MathRevolution wrote:
Que: A number 4p25q is divisible by 4 and 9; where p and q are the thousands and units digits, respectively. What is the minimum value of \(\frac{p}{ q}\)
(A) \(\frac{1}{8}\)
(B) \(\frac{1}{7}\)
(C) \(\frac{1}{6}\)
(D) \(\frac{2}{5}\)
(E) \(\frac{5}{2}\)
Solution: Divisibility property of ‘4’: A number is divisible by ‘4’ when its last two digits are divisible by ‘4’.
Therefore, ‘5q’ is divisible by 4.
So, the possible values of ‘q’ are 2 or 6 [both 52 and 56 are divisible by 4]
Divisibility property of ‘9’: A number is divisible by ‘9’ when the sum of all its digits is divisible by ‘9’
Therefore, 4 + p + 2 + 5 + q = 11 + p + q.
So, the possible values of ‘p’ and ‘q’ so that 11 + p + q is divisible by ‘9’are:
For q = 2 => 11 + p + 2 = 13 + p
=>‘p’ should be 5 [since 13 + 5 = 18 is divisible by 9]
For q = 6 => 11 + p + 6 = 17 + p
=> ‘p’ should be 1 [since 17 + 1 = 18 is divisible by 9]
So, we have two pairs for ‘p’ and ‘q’: (5, 2) and (1,6)
=> Minimum value of \(\frac{p}{q}\) = \(\frac{1}{6}\)
C is the correct answer.
Answer C _________________