Mixing
Average of all 3 digit numbers with distinct digits
$begingroup$
I tried adding up all the digits in each digits place and average them, but I got some decimals.
I don't know if I am doing it correctly
Can someone please explain this to me
elementary-number-theory
asked Jan 25 at 2:14
semicolon822semicolon822
51
$endgroup$
add a comment |
$begingroup$
I tried adding up all the digits in each digits place and average them, but I got some decimals.
I don't know if I am doing it correctly
Can someone please explain this to me
elementary-number-theory
asked Jan 25 at 2:14
semicolon822semicolon822
51
$endgroup$
1
$begingroup$
Are you to average the numbers? Then not relevant to average the digits. Also question doesn't say answer should come out a whole number, maybe it's a fraction.
$endgroup$
– coffeemath
Jan 25 at 2:20
$begingroup$
how else would i do it???? It would be tedious to add all the numbers up and average them
$endgroup$
– semicolon822
Jan 25 at 2:22
$begingroup$
wait so your telling me to add up all the 648 numbers...
$endgroup$
– semicolon822
Jan 25 at 2:26
$begingroup$
I see no other way to average them, but maybe there's a trick to adding them up. The now deleted answer didn't restrict to numbers having distinct digits.
$endgroup$
– coffeemath
Jan 25 at 2:28
$begingroup$
So show us what you got, semicolon.
$endgroup$
– Gerry Myerson
Jan 25 at 3:07
add a comment |
$begingroup$
I tried adding up all the digits in each digits place and average them, but I got some decimals.
I don't know if I am doing it correctly
Can someone please explain this to me
elementary-number-theory
asked Jan 25 at 2:14
semicolon822semicolon822
51
$endgroup$
I tried adding up all the digits in each digits place and average them, but I got some decimals.
I don't know if I am doing it correctly
Can someone please explain this to me
elementary-number-theory
elementary-number-theory
asked Jan 25 at 2:14
semicolon822semicolon822
51
asked Jan 25 at 2:14
semicolon822semicolon822
51
asked Jan 25 at 2:14
semicolon822semicolon822
51
asked Jan 25 at 2:14
semicolon822semicolon822
51
asked Jan 25 at 2:14
semicolon822semicolon822
51
51
1
$begingroup$
Are you to average the numbers? Then not relevant to average the digits. Also question doesn't say answer should come out a whole number, maybe it's a fraction.
$endgroup$
– coffeemath
Jan 25 at 2:20
$begingroup$
how else would i do it???? It would be tedious to add all the numbers up and average them
$endgroup$
– semicolon822
Jan 25 at 2:22
$begingroup$
wait so your telling me to add up all the 648 numbers...
$endgroup$
– semicolon822
Jan 25 at 2:26
$begingroup$
I see no other way to average them, but maybe there's a trick to adding them up. The now deleted answer didn't restrict to numbers having distinct digits.
$endgroup$
– coffeemath
Jan 25 at 2:28
$begingroup$
So show us what you got, semicolon.
$endgroup$
– Gerry Myerson
Jan 25 at 3:07
add a comment |
1
$begingroup$
Are you to average the numbers? Then not relevant to average the digits. Also question doesn't say answer should come out a whole number, maybe it's a fraction.
$endgroup$
– coffeemath
Jan 25 at 2:20
$begingroup$
how else would i do it???? It would be tedious to add all the numbers up and average them
$endgroup$
– semicolon822
Jan 25 at 2:22
$begingroup$
wait so your telling me to add up all the 648 numbers...
$endgroup$
– semicolon822
Jan 25 at 2:26
$begingroup$
I see no other way to average them, but maybe there's a trick to adding them up. The now deleted answer didn't restrict to numbers having distinct digits.
$endgroup$
– coffeemath
Jan 25 at 2:28
$begingroup$
So show us what you got, semicolon.
$endgroup$
– Gerry Myerson
Jan 25 at 3:07
1
1
$begingroup$
Are you to average the numbers? Then not relevant to average the digits. Also question doesn't say answer should come out a whole number, maybe it's a fraction.
$endgroup$
– coffeemath
Jan 25 at 2:20
$begingroup$
Are you to average the numbers? Then not relevant to average the digits. Also question doesn't say answer should come out a whole number, maybe it's a fraction.
