– David K
Jan 13 at 4:04
$begingroup$
We also need to discard solutions where $x_1 = y_1 = 0$ (and similarly for the two other pairs of exponents) since the problem statement requires all factors to be greater than $1.$
$endgroup$
– David K
Jan 13 at 4:04
$begingroup$
Also, stars and bars gives you factors $atimes btimes c$ and $btimes atimes c$ as two different results if $aneq b.$ The problem statement says these should only count once.
$endgroup$
– David K
Jan 13 at 4:06
$begingroup$
Also, stars and bars gives you factors $atimes btimes c$ and $btimes atimes c$ as two different results if $aneq b.$ The problem statement says these should only count once.
$endgroup$
– David K
Jan 13 at 4:06
$begingroup$
This answer gives a good starting point, however. Next you can figure out how many times unallowed factorizations were counted and how many times allowed factorizations were overcounted, and subtract all of those.
$endgroup$
– David K
Jan 13 at 4:14
$begingroup$
This answer gives a good starting point, however. Next you can figure out how many times unallowed factorizations were counted and how many times allowed factorizations were overcounted, and subtract all of those.
$endgroup$
– David K
Jan 13 at 4:14
add a comment |
$begingroup$
There are seven ascending partitions $(006)$–$(222)$ of $6$ into three $geq0$ parts. We have to choose such a partition for the three exponents $x_1leq x_2leq x_3$ of $2$ and independently a partition for the three exponents $y_j$ of $5$. For each of the $49$ possible combinations we have to find out how many different factor triples we can form by permuting the entries of the chosen $y_j$-partition. If I have done this correctly I obtain the following (symmetric) table, whereby the rows are corresponding to the $(x_1x_2x_3)$ and the columns to the $(y_1y_2y_3)$:
$$matrix{
&006&015&024&033&114&123&222cr
cr
006:quad &0&1&1&1&2&3&1cr
015:quad&1&4&4&2&3&6&1cr
024:quad&1&4&4&2&3&6&1cr
033:quad&1&2&2&1&2&3&1cr
114:quad&2&3&3&2&2&3&1cr
123:quad&3&6&6&3&3&6&1cr
222:quad&1&1&1&1&1&1&1cr}$$
Adding all entries in this table gives $114$.
answered Jan 13 at 17:49
Christian BlatterChristian Blatter
173k7113326
$endgroup$
$begingroup$
OP said the order of factors does not matter
$endgroup$
– Ross Millikan
Jan 14 at 3:26
$begingroup$
I see, I was counting a factor of $1$. So far so good, then.
$endgroup$
– Chris Custer
Jan 14 at 6:57
$begingroup$
@RossMillikan: In my counting the order does not matter. The pairing $(114, 015)$ means that we have $2^1 5^{y_1}cdot 2^1 5^{y_2}cdot2^4 5^{y_3}$ with ${y_1,y_2,y_3}={0,1,5}$.
$endgroup$
– Christian Blatter
Jan 14 at 8:40
add a comment |
$begingroup$
There are seven ascending partitions $(006)$–$(222)$ of $6$ into three $geq0$ parts. We have to choose such a partition for the three exponents $x_1leq x_2leq x_3$ of $2$ and independently a partition for the three exponents $y_j$ of $5$. For each of the $49$ possible combinations we have to find out how many different factor triples we can form by permuting the entries of the chosen $y_j$-partition. If I have done this correctly I obtain the following (symmetric) table, whereby the rows are corresponding to the $(x_1x_2x_3)$ and the columns to the $(y_1y_2y_3)$:
$$matrix{
&006&015&024&033&114&123&222cr
cr
006:quad &0&1&1&1&2&3&1cr
015:quad&1&4&4&2&3&6&1cr
024:quad&1&4&4&2&3&6&1cr
033:quad&1&2&2&1&2&3&1cr
114:quad&2&3&3&2&2&3&1cr
123:quad&3&6&6&3&3&6&1cr
222:quad&1&1&1&1&1&1&1cr}$$
Adding all entries in this table gives $114$.
answered Jan 13 at 17:49
Christian BlatterChristian Blatter
173k7113326
$endgroup$
$begingroup$
OP said the order of factors does not matter
$endgroup$
– Ross Millikan
Jan 14 at 3:26
$begingroup$
I see, I was counting a factor of $1$. So far so good, then.
