How many 5-digit numbers satisfy this criterion?
How many $5$-digit numbers have the property that the sum of its digits equals the product of its digits?
permutations
add a comment |
How many $5$-digit numbers have the property that the sum of its digits equals the product of its digits?
permutations
Examples seem to be rare. I found $11125$ and $11133$ though.
– SmileyCraft
1 hour ago
11222 is one of it.
– Heroic24
1 hour ago
I am convinced these are all up to permutations. So there are $20+10+10=40$ examples in total.
– SmileyCraft
1 hour ago
add a comment |
How many $5$-digit numbers have the property that the sum of its digits equals the product of its digits?
permutations
How many $5$-digit numbers have the property that the sum of its digits equals the product of its digits?
permutations
permutations
edited 57 mins ago
David G. Stork
9,93521232
9,93521232
asked 1 hour ago
Heroic24
1627
1627
Examples seem to be rare. I found $11125$ and $11133$ though.
– SmileyCraft
1 hour ago
11222 is one of it.
– Heroic24
1 hour ago
I am convinced these are all up to permutations. So there are $20+10+10=40$ examples in total.
– SmileyCraft
1 hour ago
add a comment |
Examples seem to be rare. I found $11125$ and $11133$ though.
– SmileyCraft
1 hour ago
11222 is one of it.
– Heroic24
1 hour ago
I am convinced these are all up to permutations. So there are $20+10+10=40$ examples in total.
– SmileyCraft
1 hour ago
Examples seem to be rare. I found $11125$ and $11133$ though.
– SmileyCraft
1 hour ago
Examples seem to be rare. I found $11125$ and $11133$ though.
– SmileyCraft
1 hour ago
11222 is one of it.
– Heroic24
1 hour ago
11222 is one of it.
– Heroic24
1 hour ago
I am convinced these are all up to permutations. So there are $20+10+10=40$ examples in total.
– SmileyCraft
1 hour ago
I am convinced these are all up to permutations. So there are $20+10+10=40$ examples in total.
– SmileyCraft
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
We want to find natural number solutions to $$x_1+x_2+cdots +x_5 = x_1cdot x_2cdots x_5.$$
Assume that $$x_1leq x_2leq x_3leq x_4 leq x_5.$$
Since $x_1+cdots +x_5 = x_1cdots x_5 < 5x_5$ (as $x_1=x_2=cdots =x_5$ is impossible) we have that $x_1x_2x_3x_4 leq 4.$ This implies that $(x_1, x_2, x_3, x_4)$ is one of the sequences
$$(1, 1, 1, 1), (1, 1, 1, 2), (1, 1, 1, 3), (1, 1, 1, 4),(1, 1, 2, 2).$$
The sequences $(1, 1, 1, 1)$ and $(1, 1, 1, 4)$ are not good. There is no number
$x_5$ for such sequences. From the remaining sequences we obtain all the solutions
$$(1, 1, 1, 2, 5), (1, 1, 1, 3, 3) text{ and } (1, 1, 2, 2, 2).$$
This diophantine equation has an unsolved problem associated with it. If $a(n)$ is the number of solutions then $a(n)=1$ when $n=114,174text{ or }444$ where $n>100.$
Oh, I should have mentioned that $n>100.$
– Hello_World
1 hour ago
add a comment |
A simple Mathematica program finds the $40$ cases:
${11125, 11133, 11152, 11215, 11222, 11251, 11313, 11331, 11512,
11521, 12115, 12122, 12151, 12212, 12221, 12511, 13113, 13131, 13311,
15112, 15121, 15211, 21115, 21122, 21151, 21212, 21221, 21511, 22112,