$endgroup$
– coffeemath
Jan 25 at 2:20
$begingroup$
how else would i do it???? It would be tedious to add all the numbers up and average them
$endgroup$
– semicolon822
Jan 25 at 2:22
$begingroup$
how else would i do it???? It would be tedious to add all the numbers up and average them
$endgroup$
– semicolon822
Jan 25 at 2:22
$begingroup$
wait so your telling me to add up all the 648 numbers...
$endgroup$
– semicolon822
Jan 25 at 2:26
$begingroup$
wait so your telling me to add up all the 648 numbers...
$endgroup$
– semicolon822
Jan 25 at 2:26
$begingroup$
I see no other way to average them, but maybe there's a trick to adding them up. The now deleted answer didn't restrict to numbers having distinct digits.
$endgroup$
– coffeemath
Jan 25 at 2:28
$begingroup$
I see no other way to average them, but maybe there's a trick to adding them up. The now deleted answer didn't restrict to numbers having distinct digits.
$endgroup$
– coffeemath
Jan 25 at 2:28
$begingroup$
So show us what you got, semicolon.
$endgroup$
– Gerry Myerson
Jan 25 at 3:07
$begingroup$
So show us what you got, semicolon.
$endgroup$
– Gerry Myerson
Jan 25 at 3:07
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The only thing that makes this hard is that the numbers can't start with $0.$ If they were allowed to start with $0,$ it would be easy, so let's do the problem on the assumption that numbers are allowed to start with $0,$ and then adjust the result.
If numbers can start with $0$, there are $10cdot9cdot8=720$ nunbers with distinct digits. By symmetry, each of the digits appears the same number of times $(72)$ in each column, and since the sum of the numbers from $0$ t $9$ is $45,$ the sum of the numbers is $$72cdot45cdot111=359640,$$ where the $111$ comes from the fact that we are summing the ones, tens, and hundreds columns.
Now we just have to subtract out the sum of the numbers that started with $0$. There are $9cdot8=72$ of these, since the last two digits must be distinct, and neither can be $0.$ So, to finish this off, we need to subtract the sum of the two-digit numbers with distinct non-zero digits, and divide by $648(720-72).$
I'll leave that to you.
answered Jan 25 at 2:58
saulspatzsaulspatz
17k31435
$endgroup$
$begingroup$
can you verify my answer I got 548.888 (repeating)
$endgroup$
– semicolon822
Jan 25 at 22:16
add a comment |
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1 Answer
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$begingroup$
The only thing that makes this hard is that the numbers can't start with $0.$ If they were allowed to start with $0,$ it would be easy, so let's do the problem on the assumption that numbers are allowed to start with $0,$ and then adjust the result.
If numbers can start with $0$, there are $10cdot9cdot8=720$ nunbers with distinct digits. By symmetry, each of the digits appears the same number of times $(72)$ in each column, and since the sum of the numbers from $0$ t $9$ is $45,$ the sum of the numbers is $$72cdot45cdot111=359640,$$ where the $111$ comes from the fact that we are summing the ones, tens, and hundreds columns.
Now we just have to subtract out the sum of the numbers that started with $0$. There are $9cdot8=72$ of these, since the last two digits must be distinct, and neither can be $0.$ So, to finish this off, we need to subtract the sum of the two-digit numbers with distinct non-zero digits, and divide by $648(720-72).$
I'll leave that to you.
answered Jan 25 at 2:58
saulspatzsaulspatz
17k31435
$endgroup$
$begingroup$
can you verify my answer I got 548.888 (repeating)
$endgroup$
– semicolon822
Jan 25 at 22:16
add a comment |
$begingroup$
The only thing that makes this hard is that the numbers can't start with $0.$ If they were allowed to start with $0,$ it would be easy, so let's do the problem on the assumption that numbers are allowed to start with $0,$ and then adjust the result.
If numbers can start with $0$, there are $10cdot9cdot8=720$ nunbers with distinct digits. By symmetry, each of the digits appears the same number of times $(72)$ in each column, and since the sum of the numbers from $0$ t $9$ is $45,$ the sum of the numbers is $$72cdot45cdot111=359640,$$ where the $111$ comes from the fact that we are summing the ones, tens, and hundreds columns.
