Find the largest three-digit number (written in decimal notation) that is divisible by 22 and such that the sum of the units digit and the tens digit is 11.

Guest Dec 21, 2019

#1**+1 **

I get 792

Let the number be a*10^2 + b*10 + c

If the number is divisible by 22, it is also divisible by 11

This means (possibly) that

a - b + c = 0

And b + c = 11 subtract these

a - 2b = -11

2b - a = 11

2b = 11 + a

So....we want a as large as possible such that b is a single digit....and a must be odd because 2b is even

The largest value of a such that b is a single digit is when a = 7

So......this means that b = 9

And

7 - 9 + 2 = 0

So...792 is the number

CPhill Dec 21, 2019