$endgroup$
– Chris Custer
Jan 14 at 6:57
$begingroup$
@RossMillikan: In my counting the order does not matter. The pairing $(114, 015)$ means that we have $2^1 5^{y_1}cdot 2^1 5^{y_2}cdot2^4 5^{y_3}$ with ${y_1,y_2,y_3}={0,1,5}$.
$endgroup$
– Christian Blatter
Jan 14 at 8:40
add a comment |
$begingroup$
There are seven ascending partitions $(006)$–$(222)$ of $6$ into three $geq0$ parts. We have to choose such a partition for the three exponents $x_1leq x_2leq x_3$ of $2$ and independently a partition for the three exponents $y_j$ of $5$. For each of the $49$ possible combinations we have to find out how many different factor triples we can form by permuting the entries of the chosen $y_j$-partition. If I have done this correctly I obtain the following (symmetric) table, whereby the rows are corresponding to the $(x_1x_2x_3)$ and the columns to the $(y_1y_2y_3)$:
$$matrix{
&006&015&024&033&114&123&222cr
cr
006:quad &0&1&1&1&2&3&1cr
015:quad&1&4&4&2&3&6&1cr
024:quad&1&4&4&2&3&6&1cr
033:quad&1&2&2&1&2&3&1cr
114:quad&2&3&3&2&2&3&1cr
123:quad&3&6&6&3&3&6&1cr
222:quad&1&1&1&1&1&1&1cr}$$
Adding all entries in this table gives $114$.
answered Jan 13 at 17:49
Christian BlatterChristian Blatter
173k7113326
$endgroup$
There are seven ascending partitions $(006)$–$(222)$ of $6$ into three $geq0$ parts. We have to choose such a partition for the three exponents $x_1leq x_2leq x_3$ of $2$ and independently a partition for the three exponents $y_j$ of $5$. For each of the $49$ possible combinations we have to find out how many different factor triples we can form by permuting the entries of the chosen $y_j$-partition. If I have done this correctly I obtain the following (symmetric) table, whereby the rows are corresponding to the $(x_1x_2x_3)$ and the columns to the $(y_1y_2y_3)$:
$$matrix{
&006&015&024&033&114&123&222cr
cr
006:quad &0&1&1&1&2&3&1cr
015:quad&1&4&4&2&3&6&1cr
024:quad&1&4&4&2&3&6&1cr
033:quad&1&2&2&1&2&3&1cr
114:quad&2&3&3&2&2&3&1cr
123:quad&3&6&6&3&3&6&1cr
222:quad&1&1&1&1&1&1&1cr}$$
Adding all entries in this table gives $114$.
answered Jan 13 at 17:49
Christian BlatterChristian Blatter
173k7113326
answered Jan 13 at 17:49
Christian BlatterChristian Blatter
173k7113326
answered Jan 13 at 17:49
Christian BlatterChristian Blatter
173k7113326
answered Jan 13 at 17:49
Christian BlatterChristian Blatter
173k7113326
173k7113326
$begingroup$
OP said the order of factors does not matter
$endgroup$
– Ross Millikan
Jan 14 at 3:26
$begingroup$
I see, I was counting a factor of $1$. So far so good, then.
$endgroup$
– Chris Custer
Jan 14 at 6:57
$begingroup$
@RossMillikan: In my counting the order does not matter. The pairing $(114, 015)$ means that we have $2^1 5^{y_1}cdot 2^1 5^{y_2}cdot2^4 5^{y_3}$ with ${y_1,y_2,y_3}={0,1,5}$.
$endgroup$
– Christian Blatter
Jan 14 at 8:40
add a comment |
$begingroup$
OP said the order of factors does not matter
$endgroup$
– Ross Millikan
Jan 14 at 3:26
$begingroup$
I see, I was counting a factor of $1$. So far so good, then.
