Mixing
AMC 2000, Problem 16
$begingroup$
A checkerboard of  $13$ rows and  $17$ columns has a number written in each square, beginning in the upper left corner, so that the first row is numbered $1,2,3,4,...,17$ , the second row $18,19,20...34$, and so on down the board. If the board is renumbered so that the left column, top to bottom, is $1,2,3,....13$ , the second column $14,15,16,17,...26$ and so on across the board, some squares have the same numbers in both numbering systems. Find the sum of the numbers in these squares (under either system).
The answer was 555, with the solution provided below
Index the rows with  $ i=1,2,3,...13$
Index the columns with $ j=1,2,3,4,...17$
For the first row number the cells $1,2,3...17$
For the second, $18,19,20..34$ and so on
So the number in row = $i$ and column = $j$ is 
$f(i,j) = 17(i-1)+j$
Similarly, numbering the same cells columnwise we find the number in row = $i$ and column = $j$ is $g(i,j)= i+13j-13$
So we need to solve
$f(i,j) =g(i,j)$
Which gives $i=(1+3j)/4$





We get the solutions $1, 56, 111, 166,221$ which sums to $555$.




The question is why would there be $17(i-1)+j $ being equated as follows? Is it necessarily unique? How did the author derived it to find the solutions?
And is there another way?
elementary-number-theory
edited Jan 22 at 1:14
Bill Dubuque
212k29195650
asked Jan 22 at 1:07
299792458299792458
527
$endgroup$
add a comment |
$begingroup$
A checkerboard of  $13$ rows and  $17$ columns has a number written in each square, beginning in the upper left corner, so that the first row is numbered $1,2,3,4,...,17$ , the second row $18,19,20...34$, and so on down the board. If the board is renumbered so that the left column, top to bottom, is $1,2,3,....13$ , the second column $14,15,16,17,...26$ and so on across the board, some squares have the same numbers in both numbering systems. Find the sum of the numbers in these squares (under either system).
The answer was 555, with the solution provided below
Index the rows with  $ i=1,2,3,...13$
Index the columns with $ j=1,2,3,4,...17$
For the first row number the cells $1,2,3...17$
For the second, $18,19,20..34$ and so on
So the number in row = $i$ and column = $j$ is 
$f(i,j) = 17(i-1)+j$
Similarly, numbering the same cells columnwise we find the number in row = $i$ and column = $j$ is $g(i,j)= i+13j-13$
So we need to solve
$f(i,j) =g(i,j)$
Which gives $i=(1+3j)/4$





We get the solutions $1, 56, 111, 166,221$ which sums to $555$.




The question is why would there be $17(i-1)+j $ being equated as follows? Is it necessarily unique? How did the author derived it to find the solutions?
And is there another way?
elementary-number-theory
edited Jan 22 at 1:14
Bill Dubuque
212k29195650
asked Jan 22 at 1:07
299792458299792458
527
$endgroup$
add a comment |
$begingroup$
A checkerboard of  $13$ rows and  $17$ columns has a number written in each square, beginning in the upper left corner, so that the first row is numbered $1,2,3,4,...,17$ , the second row $18,19,20...34$, and so on down the board. If the board is renumbered so that the left column, top to bottom, is $1,2,3,....13$ , the second column $14,15,16,17,...26$ and so on across the board, some squares have the same numbers in both numbering systems. Find the sum of the numbers in these squares (under either system).
The answer was 555, with the solution provided below
Index the rows with  $ i=1,2,3,...13$
Index the columns with $ j=1,2,3,4,...17$
For the first row number the cells $1,2,3...17$
For the second, $18,19,20..34$ and so on
So the number in row = $i$ and column = $j$ is 
$f(i,j) = 17(i-1)+j$
Similarly, numbering the same cells columnwise we find the number in row = $i$ and column = $j$ is $g(i,j)= i+13j-13$
So we need to solve
$f(i,j) =g(i,j)$
Which gives $i=(1+3j)/4$





We get the solutions $1, 56, 111, 166,221$ which sums to $555$.




The question is why would there be $17(i-1)+j $ being equated as follows? Is it necessarily unique? How did the author derived it to find the solutions?
