For any four digit number, abcd, *abcd*= (3^a)(5^b)(7^c)(11^d). What is the value of (n – m) if m and n are four-digit numbers for which *m* = (3^r)(5^s)(7^t)(11^u) and *n* = (25)(*m*)?


a 2000
b 200
c 25
d 20
e 2

Can someone please explain this question? Possibly with numbers
The correct answer is b.

Given that *abcd*= (3a)(5b)(7c)(11d)

and also given that *m* = (3r)(5s)(7t)(11u)

we can infer that the 4 digits of the number 'm' will look like this: 'rstu'

now n = 25* *m*. This can be written as the above number 'm' multiplied by 25.
This looks like: 5^2 * (rstu)

Since 's' is the power of 5, the above multiplication with 2 more powers of 5 will add into s, making 5^s in the original 'm' now equal to 5^(s+2).

So the 4 digits of 'n' are now 'r(s+2)tu'
And the 4 digits of 'm' are now 'rstu'

If you subtract m from n to get 'n-m', as the question asks, notice that 's' is in the thousands place, and all other digits are the same in both numbers.

Therefor, 'n' is greater than 'm' by 2 in the thousands place = 2000. Answer = (B).

Hope this helps!