Mixing
4 digit numbers divisible by 11
$begingroup$
Four digit numbers are formed using the
digits 1,2,3,4 (repetition is allowed). The
number of such four digit numbers divisible
by 11 is-
(1) 22 (2) 36 (3) 44 (4) 52
I know for a number to be divisible by 11 the sum of digits at even places must be equal to the sum of those at odd places. But how do I use this to get the answer?
permutations
asked Mar 26 '18 at 5:39
CodeBloodedCodeBlooded
1165
$endgroup$
add a comment |
$begingroup$
Four digit numbers are formed using the
digits 1,2,3,4 (repetition is allowed). The
number of such four digit numbers divisible
by 11 is-
(1) 22 (2) 36 (3) 44 (4) 52
I know for a number to be divisible by 11 the sum of digits at even places must be equal to the sum of those at odd places. But how do I use this to get the answer?
permutations
asked Mar 26 '18 at 5:39
CodeBloodedCodeBlooded
1165
$endgroup$
$begingroup$
I would break into cases based on what the sum of the even places is. The odd places must have the same sum. If the sum is $2$, then that can only be accomplished using two ones, giving the four digit number $1111$. If the sum of the even places is $3$, then that can be accomplished either as a $1$ followed by a $2$, or vice versa, for two options on how to fill the even slots. Similarly we will have the same two options for how to fill the odd slots. Applying rule of product we get four possibilities here, 1122, 1221, 2112, 2211. Continue
$endgroup$
– JMoravitz
Mar 26 '18 at 5:44
add a comment |
$begingroup$
Four digit numbers are formed using the
digits 1,2,3,4 (repetition is allowed). The
number of such four digit numbers divisible
by 11 is-
(1) 22 (2) 36 (3) 44 (4) 52
I know for a number to be divisible by 11 the sum of digits at even places must be equal to the sum of those at odd places. But how do I use this to get the answer?
permutations
asked Mar 26 '18 at 5:39
CodeBloodedCodeBlooded
1165
$endgroup$
Four digit numbers are formed using the
digits 1,2,3,4 (repetition is allowed). The
number of such four digit numbers divisible
by 11 is-
(1) 22 (2) 36 (3) 44 (4) 52
I know for a number to be divisible by 11 the sum of digits at even places must be equal to the sum of those at odd places. But how do I use this to get the answer?
permutations
permutations
asked Mar 26 '18 at 5:39
CodeBloodedCodeBlooded
1165
asked Mar 26 '18 at 5:39
CodeBloodedCodeBlooded
1165
asked Mar 26 '18 at 5:39
CodeBloodedCodeBlooded
1165
asked Mar 26 '18 at 5:39
CodeBloodedCodeBlooded
1165
asked Mar 26 '18 at 5:39
CodeBloodedCodeBlooded
1165
1165
$begingroup$
I would break into cases based on what the sum of the even places is. The odd places must have the same sum. If the sum is $2$, then that can only be accomplished using two ones, giving the four digit number $1111$. If the sum of the even places is $3$, then that can be accomplished either as a $1$ followed by a $2$, or vice versa, for two options on how to fill the even slots. Similarly we will have the same two options for how to fill the odd slots. Applying rule of product we get four possibilities here, 1122, 1221, 2112, 2211. Continue
$endgroup$
– JMoravitz
Mar 26 '18 at 5:44
add a comment |
$begingroup$
I would break into cases based on what the sum of the even places is. The odd places must have the same sum. If the sum is $2$, then that can only be accomplished using two ones, giving the four digit number $1111$. If the sum of the even places is $3$, then that can be accomplished either as a $1$ followed by a $2$, or vice versa, for two options on how to fill the even slots. Similarly we will have the same two options for how to fill the odd slots. Applying rule of product we get four possibilities here, 1122, 1221, 2112, 2211. Continue
$endgroup$
– JMoravitz
Mar 26 '18 at 5:44
$begingroup$
I would break into cases based on what the sum of the even places is. The odd places must have the same sum. If the sum is $2$, then that can only be accomplished using two ones, giving the four digit number $1111$. If the sum of the even places is $3$, then that can be accomplished either as a $1$ followed by a $2$, or vice versa, for two options on how to fill the even slots. Similarly we will have the same two options for how to fill the odd slots. Applying rule of product we get four possibilities here, 1122, 1221, 2112, 2211. Continue
$endgroup$
– JMoravitz
Mar 26 '18 at 5:44
$begingroup$
I would break into cases based on what the sum of the even places is. The odd places must have the same sum. If the sum is $2$, then that can only be accomplished using two ones, giving the four digit number $1111$. If the sum of the even places is $3$, then that can be accomplished either as a $1$ followed by a $2$, or vice versa, for two options on how to fill the even slots. Similarly we will have the same two options for how to fill the odd slots. Applying rule of product we get four possibilities here, 1122, 1221, 2112, 2211. Continue
$endgroup$
– JMoravitz
Mar 26 '18 at 5:44
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $abcd$ be the number.