And is there another way?
elementary-number-theory
edited Jan 22 at 1:14
Bill Dubuque
212k29195650
asked Jan 22 at 1:07
299792458299792458
527
$endgroup$
A checkerboard of  $13$ rows and  $17$ columns has a number written in each square, beginning in the upper left corner, so that the first row is numbered $1,2,3,4,...,17$ , the second row $18,19,20...34$, and so on down the board. If the board is renumbered so that the left column, top to bottom, is $1,2,3,....13$ , the second column $14,15,16,17,...26$ and so on across the board, some squares have the same numbers in both numbering systems. Find the sum of the numbers in these squares (under either system).
The answer was 555, with the solution provided below
Index the rows with  $ i=1,2,3,...13$
Index the columns with $ j=1,2,3,4,...17$
For the first row number the cells $1,2,3...17$
For the second, $18,19,20..34$ and so on
So the number in row = $i$ and column = $j$ is 
$f(i,j) = 17(i-1)+j$
Similarly, numbering the same cells columnwise we find the number in row = $i$ and column = $j$ is $g(i,j)= i+13j-13$
So we need to solve
$f(i,j) =g(i,j)$
Which gives $i=(1+3j)/4$





We get the solutions $1, 56, 111, 166,221$ which sums to $555$.




The question is why would there be $17(i-1)+j $ being equated as follows? Is it necessarily unique? How did the author derived it to find the solutions?
And is there another way?
elementary-number-theory
elementary-number-theory
edited Jan 22 at 1:14
Bill Dubuque
212k29195650
asked Jan 22 at 1:07
299792458299792458
527
edited Jan 22 at 1:14
Bill Dubuque
212k29195650
asked Jan 22 at 1:07
299792458299792458
527
edited Jan 22 at 1:14
Bill Dubuque
212k29195650
edited Jan 22 at 1:14
Bill Dubuque
212k29195650
edited Jan 22 at 1:14
Bill Dubuque
212k29195650
212k29195650
asked Jan 22 at 1:07
299792458299792458
527
asked Jan 22 at 1:07
299792458299792458
527
asked Jan 22 at 1:07
299792458299792458
527
527
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Imagine first that we are number along the rows. What number will we place in the cell in row $i,$ column $j?$ We have already full numbered $i-1$ rows, and since there are $17$ numbers in a row, in these rows we have used the first $17(i-1)$ numbers. In row $i,$ column $1$ we therefore place $17(i-1)+1,$ in row $i,$ column $2$ we place $17(i-1)+2,$ ... and in row $i,$ column $j$ we place $17(i-1)+j$. This explains the formula for $f(i,j),$ and a wholly analogous argument explains the formula for $g(i,j).$
I don't fully understand all of your questions. We equate $f(i,j)$ and $g(i,j)$ because the question asks for the cells where the two numbers are the same. We easily get the solution $$i={3j+1over4}$$ Since $i$ and $j$ are integers, we need $3j+1$ to be divisible by $4,$ which happens when $jin{1,5,9,13,17}$
Solving for the corresponding $i$ values and plugging into the the formula for $f(i,j)$ (or $g(i,j),$) should give you the five numbers that the author claims, but I haven't checked this.
answered Jan 22 at 1:39
saulspatzsaulspatz
16.7k31333
$endgroup$
$begingroup$
But wouldn’t I obtain a different answer if I had considered $17(i-2) +2j$?
$endgroup$
– 299792458
Jan 22 at 1:50
$begingroup$
Yes, but you wouldn't get a correct answer. With that formula, the cell in row $1$, column $1$ would be numbered $-15.$ The first part of the argument proves that $f(i,j)$ gives the correct numbering.
$endgroup$
– saulspatz
Jan 22 at 1:54
add a comment |
$begingroup$
Let $fleft(i,jright)$ be the integer written in each cell according to the first numbering system, such that:
begin{equation*}
fleft(i,jright)=17(i-1)+j
end{equation*}
where $1leq ileq 13$ is the row value, and $1leq jleq 17$ is the column value.
This formula means that each cell is the sum of the column value $j$ with a multiple of $17$ that depends on the row value.
The same idea applies to the second numbering system, with $gleft(i,jright)$ being the writter integer in this case, such that:
begin{equation*}
gleft(i,jright)=13(j-1)+i
end{equation*}
This second formula means that each cell is the sum of the row value $i$ with a multiple of $13$ that depends on the column value this time.