Now we just have to subtract out the sum of the numbers that started with $0$. There are $9cdot8=72$ of these, since the last two digits must be distinct, and neither can be $0.$ So, to finish this off, we need to subtract the sum of the two-digit numbers with distinct non-zero digits, and divide by $648(720-72).$
I'll leave that to you.
answered Jan 25 at 2:58
saulspatzsaulspatz
17k31435
$endgroup$
$begingroup$
can you verify my answer I got 548.888 (repeating)
$endgroup$
– semicolon822
Jan 25 at 22:16
add a comment |
$begingroup$
The only thing that makes this hard is that the numbers can't start with $0.$ If they were allowed to start with $0,$ it would be easy, so let's do the problem on the assumption that numbers are allowed to start with $0,$ and then adjust the result.
If numbers can start with $0$, there are $10cdot9cdot8=720$ nunbers with distinct digits. By symmetry, each of the digits appears the same number of times $(72)$ in each column, and since the sum of the numbers from $0$ t $9$ is $45,$ the sum of the numbers is $$72cdot45cdot111=359640,$$ where the $111$ comes from the fact that we are summing the ones, tens, and hundreds columns.
Now we just have to subtract out the sum of the numbers that started with $0$. There are $9cdot8=72$ of these, since the last two digits must be distinct, and neither can be $0.$ So, to finish this off, we need to subtract the sum of the two-digit numbers with distinct non-zero digits, and divide by $648(720-72).$
I'll leave that to you.
answered Jan 25 at 2:58
saulspatzsaulspatz
17k31435
$endgroup$
The only thing that makes this hard is that the numbers can't start with $0.$ If they were allowed to start with $0,$ it would be easy, so let's do the problem on the assumption that numbers are allowed to start with $0,$ and then adjust the result.
If numbers can start with $0$, there are $10cdot9cdot8=720$ nunbers with distinct digits. By symmetry, each of the digits appears the same number of times $(72)$ in each column, and since the sum of the numbers from $0$ t $9$ is $45,$ the sum of the numbers is $$72cdot45cdot111=359640,$$ where the $111$ comes from the fact that we are summing the ones, tens, and hundreds columns.
Now we just have to subtract out the sum of the numbers that started with $0$. There are $9cdot8=72$ of these, since the last two digits must be distinct, and neither can be $0.$ So, to finish this off, we need to subtract the sum of the two-digit numbers with distinct non-zero digits, and divide by $648(720-72).$
I'll leave that to you.
answered Jan 25 at 2:58
saulspatzsaulspatz
17k31435
answered Jan 25 at 2:58
saulspatzsaulspatz
17k31435
answered Jan 25 at 2:58
saulspatzsaulspatz
17k31435
answered Jan 25 at 2:58
saulspatzsaulspatz
17k31435
17k31435
$begingroup$
can you verify my answer I got 548.888 (repeating)
$endgroup$
– semicolon822
Jan 25 at 22:16
add a comment |
$begingroup$
can you verify my answer I got 548.888 (repeating)
$endgroup$
– semicolon822
Jan 25 at 22:16
$begingroup$
can you verify my answer I got 548.888 (repeating)
$endgroup$
– semicolon822
Jan 25 at 22:16
$begingroup$
can you verify my answer I got 548.888 (repeating)
$endgroup$
– semicolon822
Jan 25 at 22:16
add a comment |
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1
$begingroup$
Are you to average the numbers? Then not relevant to average the digits. Also question doesn't say answer should come out a whole number, maybe it's a fraction.
$endgroup$
– coffeemath
Jan 25 at 2:20
$begingroup$
how else would i do it???? It would be tedious to add all the numbers up and average them
$endgroup$
– semicolon822
Jan 25 at 2:22
$begingroup$
wait so your telling me to add up all the 648 numbers...
$endgroup$
– semicolon822
Jan 25 at 2:26
$begingroup$
I see no other way to average them, but maybe there's a trick to adding them up. The now deleted answer didn't restrict to numbers having distinct digits.
$endgroup$
– coffeemath
Jan 25 at 2:28
$begingroup$
So show us what you got, semicolon.
$endgroup$
– Gerry Myerson
Jan 25 at 3:07