22121, 22211, 25111, 31113, 31131, 31311, 33111, 51112, 51121, 51211,
52111}$
myList = Table[i, {i, 10000, 99999}];
Select[myList,
Times @@ RealDigits[#][[1]] == Total@RealDigits[#][[1]] &]
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
We want to find natural number solutions to $$x_1+x_2+cdots +x_5 = x_1cdot x_2cdots x_5.$$
Assume that $$x_1leq x_2leq x_3leq x_4 leq x_5.$$
Since $x_1+cdots +x_5 = x_1cdots x_5 < 5x_5$ (as $x_1=x_2=cdots =x_5$ is impossible) we have that $x_1x_2x_3x_4 leq 4.$ This implies that $(x_1, x_2, x_3, x_4)$ is one of the sequences
$$(1, 1, 1, 1), (1, 1, 1, 2), (1, 1, 1, 3), (1, 1, 1, 4),(1, 1, 2, 2).$$
The sequences $(1, 1, 1, 1)$ and $(1, 1, 1, 4)$ are not good. There is no number
$x_5$ for such sequences. From the remaining sequences we obtain all the solutions
$$(1, 1, 1, 2, 5), (1, 1, 1, 3, 3) text{ and } (1, 1, 2, 2, 2).$$
This diophantine equation has an unsolved problem associated with it. If $a(n)$ is the number of solutions then $a(n)=1$ when $n=114,174text{ or }444$ where $n>100.$
Oh, I should have mentioned that $n>100.$
– Hello_World
1 hour ago
add a comment |
We want to find natural number solutions to $$x_1+x_2+cdots +x_5 = x_1cdot x_2cdots x_5.$$
Assume that $$x_1leq x_2leq x_3leq x_4 leq x_5.$$
Since $x_1+cdots +x_5 = x_1cdots x_5 < 5x_5$ (as $x_1=x_2=cdots =x_5$ is impossible) we have that $x_1x_2x_3x_4 leq 4.$ This implies that $(x_1, x_2, x_3, x_4)$ is one of the sequences
$$(1, 1, 1, 1), (1, 1, 1, 2), (1, 1, 1, 3), (1, 1, 1, 4),(1, 1, 2, 2).$$
The sequences $(1, 1, 1, 1)$ and $(1, 1, 1, 4)$ are not good. There is no number
$x_5$ for such sequences. From the remaining sequences we obtain all the solutions
$$(1, 1, 1, 2, 5), (1, 1, 1, 3, 3) text{ and } (1, 1, 2, 2, 2).$$
This diophantine equation has an unsolved problem associated with it. If $a(n)$ is the number of solutions then $a(n)=1$ when $n=114,174text{ or }444$ where $n>100.$
Oh, I should have mentioned that $n>100.$
– Hello_World
1 hour ago
add a comment |
We want to find natural number solutions to $$x_1+x_2+cdots +x_5 = x_1cdot x_2cdots x_5.$$
Assume that $$x_1leq x_2leq x_3leq x_4 leq x_5.$$
Since $x_1+cdots +x_5 = x_1cdots x_5 < 5x_5$ (as $x_1=x_2=cdots =x_5$ is impossible) we have that $x_1x_2x_3x_4 leq 4.$ This implies that $(x_1, x_2, x_3, x_4)$ is one of the sequences
$$(1, 1, 1, 1), (1, 1, 1, 2), (1, 1, 1, 3), (1, 1, 1, 4),(1, 1, 2, 2).$$
The sequences $(1, 1, 1, 1)$ and $(1, 1, 1, 4)$ are not good. There is no number
$x_5$ for such sequences. From the remaining sequences we obtain all the solutions
$$(1, 1, 1, 2, 5), (1, 1, 1, 3, 3) text{ and } (1, 1, 2, 2, 2).$$
This diophantine equation has an unsolved problem associated with it. If $a(n)$ is the number of solutions then $a(n)=1$ when $n=114,174text{ or }444$ where $n>100.$
We want to find natural number solutions to $$x_1+x_2+cdots +x_5 = x_1cdot x_2cdots x_5.$$
Assume that $$x_1leq x_2leq x_3leq x_4 leq x_5.$$
Since $x_1+cdots +x_5 = x_1cdots x_5 < 5x_5$ (as $x_1=x_2=cdots =x_5$ is impossible) we have that $x_1x_2x_3x_4 leq 4.$ This implies that $(x_1, x_2, x_3, x_4)$ is one of the sequences