Anyway we are interested on coincident values on both numbering systems. then:
begin{equation*}
fleft(i,jright)=gleft(i,jright)
end{equation*}
begin{equation*}
17(i-1)+j=13(j-1)+i
end{equation*}
begin{equation*}
16i-12j=4
end{equation*}
begin{equation*}
4i-3j=1
end{equation*}
Thus, a multiple of $4$ subtracted by a multiple of $3$ that is equal to $1$, which happens to:
begin{equation*}
4-3=1rightarrowleft(i,jright)=left(1,1right)
end{equation*}
begin{equation*}
16-15=1rightarrowleft(i,jright)=left(4,5right)
end{equation*}
begin{equation*}
28-27=1rightarrowleft(i,jright)=left(7,9right)
end{equation*}
begin{equation*}
40-39=1rightarrowleft(i,jright)=left(10,13right)
end{equation*}
begin{equation*}
52-51=1rightarrowleft(i,jright)=left(13,17right)
end{equation*}
Thus, the sum of these cells is:
begin{equation*}
sum=fleft(1,1right)+fleft(4,5right)+fleft(7,9right)+fleft(10,13right)+fleft(13,17right)
end{equation*}
begin{equation*}
sum=gleft(1,1right)+gleft(4,5right)+gleft(7,9right)+gleft(10,13right)+gleft(13,17right)
end{equation*}
begin{equation*}
sum=1+56+111+166+221=555
end{equation*}
answered Jan 22 at 11:52
Cleyton MutoCleyton Muto
913410
$endgroup$
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%2f3082629%2famc-2000-problem-16%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
$begingroup$
Imagine first that we are number along the rows. What number will we place in the cell in row $i,$ column $j?$ We have already full numbered $i-1$ rows, and since there are $17$ numbers in a row, in these rows we have used the first $17(i-1)$ numbers. In row $i,$ column $1$ we therefore place $17(i-1)+1,$ in row $i,$ column $2$ we place $17(i-1)+2,$ ... and in row $i,$ column $j$ we place $17(i-1)+j$. This explains the formula for $f(i,j),$ and a wholly analogous argument explains the formula for $g(i,j).$
I don't fully understand all of your questions. We equate $f(i,j)$ and $g(i,j)$ because the question asks for the cells where the two numbers are the same. We easily get the solution $$i={3j+1over4}$$ Since $i$ and $j$ are integers, we need $3j+1$ to be divisible by $4,$ which happens when $jin{1,5,9,13,17}$
Solving for the corresponding $i$ values and plugging into the the formula for $f(i,j)$ (or $g(i,j),$) should give you the five numbers that the author claims, but I haven't checked this.
answered Jan 22 at 1:39
saulspatzsaulspatz
16.7k31333
$endgroup$
$begingroup$
But wouldn’t I obtain a different answer if I had considered $17(i-2) +2j$?
$endgroup$
– 299792458
Jan 22 at 1:50
$begingroup$
Yes, but you wouldn't get a correct answer. With that formula, the cell in row $1$, column $1$ would be numbered $-15.$ The first part of the argument proves that $f(i,j)$ gives the correct numbering.
$endgroup$
– saulspatz
Jan 22 at 1:54
add a comment |
$begingroup$
Imagine first that we are number along the rows. What number will we place in the cell in row $i,$ column $j?$ We have already full numbered $i-1$ rows, and since there are $17$ numbers in a row, in these rows we have used the first $17(i-1)$ numbers. In row $i,$ column $1$ we therefore place $17(i-1)+1,$ in row $i,$ column $2$ we place $17(i-1)+2,$ ... and in row $i,$ column $j$ we place $17(i-1)+j$. This explains the formula for $f(i,j),$ and a wholly analogous argument explains the formula for $g(i,j).$
I don't fully understand all of your questions. We equate $f(i,j)$ and $g(i,j)$ because the question asks for the cells where the two numbers are the same. We easily get the solution $$i={3j+1over4}$$ Since $i$ and $j$ are integers, we need $3j+1$ to be divisible by $4,$ which happens when $jin{1,5,9,13,17}$
Solving for the corresponding $i$ values and plugging into the the formula for $f(i,j)$ (or $g(i,j),$) should give you the five numbers that the author claims, but I haven't checked this.
answered Jan 22 at 1:39
saulspatzsaulspatz
16.7k31333
$endgroup$
$begingroup$
But wouldn’t I obtain a different answer if I had considered $17(i-2) +2j$?
$endgroup$
– 299792458
Jan 22 at 1:50
$begingroup$
Yes, but you wouldn't get a correct answer. With that formula, the cell in row $1$, column $1$ would be numbered $-15.$ The first part of the argument proves that $f(i,j)$ gives the correct numbering.