$endgroup$
– Chris Custer
Jan 14 at 6:57
$begingroup$
@RossMillikan: In my counting the order does not matter. The pairing $(114, 015)$ means that we have $2^1 5^{y_1}cdot 2^1 5^{y_2}cdot2^4 5^{y_3}$ with ${y_1,y_2,y_3}={0,1,5}$.
$endgroup$
– Christian Blatter
Jan 14 at 8:40
$begingroup$
OP said the order of factors does not matter
$endgroup$
– Ross Millikan
Jan 14 at 3:26
$begingroup$
OP said the order of factors does not matter
$endgroup$
– Ross Millikan
Jan 14 at 3:26
$begingroup$
I see, I was counting a factor of $1$. So far so good, then.
$endgroup$
– Chris Custer
Jan 14 at 6:57
$begingroup$
I see, I was counting a factor of $1$. So far so good, then.
$endgroup$
– Chris Custer
Jan 14 at 6:57
$begingroup$
@RossMillikan: In my counting the order does not matter. The pairing $(114, 015)$ means that we have $2^1 5^{y_1}cdot 2^1 5^{y_2}cdot2^4 5^{y_3}$ with ${y_1,y_2,y_3}={0,1,5}$.
$endgroup$
– Christian Blatter
Jan 14 at 8:40
$begingroup$
@RossMillikan: In my counting the order does not matter. The pairing $(114, 015)$ means that we have $2^1 5^{y_1}cdot 2^1 5^{y_2}cdot2^4 5^{y_3}$ with ${y_1,y_2,y_3}={0,1,5}$.
$endgroup$
– Christian Blatter
Jan 14 at 8:40
add a comment |
$begingroup$
I get $114$. Here's how:
Exponents on the three powers of $2$ can be (in ascending order):
all different
- $0,1,5$
- $0,2,4$
- $1,2,3$
two the same
- $0,0,6$
- $0,3,3$
- $1,1,4$
all three same
- $2,2,2$
and likewise for the three powers of $5$.
So for the "all different powers on the $2$s and all different powers on the $5$s" we have:
Example: pairing $0,1,5$ for $2$s with $0,1,5$ for $5$s:
- $(2^0 5^5) (2^1 5^0) (2^5 5^1)$
- $(2^0 5^0) (2^1 5^5) (2^4 5^1)$
- $(2^0 5^1) (2^1 5^0) (2^4 5^5)$
- etc...
But... as pointed out by @farruhota, the calculation includes cases where there is a factor of $1$ (i.e., $2^0 5^0)$ so we do not include those.
Go through all such cases and find $114$.
Here's a list:
{{2, 2, 250000}, {2, 4, 125000}, {2, 5, 100000}, {2, 8, 62500}, {2,
10, 50000}, {2, 16, 31250}, {2, 20, 25000}, {2, 25, 20000}, {2, 32,
15625}, {2, 40, 12500}, {2, 50, 10000}, {2, 80, 6250}, {2, 100,
5000}, {2, 125, 4000}, {2, 160, 3125}, {2, 200, 2500}, {2, 250,
2000}, {2, 400, 1250}, {2, 500, 1000}, {2, 625, 800}, {4, 4,
62500}, {4, 5, 50000}, {4, 8, 31250}, {4, 10, 25000}, {4, 16,
15625}, {4, 20, 12500}, {4, 25, 10000}, {4, 40, 6250}, {4, 50,
5000}, {4, 80, 3125}, {4, 100, 2500}, {4, 125, 2000}, {4, 200,
1250}, {4, 250, 1000}, {4, 400, 625}, {4, 500, 500}, {5, 5,
40000}, {5, 8, 25000}, {5, 10, 20000}, {5, 16, 12500}, {5, 20,
10000}, {5, 25, 8000}, {5, 32, 6250}, {5, 40, 5000}, {5, 50,
4000}, {5, 64, 3125}, {5, 80, 2500}, {5, 100, 2000}, {5, 125,
1600}, {5, 160, 1250}, {5, 200, 1000}, {5, 250, 800}, {5, 320,