If $a=b$ then $c=d$. There are $4*4=16$ ways that can occur. (Four options for $a$ and four options for $b$).
If $a=bpm 1$ then $c=d mp1$. And there are $2*3*3=18$ ways this can occur. (Two choices whether $a > b$ or $b > a$ and three choices from $1,2,3,4$ that are one apart.
If $a = bpm 2$ then $c = dmp 2$ and there $ 2*2*2 = 8$ ways.
And if $a = bpm 4$ then either $a = 1;b=4;c=4;d=1$ or $a=4;b=1;c=1;d=4$. $2$ ways.
So $16 + 18 + 8 + 2 = 44$ ways.
answered Mar 26 '18 at 6:09
fleabloodfleablood
71.7k22686
$endgroup$
add a comment |
$begingroup$
Let $[abcd]$ be the number. Then $[abcd] equiv [ab]+[cd] pmod{99}$ which implies $[abcd] equiv [ab]+[cd] pmod{11}$. Because $a,b,c,d in {1,2,3,4}$,
$[abcd] equiv [(a+c)(b+d)] pmod{11}$. Hence $[abcd]$ is a multiple of $11$ if and only if $a+c=b+d$.
begin{array}{|r|r|r|}
text{sum} &text{ac and bd events} & text{counts} \
hline
2 & 11 & 1\
3 & 12, 21 & 4\
4 & 13, 22, 31 & 9 \
5 & 14, 23, 32, 41 & 16\
6 & 24, 33, 42 & 9\
7 & 34, 43 & 4\
8 & 44 & 1 \
hline
text{total} && 44
end{array}
answered Mar 26 '18 at 6:52
steven gregorysteven gregory
18.2k32258
$endgroup$
add a comment |
$begingroup$
Alternatively, consider permutations of numbers made of one, two, three and four digits.
There is one number made of single number $1$, which is $1111$. Total number is:
$$1cdot {4choose 1}=4.$$
There are four numbers made of two digits $1$ and $2$, which are $1122,1221,2112,2211$. Total number is:
$$4cdot {4choose 2}=24.$$
There are four numbers made of three digits $1,2,3$, which are $1232,2123,2321,3212$. And there are two possible cases: $1,2,3$ and $2,3,4$. Total number is:
$$4cdot 2=8.$$
There are eight numbers made of four digits $1,2,3,4$, which are $1234,1342,2134,2431,3241,3421,4213,4312$. Total number is:
$$8cdot 1=8.$$
In conclusion, the grand total is:
$$4+24+8+8=44.$$
answered Mar 26 '18 at 18:45
farruhotafarruhota
20.6k2740
$endgroup$
add a comment |
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3 Answers
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3 Answers
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$begingroup$
Let $abcd$ be the number.
If $a=b$ then $c=d$. There are $4*4=16$ ways that can occur. (Four options for $a$ and four options for $b$).
