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 56 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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063537%2fhow-many-5-digit-numbers-satisfy-this-criterion%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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 57 mins ago
answered 1 hour ago
David G. Stork
9,93521232
9,93521232
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063537%2fhow-many-5-digit-numbers-satisfy-this-criterion%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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