$endgroup$
– saulspatz
Jan 22 at 1:54
add a comment |
$begingroup$
Imagine first that we are number along the rows. What number will we place in the cell in row $i,$ column $j?$ We have already full numbered $i-1$ rows, and since there are $17$ numbers in a row, in these rows we have used the first $17(i-1)$ numbers. In row $i,$ column $1$ we therefore place $17(i-1)+1,$ in row $i,$ column $2$ we place $17(i-1)+2,$ ... and in row $i,$ column $j$ we place $17(i-1)+j$. This explains the formula for $f(i,j),$ and a wholly analogous argument explains the formula for $g(i,j).$
I don't fully understand all of your questions. We equate $f(i,j)$ and $g(i,j)$ because the question asks for the cells where the two numbers are the same. We easily get the solution $$i={3j+1over4}$$ Since $i$ and $j$ are integers, we need $3j+1$ to be divisible by $4,$ which happens when $jin{1,5,9,13,17}$
Solving for the corresponding $i$ values and plugging into the the formula for $f(i,j)$ (or $g(i,j),$) should give you the five numbers that the author claims, but I haven't checked this.
answered Jan 22 at 1:39
saulspatzsaulspatz
16.7k31333
$endgroup$
Imagine first that we are number along the rows. What number will we place in the cell in row $i,$ column $j?$ We have already full numbered $i-1$ rows, and since there are $17$ numbers in a row, in these rows we have used the first $17(i-1)$ numbers. In row $i,$ column $1$ we therefore place $17(i-1)+1,$ in row $i,$ column $2$ we place $17(i-1)+2,$ ... and in row $i,$ column $j$ we place $17(i-1)+j$. This explains the formula for $f(i,j),$ and a wholly analogous argument explains the formula for $g(i,j).$
I don't fully understand all of your questions. We equate $f(i,j)$ and $g(i,j)$ because the question asks for the cells where the two numbers are the same. We easily get the solution $$i={3j+1over4}$$ Since $i$ and $j$ are integers, we need $3j+1$ to be divisible by $4,$ which happens when $jin{1,5,9,13,17}$
Solving for the corresponding $i$ values and plugging into the the formula for $f(i,j)$ (or $g(i,j),$) should give you the five numbers that the author claims, but I haven't checked this.
answered Jan 22 at 1:39
saulspatzsaulspatz
16.7k31333
answered Jan 22 at 1:39
saulspatzsaulspatz
16.7k31333
answered Jan 22 at 1:39
saulspatzsaulspatz
16.7k31333
answered Jan 22 at 1:39
saulspatzsaulspatz
16.7k31333
16.7k31333
$begingroup$
But wouldn’t I obtain a different answer if I had considered $17(i-2) +2j$?
$endgroup$
– 299792458
Jan 22 at 1:50
$begingroup$
Yes, but you wouldn't get a correct answer. With that formula, the cell in row $1$, column $1$ would be numbered $-15.$ The first part of the argument proves that $f(i,j)$ gives the correct numbering.
$endgroup$
– saulspatz
Jan 22 at 1:54
add a comment |
$begingroup$
But wouldn’t I obtain a different answer if I had considered $17(i-2) +2j$?
$endgroup$
– 299792458
Jan 22 at 1:50
$begingroup$
Yes, but you wouldn't get a correct answer. With that formula, the cell in row $1$, column $1$ would be numbered $-15.$ The first part of the argument proves that $f(i,j)$ gives the correct numbering.
$endgroup$
– saulspatz
Jan 22 at 1:54
$begingroup$
But wouldn’t I obtain a different answer if I had considered $17(i-2) +2j$?
$endgroup$
– 299792458
Jan 22 at 1:50
$begingroup$
But wouldn’t I obtain a different answer if I had considered $17(i-2) +2j$?
$endgroup$
– 299792458
Jan 22 at 1:50
$begingroup$
Yes, but you wouldn't get a correct answer. With that formula, the cell in row $1$, column $1$ would be numbered $-15.$ The first part of the argument proves that $f(i,j)$ gives the correct numbering.
$endgroup$
– saulspatz
Jan 22 at 1:54
$begingroup$
Yes, but you wouldn't get a correct answer. With that formula, the cell in row $1$, column $1$ would be numbered $-15.$ The first part of the argument proves that $f(i,j)$ gives the correct numbering.
$endgroup$
– saulspatz
Jan 22 at 1:54
add a comment |
$begingroup$
Let $fleft(i,jright)$ be the integer written in each cell according to the first numbering system, such that:
begin{equation*}
fleft(i,jright)=17(i-1)+j
end{equation*}
where $1leq ileq 13$ is the row value, and $1leq jleq 17$ is the column value.
