Numbers $n$ such that $n$ plus the sum of $n$'s digits is $313$
$begingroup$
Good morning, everyone. Here is the problem I'm faced with:
The sum of the number $n$ and the digits of $n$ equals $313$. What are the possible values of $n$?
By reasoning I concluded that it has to be a $3$ digit number and by hit and trial I found $305$ as one of the answers. How would I find all possible values of $n$?
It would be great if someone could throw light on how to proceed in a mathematical way and get the values of $n$ (as opposed to brute-forcing/trial-and-error).
I tried as below:
Let our number be $xyz$, where $x,y,z$ are the digits of the number. Then we seek solutions to
$$underbrace{100x + 10y + z}_{n} + underbrace{x+y+z}_{n's ; digits} = 313$$
Thus, simplifying, we seek integer solutions to
$$101 x + 11 y + 2 z = 313$$
I am stuck now as to how to solve this.
Thanks to all in advance.
number-theory elementary-number-theory
$endgroup$
add a comment |
$begingroup$
Good morning, everyone. Here is the problem I'm faced with:
The sum of the number $n$ and the digits of $n$ equals $313$. What are the possible values of $n$?
By reasoning I concluded that it has to be a $3$ digit number and by hit and trial I found $305$ as one of the answers. How would I find all possible values of $n$?
It would be great if someone could throw light on how to proceed in a mathematical way and get the values of $n$ (as opposed to brute-forcing/trial-and-error).
I tried as below:
Let our number be $xyz$, where $x,y,z$ are the digits of the number. Then we seek solutions to
$$underbrace{100x + 10y + z}_{n} + underbrace{x+y+z}_{n's ; digits} = 313$$
Thus, simplifying, we seek integer solutions to
$$101 x + 11 y + 2 z = 313$$
I am stuck now as to how to solve this.
Thanks to all in advance.
number-theory elementary-number-theory
$endgroup$
add a comment |
$begingroup$
Good morning, everyone. Here is the problem I'm faced with:
The sum of the number $n$ and the digits of $n$ equals $313$. What are the possible values of $n$?
By reasoning I concluded that it has to be a $3$ digit number and by hit and trial I found $305$ as one of the answers. How would I find all possible values of $n$?
It would be great if someone could throw light on how to proceed in a mathematical way and get the values of $n$ (as opposed to brute-forcing/trial-and-error).
I tried as below:
Let our number be $xyz$, where $x,y,z$ are the digits of the number. Then we seek solutions to
$$underbrace{100x + 10y + z}_{n} + underbrace{x+y+z}_{n's ; digits} = 313$$
Thus, simplifying, we seek integer solutions to
$$101 x + 11 y + 2 z = 313$$
I am stuck now as to how to solve this.
Thanks to all in advance.
number-theory elementary-number-theory
$endgroup$
Good morning, everyone. Here is the problem I'm faced with:
The sum of the number $n$ and the digits of $n$ equals $313$. What are the possible values of $n$?
By reasoning I concluded that it has to be a $3$ digit number and by hit and trial I found $305$ as one of the answers. How would I find all possible values of $n$?
It would be great if someone could throw light on how to proceed in a mathematical way and get the values of $n$ (as opposed to brute-forcing/trial-and-error).
I tried as below:
Let our number be $xyz$, where $x,y,z$ are the digits of the number. Then we seek solutions to
$$underbrace{100x + 10y + z}_{n} + underbrace{x+y+z}_{n's ; digits} = 313$$
Thus, simplifying, we seek integer solutions to
$$101 x + 11 y + 2 z = 313$$
I am stuck now as to how to solve this.
Thanks to all in advance.
number-theory elementary-number-theory
number-theory elementary-number-theory
edited Jan 13 at 7:51
Eevee Trainer
7,97021439
7,97021439
asked Jan 13 at 7:22
BlueStarBlueStar
82
82
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Going from your final conclusion,
$$101x + 11y + 2z = 313$$
it seems sufficient to simply notice that $x,y,z in {0,1,2,3,...,9}$ since the number's digits are $x,y,z$. In particular, we can eliminate further: we must have $x in {1,2}$ since the maximum value of $11y+2z=117$ (when you have $y=z=9$). The least value of $11y+2z$ is obviously $0$, so with the previous we know:
$$x in {1,2,3}$$
From here, it becomes more processes of elimination.
