Can someone please explain?

This is a tough question which maybe goes beyond the GRE scope.

So, we should recognize the common exponents base.

$W= 5^{a-p} \times 2^{b-q} \times 7^{c-r} \times 3^{d-s}$ AND $W =16=2^4$

If $W=2^4$ and at the same time has also 5 and 7 and 3 as a factor this is impossible because W contains only 4 of two unless $5^0=1$ and $3^0=1$ and $7^0=1$


$W= 5^{a-p} \times 2^{b-q} \times 7^{c-r} \times 3^{d-s} = 5^0 \times 2^{b-q} \times 7^0 \times 3^0 = 16$

From this $b-q$ must be $=4$

Moreover, since the exponents of 5, 7, and 3 are equal to zero, the differences between the thousands, tens, and units digits of K and L are zero, implying that K and L differ only in their hundreds digit.


Since the hundreds digit of K is 4 greater than that of L, the difference between K and L is $4 \times 100 = 400$. Therefore $K — L = 400$. Since Z is defined as $(K — L) 10$, we can determine that $Z = 400 ÷ 10 = 40$. The correct answer is D.

Hey Carcass, This was my query. The question shows that there is a sign of addition(+) in the question whereas you have solved the question taking a division sign.
Hence, I felt that the question was wrong.

Well, a typo is always possible. Too many candies these holydays could clog your brain

However, it was correct and the process is to boil down the equation to common base exponents

Regards