625}, {5, 400, 500}, {8, 8, 15625}, {8, 10, 12500}, {8, 20,
6250}, {8, 25, 5000}, {8, 40, 3125}, {8, 50, 2500}, {8, 100,
1250}, {8, 125, 1000}, {8, 200, 625}, {8, 250, 500}, {10, 10,
10000}, {10, 16, 6250}, {10, 20, 5000}, {10, 25, 4000}, {10, 32,
3125}, {10, 40, 2500}, {10, 50, 2000}, {10, 80, 1250}, {10, 100,
1000}, {10, 125, 800}, {10, 160, 625}, {10, 200, 500}, {10, 250,
400}, {16, 20, 3125}, {16, 25, 2500}, {16, 50, 1250}, {16, 100,
625}, {16, 125, 500}, {16, 250, 250}, {20, 20, 2500}, {20, 25,
2000}, {20, 40, 1250}, {20, 50, 1000}, {20, 80, 625}, {20, 100,
500}, {20, 125, 400}, {20, 200, 250}, {25, 25, 1600}, {25, 32,
1250}, {25, 40, 1000}, {25, 50, 800}, {25, 64, 625}, {25, 80,
500}, {25, 100, 400}, {25, 125, 320}, {25, 160, 250}, {25, 200,
200}, {32, 50, 625}, {32, 125, 250}, {40, 40, 625}, {40, 50,
500}, {40, 100, 250}, {40, 125, 200}, {50, 50, 400}, {50, 80,
250}, {50, 100, 200}, {50, 125, 160}, {64, 125, 125}, {80, 100,
125}, {100, 100, 100}}
answered Jan 13 at 5:41
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
But the choices with $(2^05^0)$ must be ignored, because by requirement each integer must be greater than $1$.
$endgroup$
– farruhota
Jan 13 at 6:55
add a comment |
$begingroup$
I get $114$. Here's how:
Exponents on the three powers of $2$ can be (in ascending order):
all different
- $0,1,5$
- $0,2,4$
- $1,2,3$
two the same
- $0,0,6$
- $0,3,3$
- $1,1,4$
all three same
- $2,2,2$
and likewise for the three powers of $5$.
So for the "all different powers on the $2$s and all different powers on the $5$s" we have:
Example: pairing $0,1,5$ for $2$s with $0,1,5$ for $5$s:
- $(2^0 5^5) (2^1 5^0) (2^5 5^1)$
- $(2^0 5^0) (2^1 5^5) (2^4 5^1)$
- $(2^0 5^1) (2^1 5^0) (2^4 5^5)$
- etc...
But... as pointed out by @farruhota, the calculation includes cases where there is a factor of $1$ (i.e., $2^0 5^0)$ so we do not include those.
Go through all such cases and find $114$.
Here's a list:
{{2, 2, 250000}, {2, 4, 125000}, {2, 5, 100000}, {2, 8, 62500}, {2,
10, 50000}, {2, 16, 31250}, {2, 20, 25000}, {2, 25, 20000}, {2, 32,
15625}, {2, 40, 12500}, {2, 50, 10000}, {2, 80, 6250}, {2, 100,
5000}, {2, 125, 4000}, {2, 160, 3125}, {2, 200, 2500}, {2, 250,
2000}, {2, 400, 1250}, {2, 500, 1000}, {2, 625, 800}, {4, 4,
62500}, {4, 5, 50000}, {4, 8, 31250}, {4, 10, 25000}, {4, 16,
15625}, {4, 20, 12500}, {4, 25, 10000}, {4, 40, 6250}, {4, 50,
5000}, {4, 80, 3125}, {4, 100, 2500}, {4, 125, 2000}, {4, 200,
1250}, {4, 250, 1000}, {4, 400, 625}, {4, 500, 500}, {5, 5,
40000}, {5, 8, 25000}, {5, 10, 20000}, {5, 16, 12500}, {5, 20,
10000}, {5, 25, 8000}, {5, 32, 6250}, {5, 40, 5000}, {5, 50,
4000}, {5, 64, 3125}, {5, 80, 2500}, {5, 100, 2000}, {5, 125,
