Hi All,
While this prompt is 'complex-looking', it's based on some relatively simple concepts (once you unravel the information that you've been given).
To start, we're told that each DIGIT of a positive 3-digit number is defined by...
f(N) = (2^X)(3^Y)(5^Z) where X, Y and Z are the hundreds, tens and units digits of N.
For example, IF... N = 410, then f(410) = (2^4)(3^1)(5^0) = (16)(3)(1) = 48
We're told that M and V are 3-digit numbers such that f(M) = 9(f(V)). We're asked for the value of M-V.
Given how the function multiplies 'powers of primes', for one 3-digit number to be 9 TIMES another, the larger number MUST have "two more 3s" than the smaller number. For that to occur, the tens digit of the larger number must be 2 greater than the tens digit of the smaller number. The other two digits (hundreds and units) must be the SAME.
For example: 240 and 220
f(240) = (2^2)(3^4)(5^0) = (4)(81)(1) = 324
f(220) = (2^2)(3^2)(5^0) = (4)(9)(1) = 36
324 = 9 times 36
Thus, the difference between the M and V comes down to the tens digits - and while there are several potential options here - they ALL differ by 20.
Final Answer:
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