Given: \(w=\frac{5^a*2^b*7^c*3^d}{5^p*2^q*7^r*3^s}=16\) --> \(w=5^{a-p}*2^{b-q}*7^{c-r}*3^{d-s}=2^4\) --> the powers of 3, 5, and 7 must be zero and the power of 2 must be 4: \(a=p\), \(b-q=4\), \(c=r\) and \(d=s\)

Now, as thousands, tens, and units digits in K and L are equal and the difference between hundreds' digits is 4, then K-L=400 (for example K=1923 and L=1523 --> K-L=1923-1523=400).

Z=(K-L)/10=400/10=40.

Answer: D.

Also discussed here: functions-concepts-testing-91004.html
Similar questions:
the-function-f-is-defined-for-each-positive-three-digit-100847.html
for-any-four-digit-number-abcd-abcd-3-a-5-b-7-c-11-d-126522.html

Hope it helps.
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Bunuel thanks - you super star.
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Hi Bunuel ,

Can you move this question to the quant section

Done: the question is moved to PS forum. Thank you.
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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[quote="enigma123"]K and L are each four-digit positive integers with thousands, hundreds, tens, and units digits defined as a, b, c, and d, respectively, for the number K, and p, q, r, and s, respectively, for the number L. For numbers K and L, the function W is defined as \(5^a 2^b 7^c 3^d\) ÷ \(5^p 2^q 7^r 3^s\). The function Z is defined as (K – L) ÷ 10. If W = 16, what is the value of Z?

(A) 16
(B) 20
(C) 25
(D) 40
(E) It cannot be determined from the information given.

Given:
K = abcd = 1000a + 100b + 10c + d
L = pqrs = 1000p + 100q + 10r + s
W = \(5^a 2^b 7^c 3^d\) ÷ \(5^p 2^q 7^r 3^s\) = \(5^{a-p} 2^{b-q} 3^{c-r} 5^{d-s}\) = 16 = \(2^4\)

W can 16 only when W carries the powers of 2 only.
Hence b - q = 4 (i)
And the rest of the powers will be 0.
a= p, c = r, d = s (ii)

Required: Z = (K – L) ÷ 10 =?
Z = (abcd - pqrs)÷10 = (1000a + 100b + 10c + d) - (1000p + 100q + 10r + s) ÷ 10
Z = 1000 (a - p) + 100(b - q) + 10 (c - r) + 10 (d - s) ÷ 10
From equations (i) and (ii)
Z = 100(b-q) ÷ 10 = 100*4 ÷ 10= 40
Option D

This question is tricky in the sense that it is a simple recognition of 2^4=16 that is buried under quite a bit of language.

W can be rewritten to the following: 16=5^(a-p) x 2^(b-q) x 7^(c-r) x 3^(d-s)

K and L are four digit numbers, so we need to have at least 1 in a and p's position. Thus, we will let a=1, p=1

b-q needs to equal 4, so we can assign any single digits that would give us that result. b=9, q=5

The rest of the numbers can be either 0's or 1's. Doesn't matter as we will be subtracting them off when we plug values into the Z equation.

K = 1911
L = 1511

Z = (K-L)/40 = 400/10 = 40

Agreed that this question's biggest challenge is the wordiness...

The way I approached it:

1) f(W) shows us the division of the prime factorization of the 2 numbers, K and L. That f(W) = 16 means there's 2^4 more in the numerator K, i.e. that b-4 = q. This also means that all the other variables are equal.

2) f(Z) = (F - L)/10 ... since all the variables are equal except b and q (the hundreds place), it will be some number where the difference between them is 400 (since b is greater than q by 4). 400/10 = 40, D.

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