Suppose $x=1$. Then $11y+2z = 313-101 = 212$. However the maximum value noted earlier for the former expression is $117$, so the one's digit ($x$) definitely cannot be $1$.
Suppose $x=2$. Then $11y + 2z = 313 - 202 = 111$.
Try $y=9$. This yields $z=6$.
Try $y=8$, which yields $z = 11.5 > 10$, so we can immediately conclude that $z=2,y=9$ is the only solution here. We need not try more $y$'s.
Suppose $x=3$. then $11y + 2z = 313 - 303 = 10$.
As is easily deducible, this immediately implies $y=0$ and $z=5$. Thus, that would be the only solution to this case.
Thus, we end up with two solutions:
$$x=3,y=0,z=5 implies 305 ; text{ is a solution} $$
$$x=2,y=9,z=6 implies 296 ; text{ is a solution} $$
Throw in some details to show numbers with fewer than $3$ or more than $3$ digits cannot be solutions (should be pretty easy), and you have thus shown these are the only base-$10$ solutions to
$$n + s(n) = 313$$
where $s(n)$ denotes the sum of the digits of $n$.
$endgroup$
add a comment |
$begingroup$
The greatest possible sum of digits for a positive integer less than $313$ is $20$ (from $299$) so the minimum value you need to test is $293$. The sum of digits in this range is at least $3$ ($300$) and is at least $4$ for numbers other than $300$ so the greatest number you need to test is $309$
Either you are less than $300$ when, with units digit $a$ you have $290+2a+11=313$ and $a=6$, or greater than $300$ when $300+2a+3=313$, and $a=5$.
$endgroup$
add a comment |
$begingroup$
You are on the right track with:
$$overline{xyz}+x+y+z=101x+11y+2z=313.$$
$xne 1$, because otherwise $11y+2zle 117<212$.
$x=2$, then:
$$z=frac{111-11y}{2} Rightarrow y=9, z=6 Rightarrow overline{xyz}=296.$$
$endgroup$
add a comment |
$begingroup$
$$101x+11y+2z=313$$
where $x,y,z$ are numbers between $0$ and $9$.
If $xle 1$, then $101x+11y + 2z le 101+11(9)+2(9)=101+13(9)=218 $
If $x=2$, then we have $$11y+2z=111$$
Since $zle 9$, we have $$11y =111-2z ge 93$$
Hence, we must have $y=9$. $$2z=111-11(9)=111-99=12.$$
Hence $296$ is a solution.
If $x=3$, then we have $11y+2z=10$ which forces $y=0, z=5$.
The solutions are $305$ and $296$.
$endgroup$
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%2f3071768%2fnumbers-n-such-that-n-plus-the-sum-of-ns-digits-is-313%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Going from your final conclusion,
$$101x + 11y + 2z = 313$$
it seems sufficient to simply notice that $x,y,z in {0,1,2,3,...,9}$ since the number's digits are $x,y,z$. In particular, we can eliminate further: we must have $x in {1,2}$ since the maximum value of $11y+2z=117$ (when you have $y=z=9$). The least value of $11y+2z$ is obviously $0$, so with the previous we know:
$$x in {1,2,3}$$
From here, it becomes more processes of elimination.
Suppose $x=1$. Then $11y+2z = 313-101 = 212$. However the maximum value noted earlier for the former expression is $117$, so the one's digit ($x$) definitely cannot be $1$.
Suppose $x=2$. Then $11y + 2z = 313 - 202 = 111$.
Try $y=9$. This yields $z=6$.
Try $y=8$, which yields $z = 11.5 > 10$, so we can immediately conclude that $z=2,y=9$ is the only solution here. We need not try more $y$'s.
