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.
+128475 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