$$(1, 1, 1, 1), (1, 1, 1, 2), (1, 1, 1, 3), (1, 1, 1, 4),(1, 1, 2, 2).$$
The sequences $(1, 1, 1, 1)$ and $(1, 1, 1, 4)$ are not good. There is no number
$x_5$ for such sequences. From the remaining sequences we obtain all the solutions
$$(1, 1, 1, 2, 5), (1, 1, 1, 3, 3) text{ and } (1, 1, 2, 2, 2).$$
This diophantine equation has an unsolved problem associated with it. If $a(n)$ is the number of solutions then $a(n)=1$ when $n=114,174text{ or }444$ where $n>100.$
edited 1 hour ago
answered 1 hour ago
Hello_World
4,02821630
4,02821630
Oh, I should have mentioned that $n>100.$
– Hello_World
1 hour ago
add a comment |
Oh, I should have mentioned that $n>100.$
– Hello_World
1 hour ago
Oh, I should have mentioned that $n>100.$
– Hello_World
1 hour ago
Oh, I should have mentioned that $n>100.$
– Hello_World
1 hour ago
add a comment |
A simple Mathematica program finds the $40$ cases:
${11125, 11133, 11152, 11215, 11222, 11251, 11313, 11331, 11512,
11521, 12115, 12122, 12151, 12212, 12221, 12511, 13113, 13131, 13311,
15112, 15121, 15211, 21115, 21122, 21151, 21212, 21221, 21511, 22112,
22121, 22211, 25111, 31113, 31131, 31311, 33111, 51112, 51121, 51211,
52111}$
myList = Table[i, {i, 10000, 99999}];
Select[myList,
Times @@ RealDigits[#][[1]] == Total@RealDigits[#][[1]] &]
add a comment |
A simple Mathematica program finds the $40$ cases:
${11125, 11133, 11152, 11215, 11222, 11251, 11313, 11331, 11512,
11521, 12115, 12122, 12151, 12212, 12221, 12511, 13113, 13131, 13311,
15112, 15121, 15211, 21115, 21122, 21151, 21212, 21221, 21511, 22112,
22121, 22211, 25111, 31113, 31131, 31311, 33111, 51112, 51121, 51211,
52111}$
myList = Table[i, {i, 10000, 99999}];
Select[myList,
Times @@ RealDigits[#][[1]] == Total@RealDigits[#][[1]] &]
add a comment |
A simple Mathematica program finds the $40$ cases:
${11125, 11133, 11152, 11215, 11222, 11251, 11313, 11331, 11512,
11521, 12115, 12122, 12151, 12212, 12221, 12511, 13113, 13131, 13311,
15112, 15121, 15211, 21115, 21122, 21151, 21212, 21221, 21511, 22112,
22121, 22211, 25111, 31113, 31131, 31311, 33111, 51112, 51121, 51211,
52111}$
myList = Table[i, {i, 10000, 99999}];
Select[myList,
Times @@ RealDigits[#][[1]] == Total@RealDigits[#][[1]] &]
A simple Mathematica program finds the $40$ cases:
${11125, 11133, 11152, 11215, 11222, 11251, 11313, 11331, 11512,
11521, 12115, 12122, 12151, 12212, 12221, 12511, 13113, 13131, 13311,
15112, 15121, 15211, 21115, 21122, 21151, 21212, 21221, 21511, 22112,
22121, 22211, 25111, 31113, 31131, 31311, 33111, 51112, 51121, 51211,
52111}$
myList = Table[i, {i, 10000, 99999}];
Select[myList,
Times @@ RealDigits[#][[1]] == Total@RealDigits[#][[1]] &]
edited 58 mins ago
answered 1 hour ago
David G. Stork
9,93521232
9,93521232
add a comment |
add a comment |
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Examples seem to be rare. I found $11125$ and $11133$ though.
– SmileyCraft
1 hour ago
11222 is one of it.
– Heroic24
1 hour ago
I am convinced these are all up to permutations. So there are $20+10+10=40$ examples in total.
– SmileyCraft
1 hour ago