Suppose $x=3$. then $11y + 2z = 313 - 303 = 10$.
As is easily deducible, this immediately implies $y=0$ and $z=5$. Thus, that would be the only solution to this case.
Thus, we end up with two solutions:
$$x=3,y=0,z=5 implies 305 ; text{ is a solution} $$
$$x=2,y=9,z=6 implies 296 ; text{ is a solution} $$
Throw in some details to show numbers with fewer than $3$ or more than $3$ digits cannot be solutions (should be pretty easy), and you have thus shown these are the only base-$10$ solutions to
$$n + s(n) = 313$$
where $s(n)$ denotes the sum of the digits of $n$.
$endgroup$
add a comment |
$begingroup$
Going from your final conclusion,
$$101x + 11y + 2z = 313$$
it seems sufficient to simply notice that $x,y,z in {0,1,2,3,...,9}$ since the number's digits are $x,y,z$. In particular, we can eliminate further: we must have $x in {1,2}$ since the maximum value of $11y+2z=117$ (when you have $y=z=9$). The least value of $11y+2z$ is obviously $0$, so with the previous we know:
$$x in {1,2,3}$$
From here, it becomes more processes of elimination.
Suppose $x=1$. Then $11y+2z = 313-101 = 212$. However the maximum value noted earlier for the former expression is $117$, so the one's digit ($x$) definitely cannot be $1$.
Suppose $x=2$. Then $11y + 2z = 313 - 202 = 111$.
Try $y=9$. This yields $z=6$.
Try $y=8$, which yields $z = 11.5 > 10$, so we can immediately conclude that $z=2,y=9$ is the only solution here. We need not try more $y$'s.
Suppose $x=3$. then $11y + 2z = 313 - 303 = 10$.
As is easily deducible, this immediately implies $y=0$ and $z=5$. Thus, that would be the only solution to this case.
Thus, we end up with two solutions:
$$x=3,y=0,z=5 implies 305 ; text{ is a solution} $$
$$x=2,y=9,z=6 implies 296 ; text{ is a solution} $$
Throw in some details to show numbers with fewer than $3$ or more than $3$ digits cannot be solutions (should be pretty easy), and you have thus shown these are the only base-$10$ solutions to
$$n + s(n) = 313$$
where $s(n)$ denotes the sum of the digits of $n$.
$endgroup$
add a comment |
$begingroup$
Going from your final conclusion,
$$101x + 11y + 2z = 313$$
it seems sufficient to simply notice that $x,y,z in {0,1,2,3,...,9}$ since the number's digits are $x,y,z$. In particular, we can eliminate further: we must have $x in {1,2}$ since the maximum value of $11y+2z=117$ (when you have $y=z=9$). The least value of $11y+2z$ is obviously $0$, so with the previous we know:
$$x in {1,2,3}$$
From here, it becomes more processes of elimination.
Suppose $x=1$. Then $11y+2z = 313-101 = 212$. However the maximum value noted earlier for the former expression is $117$, so the one's digit ($x$) definitely cannot be $1$.
Suppose $x=2$. Then $11y + 2z = 313 - 202 = 111$.
Try $y=9$. This yields $z=6$.
Try $y=8$, which yields $z = 11.5 > 10$, so we can immediately conclude that $z=2,y=9$ is the only solution here. We need not try more $y$'s.
Suppose $x=3$. then $11y + 2z = 313 - 303 = 10$.
As is easily deducible, this immediately implies $y=0$ and $z=5$. Thus, that would be the only solution to this case.
Thus, we end up with two solutions:
$$x=3,y=0,z=5 implies 305 ; text{ is a solution} $$
$$x=2,y=9,z=6 implies 296 ; text{ is a solution} $$
Throw in some details to show numbers with fewer than $3$ or more than $3$ digits cannot be solutions (should be pretty easy), and you have thus shown these are the only base-$10$ solutions to
$$n + s(n) = 313$$
where $s(n)$ denotes the sum of the digits of $n$.