If $a=bpm 1$ then $c=d mp1$. And there are $2*3*3=18$ ways this can occur. (Two choices whether $a > b$ or $b > a$ and three choices from $1,2,3,4$ that are one apart.
If $a = bpm 2$ then $c = dmp 2$ and there $ 2*2*2 = 8$ ways.
And if $a = bpm 4$ then either $a = 1;b=4;c=4;d=1$ or $a=4;b=1;c=1;d=4$. $2$ ways.
So $16 + 18 + 8 + 2 = 44$ ways.
answered Mar 26 '18 at 6:09
fleabloodfleablood
71.7k22686
$endgroup$
add a comment |
$begingroup$
Let $abcd$ be the number.
If $a=b$ then $c=d$. There are $4*4=16$ ways that can occur. (Four options for $a$ and four options for $b$).
If $a=bpm 1$ then $c=d mp1$. And there are $2*3*3=18$ ways this can occur. (Two choices whether $a > b$ or $b > a$ and three choices from $1,2,3,4$ that are one apart.
If $a = bpm 2$ then $c = dmp 2$ and there $ 2*2*2 = 8$ ways.
And if $a = bpm 4$ then either $a = 1;b=4;c=4;d=1$ or $a=4;b=1;c=1;d=4$. $2$ ways.
So $16 + 18 + 8 + 2 = 44$ ways.
answered Mar 26 '18 at 6:09
fleabloodfleablood
71.7k22686
$endgroup$
add a comment |
$begingroup$
Let $abcd$ be the number.
If $a=b$ then $c=d$. There are $4*4=16$ ways that can occur. (Four options for $a$ and four options for $b$).
If $a=bpm 1$ then $c=d mp1$. And there are $2*3*3=18$ ways this can occur. (Two choices whether $a > b$ or $b > a$ and three choices from $1,2,3,4$ that are one apart.
If $a = bpm 2$ then $c = dmp 2$ and there $ 2*2*2 = 8$ ways.
And if $a = bpm 4$ then either $a = 1;b=4;c=4;d=1$ or $a=4;b=1;c=1;d=4$. $2$ ways.
So $16 + 18 + 8 + 2 = 44$ ways.
answered Mar 26 '18 at 6:09
fleabloodfleablood
71.7k22686
$endgroup$
Let $abcd$ be the number.
If $a=b$ then $c=d$. There are $4*4=16$ ways that can occur. (Four options for $a$ and four options for $b$).
If $a=bpm 1$ then $c=d mp1$. And there are $2*3*3=18$ ways this can occur. (Two choices whether $a > b$ or $b > a$ and three choices from $1,2,3,4$ that are one apart.
If $a = bpm 2$ then $c = dmp 2$ and there $ 2*2*2 = 8$ ways.
And if $a = bpm 4$ then either $a = 1;b=4;c=4;d=1$ or $a=4;b=1;c=1;d=4$. $2$ ways.
So $16 + 18 + 8 + 2 = 44$ ways.
answered Mar 26 '18 at 6:09
fleabloodfleablood
71.7k22686
answered Mar 26 '18 at 6:09
fleabloodfleablood
71.7k22686
answered Mar 26 '18 at 6:09
fleabloodfleablood
71.7k22686
answered Mar 26 '18 at 6:09
fleabloodfleablood
71.7k22686
71.7k22686
add a comment |
add a comment |
$begingroup$
Let $[abcd]$ be the number. Then $[abcd] equiv [ab]+[cd] pmod{99}$ which implies $[abcd] equiv [ab]+[cd] pmod{11}$. Because $a,b,c,d in {1,2,3,4}$,
$[abcd] equiv [(a+c)(b+d)] pmod{11}$. Hence $[abcd]$ is a multiple of $11$ if and only if $a+c=b+d$.