This formula means that each cell is the sum of the column value $j$ with a multiple of $17$ that depends on the row value.
The same idea applies to the second numbering system, with $gleft(i,jright)$ being the writter integer in this case, such that:
begin{equation*}
gleft(i,jright)=13(j-1)+i
end{equation*}
This second formula means that each cell is the sum of the row value $i$ with a multiple of $13$ that depends on the column value this time.
Anyway we are interested on coincident values on both numbering systems. then:
begin{equation*}
fleft(i,jright)=gleft(i,jright)
end{equation*}
begin{equation*}
17(i-1)+j=13(j-1)+i
end{equation*}
begin{equation*}
16i-12j=4
end{equation*}
begin{equation*}
4i-3j=1
end{equation*}
Thus, a multiple of $4$ subtracted by a multiple of $3$ that is equal to $1$, which happens to:
begin{equation*}
4-3=1rightarrowleft(i,jright)=left(1,1right)
end{equation*}
begin{equation*}
16-15=1rightarrowleft(i,jright)=left(4,5right)
end{equation*}
begin{equation*}
28-27=1rightarrowleft(i,jright)=left(7,9right)
end{equation*}
begin{equation*}
40-39=1rightarrowleft(i,jright)=left(10,13right)
end{equation*}
begin{equation*}
52-51=1rightarrowleft(i,jright)=left(13,17right)
end{equation*}
Thus, the sum of these cells is:
begin{equation*}
sum=fleft(1,1right)+fleft(4,5right)+fleft(7,9right)+fleft(10,13right)+fleft(13,17right)
end{equation*}
begin{equation*}
sum=gleft(1,1right)+gleft(4,5right)+gleft(7,9right)+gleft(10,13right)+gleft(13,17right)
end{equation*}
begin{equation*}
sum=1+56+111+166+221=555
end{equation*}
answered Jan 22 at 11:52
Cleyton MutoCleyton Muto
913410
$endgroup$
add a comment |
$begingroup$
Let $fleft(i,jright)$ be the integer written in each cell according to the first numbering system, such that:
begin{equation*}
fleft(i,jright)=17(i-1)+j
end{equation*}
where $1leq ileq 13$ is the row value, and $1leq jleq 17$ is the column value.
This formula means that each cell is the sum of the column value $j$ with a multiple of $17$ that depends on the row value.
The same idea applies to the second numbering system, with $gleft(i,jright)$ being the writter integer in this case, such that:
begin{equation*}
gleft(i,jright)=13(j-1)+i
end{equation*}
This second formula means that each cell is the sum of the row value $i$ with a multiple of $13$ that depends on the column value this time.
Anyway we are interested on coincident values on both numbering systems. then:
begin{equation*}
fleft(i,jright)=gleft(i,jright)
end{equation*}
begin{equation*}
17(i-1)+j=13(j-1)+i
end{equation*}
begin{equation*}
16i-12j=4
end{equation*}
begin{equation*}
4i-3j=1
end{equation*}
Thus, a multiple of $4$ subtracted by a multiple of $3$ that is equal to $1$, which happens to:
begin{equation*}
4-3=1rightarrowleft(i,jright)=left(1,1right)
end{equation*}
begin{equation*}
16-15=1rightarrowleft(i,jright)=left(4,5right)
end{equation*}
begin{equation*}
28-27=1rightarrowleft(i,jright)=left(7,9right)
end{equation*}
begin{equation*}
40-39=1rightarrowleft(i,jright)=left(10,13right)
end{equation*}
begin{equation*}
52-51=1rightarrowleft(i,jright)=left(13,17right)
end{equation*}
Thus, the sum of these cells is:
begin{equation*}
sum=fleft(1,1right)+fleft(4,5right)+fleft(7,9right)+fleft(10,13right)+fleft(13,17right)
end{equation*}
begin{equation*}
sum=gleft(1,1right)+gleft(4,5right)+gleft(7,9right)+gleft(10,13right)+gleft(13,17right)
end{equation*}
begin{equation*}
sum=1+56+111+166+221=555
end{equation*}
answered Jan 22 at 11:52
Cleyton MutoCleyton Muto
913410
$endgroup$
add a comment |
$begingroup$
Let $fleft(i,jright)$ be the integer written in each cell according to the first numbering system, such that:
begin{equation*}
fleft(i,jright)=17(i-1)+j
end{equation*}
where $1leq ileq 13$ is the row value, and $1leq jleq 17$ is the column value.