$endgroup$
Going from your final conclusion,
$$101x + 11y + 2z = 313$$
it seems sufficient to simply notice that $x,y,z in {0,1,2,3,...,9}$ since the number's digits are $x,y,z$. In particular, we can eliminate further: we must have $x in {1,2}$ since the maximum value of $11y+2z=117$ (when you have $y=z=9$). The least value of $11y+2z$ is obviously $0$, so with the previous we know:
$$x in {1,2,3}$$
From here, it becomes more processes of elimination.
Suppose $x=1$. Then $11y+2z = 313-101 = 212$. However the maximum value noted earlier for the former expression is $117$, so the one's digit ($x$) definitely cannot be $1$.
Suppose $x=2$. Then $11y + 2z = 313 - 202 = 111$.
Try $y=9$. This yields $z=6$.
Try $y=8$, which yields $z = 11.5 > 10$, so we can immediately conclude that $z=2,y=9$ is the only solution here. We need not try more $y$'s.
Suppose $x=3$. then $11y + 2z = 313 - 303 = 10$.
As is easily deducible, this immediately implies $y=0$ and $z=5$. Thus, that would be the only solution to this case.
Thus, we end up with two solutions:
$$x=3,y=0,z=5 implies 305 ; text{ is a solution} $$
$$x=2,y=9,z=6 implies 296 ; text{ is a solution} $$
Throw in some details to show numbers with fewer than $3$ or more than $3$ digits cannot be solutions (should be pretty easy), and you have thus shown these are the only base-$10$ solutions to
$$n + s(n) = 313$$
where $s(n)$ denotes the sum of the digits of $n$.
edited Jan 13 at 7:52
answered Jan 13 at 7:46
Eevee TrainerEevee Trainer
7,97021439
7,97021439
add a comment |
add a comment |
$begingroup$
The greatest possible sum of digits for a positive integer less than $313$ is $20$ (from $299$) so the minimum value you need to test is $293$. The sum of digits in this range is at least $3$ ($300$) and is at least $4$ for numbers other than $300$ so the greatest number you need to test is $309$
Either you are less than $300$ when, with units digit $a$ you have $290+2a+11=313$ and $a=6$, or greater than $300$ when $300+2a+3=313$, and $a=5$.
$endgroup$
add a comment |
$begingroup$
The greatest possible sum of digits for a positive integer less than $313$ is $20$ (from $299$) so the minimum value you need to test is $293$. The sum of digits in this range is at least $3$ ($300$) and is at least $4$ for numbers other than $300$ so the greatest number you need to test is $309$
Either you are less than $300$ when, with units digit $a$ you have $290+2a+11=313$ and $a=6$, or greater than $300$ when $300+2a+3=313$, and $a=5$.
$endgroup$
add a comment |
$begingroup$
The greatest possible sum of digits for a positive integer less than $313$ is $20$ (from $299$) so the minimum value you need to test is $293$. The sum of digits in this range is at least $3$ ($300$) and is at least $4$ for numbers other than $300$ so the greatest number you need to test is $309$
Either you are less than $300$ when, with units digit $a$ you have $290+2a+11=313$ and $a=6$, or greater than $300$ when $300+2a+3=313$, and $a=5$.
$endgroup$
The greatest possible sum of digits for a positive integer less than $313$ is $20$ (from $299$) so the minimum value you need to test is $293$. The sum of digits in this range is at least $3$ ($300$) and is at least $4$ for numbers other than $300$ so the greatest number you need to test is $309$
Either you are less than $300$ when, with units digit $a$ you have $290+2a+11=313$ and $a=6$, or greater than $300$ when $300+2a+3=313$, and $a=5$.
answered Jan 13 at 8:26
Mark BennetMark Bennet
81.6k984181
81.6k984181
add a comment |
add a comment |
$begingroup$
You are on the right track with:
$$overline{xyz}+x+y+z=101x+11y+2z=313.$$
$xne 1$, because otherwise $11y+2zle 117<212$.
