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Difference between revisions of "1995 AIME Problems/Problem 8"

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We know that <math>x \equiv 0 \mod y</math> and  <math>x+1 \equiv 0 \mod y+1</math>.
 
We know that <math>x \equiv 0 \mod y</math> and  <math>x+1 \equiv 0 \mod y+1</math>.
  
−
Write <math>x</math> and <math>ky</math> for some integer <math>k</math>. Then, <math>ky+1 \equiv 0\mod y+1</math>. We can add <math>k</math> to each side in order to factor out a <math>y+1</math>. So, <math>ky+k+1 \equiv k \mod y+1</math> or <math>k(y+1)+1 \equiv k \mod y+1</math>. We know that <math>k(y+1) \equiv 0 \mod y+1</math>. We finally achieve the congruence <math>1-k \equiv 0 \mod y+1</math>.  
+
Write <math>x</math> as <math>ky</math> for some integer <math>k</math>. Then, <math>ky+1 \equiv 0\mod y+1</math>. We can add <math>k</math> to each side in order to factor out a <math>y+1</math>. So, <math>ky+k+1 \equiv k \mod y+1</math> or <math>k(y+1)+1 \equiv k \mod y+1</math>. We know that <math>k(y+1) \equiv 0 \mod y+1</math>. We finally achieve the congruence <math>1-k \equiv 0 \mod y+1</math>.  
  
 
We can now write <math>k</math> as <math>(y+1)a+1</math>. Plugging this back in, if we have a value for <math>y</math>, then <math>x = ky = ((y+1)a+1)y = y(y+1)a+y</math>. We only have to check values of <math>y</math> when <math>y(y+1)<100</math>. This yields the equations <math>x = 2a+1, 6a+2, 12a+3, 20a+4, 30a+5, 42a+6, 56a+7, 72a+8, 90a+9</math>.  
 
We can now write <math>k</math> as <math>(y+1)a+1</math>. Plugging this back in, if we have a value for <math>y</math>, then <math>x = ky = ((y+1)a+1)y = y(y+1)a+y</math>. We only have to check values of <math>y</math> when <math>y(y+1)<100</math>. This yields the equations <math>x = 2a+1, 6a+2, 12a+3, 20a+4, 30a+5, 42a+6, 56a+7, 72a+8, 90a+9</math>.  

Revision as of 22:11, 30 January 2023

Problem

For how many ordered pairs of positive integers $(x,y),$ with $y<x\le 100,$ are both $\frac xy$ and $\frac{x+1}{y+1}$ integers?

Solution 1

Since $y|x$, $y+1|x+1$, then $\text{gcd}\,(y,x)=y$ (the bars indicate divisibility) and $\text{gcd}\,(y+1,x+1)=y+1$. By the Euclidean algorithm, these can be rewritten respectively as $\text{gcd}\,(y,x-y)=y$ and $\text{gcd}\, (y+1,x-y)=y+1$, which implies that both $y,y+1 | x-y$. Also, as $\text{gcd}\,(y,y+1) = 1$, it follows that $y(y+1)|x-y$. [1]

Thus, for a given value of $y$, we need the number of multiples of $y(y+1)$ from $0$ to $100-y$ (as $x \le 100$). It follows that there are $\left\lfloor\frac{100-y}{y(y+1)} \right\rfloor$ satisfactory positive integers for all integers $y \le 100$. The answer is

\[\sum_{y=1}^{99} \left\lfloor\frac{100-y}{y(y+1)} \right\rfloor = 49 + 16 + 8 + 4 + 3 + 2 + 1 + 1 + 1 = \boxed{085}.\]



^ Another way of stating this is to note that if $\frac{x}{y}$ and $\frac{x+1}{y+1}$ are integers, then $\frac{x}{y} - 1 = \frac{x-y}{y}$ and $\frac{x+1}{y+1} - 1 = \frac{x-y}{y+1}$ must be integers. Since $y$ and $y+1$ cannot share common prime factors, it follows that $\frac{x-y}{y(y+1)}$ must also be an integer.

Solution 2

We know that $x \equiv 0 \mod y$ and $x+1 \equiv 0 \mod y+1$.

Write $x$ as $ky$ for some integer $k$. Then, $ky+1 \equiv 0\mod y+1$. We can add $k$ to each side in order to factor out a $y+1$. So, $ky+k+1 \equiv k \mod y+1$ or $k(y+1)+1 \equiv k \mod y+1$. We know that $k(y+1) \equiv 0 \mod y+1$. We finally achieve the congruence $1-k \equiv 0 \mod y+1$.

We can now write $k$ as $(y+1)a+1$. Plugging this back in, if we have a value for $y$, then $x = ky = ((y+1)a+1)y = y(y+1)a+y$. We only have to check values of $y$ when $y(y+1)<100$. This yields the equations $x = 2a+1, 6a+2, 12a+3, 20a+4, 30a+5, 42a+6, 56a+7, 72a+8, 90a+9$.

Finding all possible values of $a$ such that $y<x<100$, we get $49 + 16 + 8 + 4 + 3 + 2 + 1 + 1 + 1 = \boxed{085}.$

See also

1995 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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