This formula means that each cell is the sum of the column value $j$ with a multiple of $17$ that depends on the row value.
The same idea applies to the second numbering system, with $gleft(i,jright)$ being the writter integer in this case, such that:
begin{equation*}
gleft(i,jright)=13(j-1)+i
end{equation*}
This second formula means that each cell is the sum of the row value $i$ with a multiple of $13$ that depends on the column value this time.
Anyway we are interested on coincident values on both numbering systems. then:
begin{equation*}
fleft(i,jright)=gleft(i,jright)
end{equation*}
begin{equation*}
17(i-1)+j=13(j-1)+i
end{equation*}
begin{equation*}
16i-12j=4
end{equation*}
begin{equation*}
4i-3j=1
end{equation*}
Thus, a multiple of $4$ subtracted by a multiple of $3$ that is equal to $1$, which happens to:
begin{equation*}
4-3=1rightarrowleft(i,jright)=left(1,1right)
end{equation*}
begin{equation*}
16-15=1rightarrowleft(i,jright)=left(4,5right)
end{equation*}
begin{equation*}
28-27=1rightarrowleft(i,jright)=left(7,9right)
end{equation*}
begin{equation*}
40-39=1rightarrowleft(i,jright)=left(10,13right)
end{equation*}
begin{equation*}
52-51=1rightarrowleft(i,jright)=left(13,17right)
end{equation*}
Thus, the sum of these cells is:
begin{equation*}
sum=fleft(1,1right)+fleft(4,5right)+fleft(7,9right)+fleft(10,13right)+fleft(13,17right)
end{equation*}
begin{equation*}
sum=gleft(1,1right)+gleft(4,5right)+gleft(7,9right)+gleft(10,13right)+gleft(13,17right)
end{equation*}
begin{equation*}
sum=1+56+111+166+221=555
end{equation*}
answered Jan 22 at 11:52
Cleyton MutoCleyton Muto
913410
$endgroup$
Let $fleft(i,jright)$ be the integer written in each cell according to the first numbering system, such that:
begin{equation*}
fleft(i,jright)=17(i-1)+j
end{equation*}
where $1leq ileq 13$ is the row value, and $1leq jleq 17$ is the column value.
This formula means that each cell is the sum of the column value $j$ with a multiple of $17$ that depends on the row value.
The same idea applies to the second numbering system, with $gleft(i,jright)$ being the writter integer in this case, such that:
begin{equation*}
gleft(i,jright)=13(j-1)+i
end{equation*}
This second formula means that each cell is the sum of the row value $i$ with a multiple of $13$ that depends on the column value this time.
Anyway we are interested on coincident values on both numbering systems. then:
begin{equation*}
fleft(i,jright)=gleft(i,jright)
end{equation*}
begin{equation*}
17(i-1)+j=13(j-1)+i
end{equation*}
begin{equation*}
16i-12j=4
end{equation*}
begin{equation*}
4i-3j=1
end{equation*}
Thus, a multiple of $4$ subtracted by a multiple of $3$ that is equal to $1$, which happens to:
begin{equation*}
4-3=1rightarrowleft(i,jright)=left(1,1right)
end{equation*}
begin{equation*}
16-15=1rightarrowleft(i,jright)=left(4,5right)
end{equation*}
begin{equation*}
28-27=1rightarrowleft(i,jright)=left(7,9right)
end{equation*}
begin{equation*}
40-39=1rightarrowleft(i,jright)=left(10,13right)
end{equation*}
begin{equation*}
52-51=1rightarrowleft(i,jright)=left(13,17right)
end{equation*}
Thus, the sum of these cells is:
begin{equation*}
sum=fleft(1,1right)+fleft(4,5right)+fleft(7,9right)+fleft(10,13right)+fleft(13,17right)
end{equation*}
begin{equation*}
sum=gleft(1,1right)+gleft(4,5right)+gleft(7,9right)+gleft(10,13right)+gleft(13,17right)
end{equation*}
begin{equation*}
sum=1+56+111+166+221=555
end{equation*}
answered Jan 22 at 11:52
Cleyton MutoCleyton Muto
913410
answered Jan 22 at 11:52
Cleyton MutoCleyton Muto
913410
answered Jan 22 at 11:52
Cleyton MutoCleyton Muto
913410
answered Jan 22 at 11:52
Cleyton MutoCleyton Muto
913410
913410
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%2f3082629%2famc-2000-problem-16%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