$x=2$, then:
$$z=frac{111-11y}{2} Rightarrow y=9, z=6 Rightarrow overline{xyz}=296.$$
$endgroup$
add a comment |
$begingroup$
You are on the right track with:
$$overline{xyz}+x+y+z=101x+11y+2z=313.$$
$xne 1$, because otherwise $11y+2zle 117<212$.
$x=2$, then:
$$z=frac{111-11y}{2} Rightarrow y=9, z=6 Rightarrow overline{xyz}=296.$$
$endgroup$
add a comment |
$begingroup$
You are on the right track with:
$$overline{xyz}+x+y+z=101x+11y+2z=313.$$
$xne 1$, because otherwise $11y+2zle 117<212$.
$x=2$, then:
$$z=frac{111-11y}{2} Rightarrow y=9, z=6 Rightarrow overline{xyz}=296.$$
$endgroup$
You are on the right track with:
$$overline{xyz}+x+y+z=101x+11y+2z=313.$$
$xne 1$, because otherwise $11y+2zle 117<212$.
$x=2$, then:
$$z=frac{111-11y}{2} Rightarrow y=9, z=6 Rightarrow overline{xyz}=296.$$
answered Jan 13 at 7:35
farruhotafarruhota
21.1k2841
21.1k2841
add a comment |
add a comment |
$begingroup$
$$101x+11y+2z=313$$
where $x,y,z$ are numbers between $0$ and $9$.
If $xle 1$, then $101x+11y + 2z le 101+11(9)+2(9)=101+13(9)=218 $
If $x=2$, then we have $$11y+2z=111$$
Since $zle 9$, we have $$11y =111-2z ge 93$$
Hence, we must have $y=9$. $$2z=111-11(9)=111-99=12.$$
Hence $296$ is a solution.
If $x=3$, then we have $11y+2z=10$ which forces $y=0, z=5$.
The solutions are $305$ and $296$.
$endgroup$
add a comment |
$begingroup$
$$101x+11y+2z=313$$
where $x,y,z$ are numbers between $0$ and $9$.
If $xle 1$, then $101x+11y + 2z le 101+11(9)+2(9)=101+13(9)=218 $
If $x=2$, then we have $$11y+2z=111$$
Since $zle 9$, we have $$11y =111-2z ge 93$$
Hence, we must have $y=9$. $$2z=111-11(9)=111-99=12.$$
Hence $296$ is a solution.
If $x=3$, then we have $11y+2z=10$ which forces $y=0, z=5$.
The solutions are $305$ and $296$.
$endgroup$
add a comment |
$begingroup$
$$101x+11y+2z=313$$
where $x,y,z$ are numbers between $0$ and $9$.
If $xle 1$, then $101x+11y + 2z le 101+11(9)+2(9)=101+13(9)=218 $
If $x=2$, then we have $$11y+2z=111$$
Since $zle 9$, we have $$11y =111-2z ge 93$$
Hence, we must have $y=9$. $$2z=111-11(9)=111-99=12.$$
Hence $296$ is a solution.
If $x=3$, then we have $11y+2z=10$ which forces $y=0, z=5$.
The solutions are $305$ and $296$.
$endgroup$
$$101x+11y+2z=313$$
where $x,y,z$ are numbers between $0$ and $9$.
If $xle 1$, then $101x+11y + 2z le 101+11(9)+2(9)=101+13(9)=218 $
If $x=2$, then we have $$11y+2z=111$$
Since $zle 9$, we have $$11y =111-2z ge 93$$
Hence, we must have $y=9$. $$2z=111-11(9)=111-99=12.$$
Hence $296$ is a solution.
If $x=3$, then we have $11y+2z=10$ which forces $y=0, z=5$.
The solutions are $305$ and $296$.
answered Jan 13 at 7:36
Siong Thye GohSiong Thye Goh
103k1468119
103k1468119
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.
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%2f3071768%2fnumbers-n-such-that-n-plus-the-sum-of-ns-digits-is-313%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