1600}, {5, 160, 1250}, {5, 200, 1000}, {5, 250, 800}, {5, 320,
625}, {5, 400, 500}, {8, 8, 15625}, {8, 10, 12500}, {8, 20,
6250}, {8, 25, 5000}, {8, 40, 3125}, {8, 50, 2500}, {8, 100,
1250}, {8, 125, 1000}, {8, 200, 625}, {8, 250, 500}, {10, 10,
10000}, {10, 16, 6250}, {10, 20, 5000}, {10, 25, 4000}, {10, 32,
3125}, {10, 40, 2500}, {10, 50, 2000}, {10, 80, 1250}, {10, 100,
1000}, {10, 125, 800}, {10, 160, 625}, {10, 200, 500}, {10, 250,
400}, {16, 20, 3125}, {16, 25, 2500}, {16, 50, 1250}, {16, 100,
625}, {16, 125, 500}, {16, 250, 250}, {20, 20, 2500}, {20, 25,
2000}, {20, 40, 1250}, {20, 50, 1000}, {20, 80, 625}, {20, 100,
500}, {20, 125, 400}, {20, 200, 250}, {25, 25, 1600}, {25, 32,
1250}, {25, 40, 1000}, {25, 50, 800}, {25, 64, 625}, {25, 80,
500}, {25, 100, 400}, {25, 125, 320}, {25, 160, 250}, {25, 200,
200}, {32, 50, 625}, {32, 125, 250}, {40, 40, 625}, {40, 50,
500}, {40, 100, 250}, {40, 125, 200}, {50, 50, 400}, {50, 80,
250}, {50, 100, 200}, {50, 125, 160}, {64, 125, 125}, {80, 100,
125}, {100, 100, 100}}
answered Jan 13 at 5:41
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
But the choices with $(2^05^0)$ must be ignored, because by requirement each integer must be greater than $1$.
$endgroup$
– farruhota
Jan 13 at 6:55
add a comment |
$begingroup$
I get $114$. Here's how:
Exponents on the three powers of $2$ can be (in ascending order):
all different
- $0,1,5$
- $0,2,4$
- $1,2,3$
two the same
- $0,0,6$
- $0,3,3$
- $1,1,4$
all three same
- $2,2,2$
and likewise for the three powers of $5$.
So for the "all different powers on the $2$s and all different powers on the $5$s" we have:
Example: pairing $0,1,5$ for $2$s with $0,1,5$ for $5$s:
- $(2^0 5^5) (2^1 5^0) (2^5 5^1)$
- $(2^0 5^0) (2^1 5^5) (2^4 5^1)$
- $(2^0 5^1) (2^1 5^0) (2^4 5^5)$
- etc...
But... as pointed out by @farruhota, the calculation includes cases where there is a factor of $1$ (i.e., $2^0 5^0)$ so we do not include those.
Go through all such cases and find $114$.
Here's a list:
{{2, 2, 250000}, {2, 4, 125000}, {2, 5, 100000}, {2, 8, 62500}, {2,
10, 50000}, {2, 16, 31250}, {2, 20, 25000}, {2, 25, 20000}, {2, 32,
15625}, {2, 40, 12500}, {2, 50, 10000}, {2, 80, 6250}, {2, 100,
5000}, {2, 125, 4000}, {2, 160, 3125}, {2, 200, 2500}, {2, 250,
2000}, {2, 400, 1250}, {2, 500, 1000}, {2, 625, 800}, {4, 4,
62500}, {4, 5, 50000}, {4, 8, 31250}, {4, 10, 25000}, {4, 16,
15625}, {4, 20, 12500}, {4, 25, 10000}, {4, 40, 6250}, {4, 50,
5000}, {4, 80, 3125}, {4, 100, 2500}, {4, 125, 2000}, {4, 200,
1250}, {4, 250, 1000}, {4, 400, 625}, {4, 500, 500}, {5, 5,
40000}, {5, 8, 25000}, {5, 10, 20000}, {5, 16, 12500}, {5, 20,
10000}, {5, 25, 8000}, {5, 32, 6250}, {5, 40, 5000}, {5, 50,
4000}, {5, 64, 3125}, {5, 80, 2500}, {5, 100, 2000}, {5, 125,
