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Problem 790

AMC 10/12, early questions

Number theory Difficulty 3.7 Find the answer CEMC Cayley · Canada · 2022

There are exactly four ordered pairs of positive integers $(x,y)$ that satisfy the equation $20x+11y=881$ . Mehdi writes down the four
values of $y$ and adds the smallest
and largest of these values. What is this sum?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

Since $20x + 11y = 881$ , then
$20x = 881 - 11y$ and $11y = 881 - 20x$ .

Since $x$ is an integer, then $20x$ is a multiple of 10 and so the units
digit of $20x$ is 0 which means that
the units digit of $881-20x$ is 1,
and so the units digit of $11y$ is
1.

Since the units digit of $11y$ is 1,
then the units digit of $y$ is
1.

Since $20x$ is positive, then $11y = 881-20x$ is smaller than $881$ , which means that $\text{11y lt; 881}$ and so $\text{y lt; 881 11 80.1}$ .

Thus, the possible values of $y$ are
1, 11, 21, 31, 41, 51, 61, 71.

We check each of these:

$y$	$11y = 881 - 20x$	$20x$	$x$
1	11	870	Not an integer
11	121	760	38
21	231	650	Not an integer
31	341	540	27
41	451	430	Not an integer
51	561	320	16
61	671	210	Not an integer
71	781	100	5

Therefore, the sum of the smallest and largest of the permissible
values of $y$ is $11 + 71 = 82$ .

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.