begin{array}{|r|r|r|}
text{sum} &text{ac and bd events} & text{counts} \
hline
2 & 11 & 1\
3 & 12, 21 & 4\
4 & 13, 22, 31 & 9 \
5 & 14, 23, 32, 41 & 16\
6 & 24, 33, 42 & 9\
7 & 34, 43 & 4\
8 & 44 & 1 \
hline
text{total} && 44
end{array}
answered Mar 26 '18 at 6:52
steven gregorysteven gregory
18.2k32258
$endgroup$
add a comment |
$begingroup$
Let $[abcd]$ be the number. Then $[abcd] equiv [ab]+[cd] pmod{99}$ which implies $[abcd] equiv [ab]+[cd] pmod{11}$. Because $a,b,c,d in {1,2,3,4}$,
$[abcd] equiv [(a+c)(b+d)] pmod{11}$. Hence $[abcd]$ is a multiple of $11$ if and only if $a+c=b+d$.
begin{array}{|r|r|r|}
text{sum} &text{ac and bd events} & text{counts} \
hline
2 & 11 & 1\
3 & 12, 21 & 4\
4 & 13, 22, 31 & 9 \
5 & 14, 23, 32, 41 & 16\
6 & 24, 33, 42 & 9\
7 & 34, 43 & 4\
8 & 44 & 1 \
hline
text{total} && 44
end{array}
answered Mar 26 '18 at 6:52
steven gregorysteven gregory
18.2k32258
$endgroup$
add a comment |
$begingroup$
Let $[abcd]$ be the number. Then $[abcd] equiv [ab]+[cd] pmod{99}$ which implies $[abcd] equiv [ab]+[cd] pmod{11}$. Because $a,b,c,d in {1,2,3,4}$,
$[abcd] equiv [(a+c)(b+d)] pmod{11}$. Hence $[abcd]$ is a multiple of $11$ if and only if $a+c=b+d$.
begin{array}{|r|r|r|}
text{sum} &text{ac and bd events} & text{counts} \
hline
2 & 11 & 1\
3 & 12, 21 & 4\
4 & 13, 22, 31 & 9 \
5 & 14, 23, 32, 41 & 16\
6 & 24, 33, 42 & 9\
7 & 34, 43 & 4\
8 & 44 & 1 \
hline
text{total} && 44
end{array}
answered Mar 26 '18 at 6:52
steven gregorysteven gregory
18.2k32258
$endgroup$
Let $[abcd]$ be the number. Then $[abcd] equiv [ab]+[cd] pmod{99}$ which implies $[abcd] equiv [ab]+[cd] pmod{11}$. Because $a,b,c,d in {1,2,3,4}$,
$[abcd] equiv [(a+c)(b+d)] pmod{11}$. Hence $[abcd]$ is a multiple of $11$ if and only if $a+c=b+d$.
begin{array}{|r|r|r|}
text{sum} &text{ac and bd events} & text{counts} \
hline
2 & 11 & 1\
3 & 12, 21 & 4\
4 & 13, 22, 31 & 9 \
5 & 14, 23, 32, 41 & 16\
6 & 24, 33, 42 & 9\
7 & 34, 43 & 4\
8 & 44 & 1 \
hline
text{total} && 44
end{array}
answered Mar 26 '18 at 6:52
steven gregorysteven gregory
18.2k32258
answered Mar 26 '18 at 6:52
steven gregorysteven gregory
18.2k32258
answered Mar 26 '18 at 6:52
steven gregorysteven gregory
18.2k32258
answered Mar 26 '18 at 6:52
steven gregorysteven gregory
18.2k32258
18.2k32258
add a comment |
add a comment |
$begingroup$
Alternatively, consider permutations of numbers made of one, two, three and four digits.
