Vassil of Lawnswood High School in Leeds sent in a good solution. He worked systematically through the possible values of c using the rearrangement c(c^2 -1) = b(10-b) + 99a to establish the range of possible values for a and then testing to find values of b which fitted.

Many thanks to Sue Liu who is in S6 at Madras College in St Andrews for the followingbeautifully neat solution.

To find all 3-digit numbers abc (in base 10) such that


a + b2 + c3 = 100a + 10b + c
Rearranging gives
c3 - c - 99a = b (10 - b)

c(c+1)(c-1) - 99a = b (10 - b)

For any three consecutive integers one of them is divisible by 3. Since 3 divides 99 it follows that 3 divides b(10-b). Since 3 is a prime this limits the possible choices of b:

Either

b=0 10-b=10 b(10-b) = 0
b=3 10-b=7 or b=7 10-b=3 b(10-b) = 21
b=6 10-b=4 or b=4 10-b=6 b(10-b) = 24
b=9 10-b=1 or b=1 10-b=9 b(10-b) = 9

Hence the possible values of b(10-b) are 0, 9, 21 and 24.

We now have to fi,nd a multiple of 99 which when subtracted from a product of 3 consecutive natural numbers gives 0, 9, 21 or 24.
Since a is at least 1 c(c+1)(c-1) is at least 99 so c is at least 5.

c c(c+1)(c-1) a 99a c(c+1)(c-1)-99a
5 120 1 99 21
6 210 2 198 12
7 336 3 297 39
8 504 5 495 9
9 720 7 693 27

(Since 0, 9, 21, 24 < 99 only the multiple of 99 which is closest to (c+1)c(c-1) needs to be checked.)

From the table we can see that the following are the possibilities for a b and c:

giving the solutions 135, 175, 518 and 598.