A*C*AC = CCCwhat is value of C?

Answer 1

A*C*(10A+C) = 111*C
A*(10A+C) = 111
10A^2+A*C = 111
C = 111/A - (10)A

A must be a factor of 111 which means that it must be 3
C = 37 - (10)3
C = 7

3*7*37 = 777

Answer 2

C = A^2

Answer 3

HAHA Damn i missed that ^^^ so much better than mine

Assuming a and c are digits

a * c * (10a+c) = 111*c

10a^2*c+a*c^2 = 111*c

10a^2c+a*c^2-111*c=0

c * (10a^2 + a*c - 111) = 0

The roots of the part int he brackets are

[-c +/- sqrt(c^2+4*10*111)]/20
=[-c +/- sqrt(c^2+4440)]/20

Obviously if c = 0 we have a solution for all digits a

but if c!=0 then there must be a digit root 'a' for a digit c.

Using some calculation (in a program) we can find that if c = 7 then a = 3

This is the only solution other than c = 0 a = {1,2,3,..,9}

A simple way to do it analytically is to find squares s^2 above [[sqrt(4440)]+1 =67 such that x^2-a^2=4440.

Answer 4

For all real numbers A and C:

A²C² = C³

C³ - A²C² = 0

C² (C - A²) = 0

C = 0 or C = A²

{0, A²}

Answer 5

if * means multiply, then C = A^2