[Guest]

Substitute each of the capital letters in bold by a different base ten digit from 0 to 9 to satisfy this alphametic equation. None of the numbers can contain any leading zero.

(WHAT)⁴ = (NUMBER)³

[Guest]

T=2 and R=6

haven't finished the math yet

Edited July 20, 2009 by Snowman63

[Guest]

WHAT=5832 and NUMBER=104976

[Guest]

Nice job on the answer, but can you provide a brief outline of the methodology leading to it?

Ideally I would like to post the answer within this week, and I am interested to know if my method matches with that of yours.

Edited July 21, 2009 by K Sengupta

[Guest]

Nice job on the answer, but can you provide a brief outline of the methodology leading to it?

Ideally I would like to post the answer within this week, and I am interested to know if my method matches with that of yours.

The way I thought of it was like this: in order for the fourth multiple of an integer to equal the cube of another, they must share some common factor x, such that both sides of the equation satisfy x¹², where 12 is the lowest common multiple of the two exponentials. I realized this after looking at 100³ and 1000² which both equal 10⁶.

The equation can now be written as (x³)⁴ = (x⁴)³ = x¹². Based on the number of digits in WHAT, x must between 10 and 21. Based on the number of digits in NUMBER, x must be between 18 and 31. The only numbers x could be are then 18, 19, 20 or 21. From that it's pretty easy to find the answer.

Edited July 21, 2009 by Catpie

[Guest]

Well done Catpie. This indeed matches precisely with the methodology known to me and accordingly completes the solution to the given puzzle.

Edited July 23, 2009 by K Sengupta