1600}, {5, 160, 1250}, {5, 200, 1000}, {5, 250, 800}, {5, 320,
625}, {5, 400, 500}, {8, 8, 15625}, {8, 10, 12500}, {8, 20,
6250}, {8, 25, 5000}, {8, 40, 3125}, {8, 50, 2500}, {8, 100,
1250}, {8, 125, 1000}, {8, 200, 625}, {8, 250, 500}, {10, 10,
10000}, {10, 16, 6250}, {10, 20, 5000}, {10, 25, 4000}, {10, 32,
3125}, {10, 40, 2500}, {10, 50, 2000}, {10, 80, 1250}, {10, 100,
1000}, {10, 125, 800}, {10, 160, 625}, {10, 200, 500}, {10, 250,
400}, {16, 20, 3125}, {16, 25, 2500}, {16, 50, 1250}, {16, 100,
625}, {16, 125, 500}, {16, 250, 250}, {20, 20, 2500}, {20, 25,
2000}, {20, 40, 1250}, {20, 50, 1000}, {20, 80, 625}, {20, 100,
500}, {20, 125, 400}, {20, 200, 250}, {25, 25, 1600}, {25, 32,
1250}, {25, 40, 1000}, {25, 50, 800}, {25, 64, 625}, {25, 80,
500}, {25, 100, 400}, {25, 125, 320}, {25, 160, 250}, {25, 200,
200}, {32, 50, 625}, {32, 125, 250}, {40, 40, 625}, {40, 50,
500}, {40, 100, 250}, {40, 125, 200}, {50, 50, 400}, {50, 80,
250}, {50, 100, 200}, {50, 125, 160}, {64, 125, 125}, {80, 100,
125}, {100, 100, 100}}
answered Jan 13 at 5:41
David G. StorkDavid G. Stork
11k41432
$endgroup$
I get $114$. Here's how:
Exponents on the three powers of $2$ can be (in ascending order):
all different
- $0,1,5$
- $0,2,4$
- $1,2,3$
two the same
- $0,0,6$
- $0,3,3$
- $1,1,4$
all three same
- $2,2,2$
and likewise for the three powers of $5$.
So for the "all different powers on the $2$s and all different powers on the $5$s" we have:
Example: pairing $0,1,5$ for $2$s with $0,1,5$ for $5$s:
- $(2^0 5^5) (2^1 5^0) (2^5 5^1)$
- $(2^0 5^0) (2^1 5^5) (2^4 5^1)$
- $(2^0 5^1) (2^1 5^0) (2^4 5^5)$
- etc...
But... as pointed out by @farruhota, the calculation includes cases where there is a factor of $1$ (i.e., $2^0 5^0)$ so we do not include those.
Go through all such cases and find $114$.
Here's a list:
{{2, 2, 250000}, {2, 4, 125000}, {2, 5, 100000}, {2, 8, 62500}, {2,
10, 50000}, {2, 16, 31250}, {2, 20, 25000}, {2, 25, 20000}, {2, 32,
15625}, {2, 40, 12500}, {2, 50, 10000}, {2, 80, 6250}, {2, 100,
5000}, {2, 125, 4000}, {2, 160, 3125}, {2, 200, 2500}, {2, 250,
2000}, {2, 400, 1250}, {2, 500, 1000}, {2, 625, 800}, {4, 4,
62500}, {4, 5, 50000}, {4, 8, 31250}, {4, 10, 25000}, {4, 16,
15625}, {4, 20, 12500}, {4, 25, 10000}, {4, 40, 6250}, {4, 50,
5000}, {4, 80, 3125}, {4, 100, 2500}, {4, 125, 2000}, {4, 200,
1250}, {4, 250, 1000}, {4, 400, 625}, {4, 500, 500}, {5, 5,
40000}, {5, 8, 25000}, {5, 10, 20000}, {5, 16, 12500}, {5, 20,
10000}, {5, 25, 8000}, {5, 32, 6250}, {5, 40, 5000}, {5, 50,
4000}, {5, 64, 3125}, {5, 80, 2500}, {5, 100, 2000}, {5, 125,
1600}, {5, 160, 1250}, {5, 200, 1000}, {5, 250, 800}, {5, 320,
625}, {5, 400, 500}, {8, 8, 15625}, {8, 10, 12500}, {8, 20,
