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Que: A number 42,p5q is divisible by both 4 and 8; where p and q are the thousands and units digits...........

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Que: A number 42,p5q is divisible by both 4 and 8; where p and q are the thousands and units digits...........

by Max@Math Revolution » Sat Jun 05, 2021 9:07 pm

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Que: A number 42,p5q is divisible by both 4 and 8; where p and q are the thousands and units digits, respectively. What is the minimum possible value of \(\frac{p}{q}\)?

(A) \(\frac{1}{2}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{1}{6}\)
(D) \(\frac{2}{3}\)
(E) \(\frac{1}{8}\)
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Re: Que: A number 42,p5q is divisible by both 4 and 8; where p and q are the thousands and units digits...........

by Max@Math Revolution » Sun Jun 06, 2021 9:09 pm
Solution: A number, here 42p5q is divisible by ‘4’ if the number formed by the last two digits of the number is divisible by 4.

Thus, the number 5q is divisible by 4.This is possible when q = 2(52) or 6(56).

Here 42p5q is divisible by ‘8’ if the number formed by the last three digits of the number p5q is divisible by 8.

Thus, the number p5q is divisible by 8

If q = 2:

=> p = 1(152), 3(352), 5(552), 7(752), 9(952) is divisible by 8.

If q = 6:

=> p = 2(256), 4(456), 6(656), 8(856) is divisible by 8.

Therefore, we have following possible combinations:

=> p = 1,3,5,7,9 , q = 2 then minimum \(\frac{p}{q}=\frac{1}{2}\)

=> p = 2,4,6,8, q = 6 then minimum \(\frac{p}{q}=\frac{2}{6}=\frac{1}{3}\)

Thus, the minimum value of \(\frac{p}{q}\) is \(\frac{1}{3}\).

Therefore, B is the correct answer.

Answer B
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