There is one number made of single number $1$, which is $1111$. Total number is:
$$1cdot {4choose 1}=4.$$
There are four numbers made of two digits $1$ and $2$, which are $1122,1221,2112,2211$. Total number is:
$$4cdot {4choose 2}=24.$$
There are four numbers made of three digits $1,2,3$, which are $1232,2123,2321,3212$. And there are two possible cases: $1,2,3$ and $2,3,4$. Total number is:
$$4cdot 2=8.$$
There are eight numbers made of four digits $1,2,3,4$, which are $1234,1342,2134,2431,3241,3421,4213,4312$. Total number is:
$$8cdot 1=8.$$
In conclusion, the grand total is:
$$4+24+8+8=44.$$
answered Mar 26 '18 at 18:45
farruhotafarruhota
20.6k2740
$endgroup$
add a comment |
$begingroup$
Alternatively, consider permutations of numbers made of one, two, three and four digits.
There is one number made of single number $1$, which is $1111$. Total number is:
$$1cdot {4choose 1}=4.$$
There are four numbers made of two digits $1$ and $2$, which are $1122,1221,2112,2211$. Total number is:
$$4cdot {4choose 2}=24.$$
There are four numbers made of three digits $1,2,3$, which are $1232,2123,2321,3212$. And there are two possible cases: $1,2,3$ and $2,3,4$. Total number is:
$$4cdot 2=8.$$
There are eight numbers made of four digits $1,2,3,4$, which are $1234,1342,2134,2431,3241,3421,4213,4312$. Total number is:
$$8cdot 1=8.$$
In conclusion, the grand total is:
$$4+24+8+8=44.$$
answered Mar 26 '18 at 18:45
farruhotafarruhota
20.6k2740
$endgroup$
add a comment |
$begingroup$
Alternatively, consider permutations of numbers made of one, two, three and four digits.
There is one number made of single number $1$, which is $1111$. Total number is:
$$1cdot {4choose 1}=4.$$
There are four numbers made of two digits $1$ and $2$, which are $1122,1221,2112,2211$. Total number is:
$$4cdot {4choose 2}=24.$$
There are four numbers made of three digits $1,2,3$, which are $1232,2123,2321,3212$. And there are two possible cases: $1,2,3$ and $2,3,4$. Total number is:
$$4cdot 2=8.$$
There are eight numbers made of four digits $1,2,3,4$, which are $1234,1342,2134,2431,3241,3421,4213,4312$. Total number is:
$$8cdot 1=8.$$
In conclusion, the grand total is:
$$4+24+8+8=44.$$
answered Mar 26 '18 at 18:45
farruhotafarruhota
20.6k2740
$endgroup$
Alternatively, consider permutations of numbers made of one, two, three and four digits.
There is one number made of single number $1$, which is $1111$. Total number is:
$$1cdot {4choose 1}=4.$$
There are four numbers made of two digits $1$ and $2$, which are $1122,1221,2112,2211$. Total number is:
$$4cdot {4choose 2}=24.$$
There are four numbers made of three digits $1,2,3$, which are $1232,2123,2321,3212$. And there are two possible cases: $1,2,3$ and $2,3,4$. Total number is:
$$4cdot 2=8.$$
There are eight numbers made of four digits $1,2,3,4$, which are $1234,1342,2134,2431,3241,3421,4213,4312$. Total number is:
$$8cdot 1=8.$$
In conclusion, the grand total is:
$$4+24+8+8=44.$$
answered Mar 26 '18 at 18:45
farruhotafarruhota
20.6k2740
answered Mar 26 '18 at 18:45
farruhotafarruhota
20.6k2740
answered Mar 26 '18 at 18:45
farruhotafarruhota
20.6k2740
answered Mar 26 '18 at 18:45
farruhotafarruhota
20.6k2740
20.6k2740
add a comment |
add a comment |
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$begingroup$
I would break into cases based on what the sum of the even places is. The odd places must have the same sum. If the sum is $2$, then that can only be accomplished using two ones, giving the four digit number $1111$. If the sum of the even places is $3$, then that can be accomplished either as a $1$ followed by a $2$, or vice versa, for two options on how to fill the even slots. Similarly we will have the same two options for how to fill the odd slots. Applying rule of product we get four possibilities here, 1122, 1221, 2112, 2211. Continue
$endgroup$
– JMoravitz
Mar 26 '18 at 5:44