6250}, {8, 25, 5000}, {8, 40, 3125}, {8, 50, 2500}, {8, 100,
1250}, {8, 125, 1000}, {8, 200, 625}, {8, 250, 500}, {10, 10,
10000}, {10, 16, 6250}, {10, 20, 5000}, {10, 25, 4000}, {10, 32,
3125}, {10, 40, 2500}, {10, 50, 2000}, {10, 80, 1250}, {10, 100,
1000}, {10, 125, 800}, {10, 160, 625}, {10, 200, 500}, {10, 250,
400}, {16, 20, 3125}, {16, 25, 2500}, {16, 50, 1250}, {16, 100,
625}, {16, 125, 500}, {16, 250, 250}, {20, 20, 2500}, {20, 25,
2000}, {20, 40, 1250}, {20, 50, 1000}, {20, 80, 625}, {20, 100,
500}, {20, 125, 400}, {20, 200, 250}, {25, 25, 1600}, {25, 32,
1250}, {25, 40, 1000}, {25, 50, 800}, {25, 64, 625}, {25, 80,
500}, {25, 100, 400}, {25, 125, 320}, {25, 160, 250}, {25, 200,
200}, {32, 50, 625}, {32, 125, 250}, {40, 40, 625}, {40, 50,
500}, {40, 100, 250}, {40, 125, 200}, {50, 50, 400}, {50, 80,
250}, {50, 100, 200}, {50, 125, 160}, {64, 125, 125}, {80, 100,
125}, {100, 100, 100}}
answered Jan 13 at 5:41
David G. StorkDavid G. Stork
11k41432
edited Jan 14 at 3:07
answered Jan 13 at 5:41
David G. StorkDavid G. Stork
11k41432
answered Jan 13 at 5:41
David G. StorkDavid G. Stork
11k41432
answered Jan 13 at 5:41
David G. StorkDavid G. Stork
11k41432
11k41432
$begingroup$
But the choices with $(2^05^0)$ must be ignored, because by requirement each integer must be greater than $1$.
$endgroup$
– farruhota
Jan 13 at 6:55
add a comment |
$begingroup$
But the choices with $(2^05^0)$ must be ignored, because by requirement each integer must be greater than $1$.
$endgroup$
– farruhota
Jan 13 at 6:55
$begingroup$
But the choices with $(2^05^0)$ must be ignored, because by requirement each integer must be greater than $1$.
$endgroup$
– farruhota
Jan 13 at 6:55
$begingroup$
But the choices with $(2^05^0)$ must be ignored, because by requirement each integer must be greater than $1$.
$endgroup$
– farruhota
Jan 13 at 6:55
add a comment |
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5
$begingroup$
Show us the ways you've tried. Then we'll be able to give you relevant advice.
$endgroup$
– saulspatz
Jan 13 at 1:49
2
$begingroup$
$3121$ is not a factor of $1000$. It should be $1000=2^35^3$ and $1000000=2^65^6$
$endgroup$
– Ross Millikan
Jan 13 at 2:03
$begingroup$
Sorry..I have made a mistake. It should be $3125$ and $3125 = 5 times 5 times 5 times 5 times 5$
$endgroup$
– T. A.
Jan 13 at 2:05
$begingroup$
Where does $3121$ come from? That's obvious not a factor of $1000000=10^=2^6cdot5^6$
$endgroup$
– saulspatz
Jan 13 at 2:06
$begingroup$
$1000000 = 2 times 2 times 2 times 2 times 2 times 2 times 5 times 5 times 5 times 5 times 5 times 5$. I tried to find out in how many ways we can create three groups using 2,2,2,2,2,2,5,5,5,5,5,5. But failed to do it.
$endgroup$
– T. A.
Jan 13 at 2:11