Question

quick help
anyone help please
There are two 3-digit numbers n having the property that n is divisible by 11 and n/11 is equal to the sum of the squares of the digits of n.

Find both values of n. You may submit them in either order.

Answer 1

@phi @Preetha @e.mccormick @SithsAndGiggles @TylerD @badatmath_01

Answer 2

@amistre64

Answer 3

@mathstudent55

Answer 4

n = 11k
since n > 99 and n < 1000,
k is between 10 and 90. (110 < n < 990)
k thus has 2 digits, so let k = a*10 +b
n = 11*k = 11*(a*10 + b)
=a*110 + 11*b
= a*100 + 10*a*b + 1*b
assuming temporarily that a*b*10 < 100 (so that a*b is the second digit of n):
k = a^2 + (a*b)^2 + b^2
11k = n, so:
11*(a^2 + (a*b)^2 + b^2) = a*100 + 10*a*b + 1*b
letting b = 0, solving quadratic for a gives a = 5
so:
k = 50 -> n = 550 for one solution, the only such solution that a*b < 10. So then, knowing that a*b > 10, we can deduce that b is greater than 1. We also note that the 10s digit of n must be zero, as b is less than 10. As such, the sum of the digits must therefore be 11, which leads us to the set:
(2,9);(3,8);(4,7);(5,6). Squaring and adding gives:
85,73,65,61. multiplying by 11:
935,803,715,671
of which 803 is the obvious answer, being the only one containing 0, 3, and 8.
Therefor, we have k = 50 and k = 73. giving:
n = 550 and n = 803