chloe2m wrote:
Bunuel wrote:
banksy wrote:
p and q are integers. If p is divisible by 10^q and cannot be divisible by 10^(q + 1), what is the value of q?
(1) p is divisible by 2^5, but is not divisible by 2^6.
(2) p is divisible by 5^6, but is not divisible by 5^7
p is divisible by 10^q and cannot be divisible by 10^(q + 1) means that # of trailing zeros of p is q (p ends with q zeros).
(1) p is divisible by 2^5, but is not divisible by 2^6 --> # of trailing zeros, q, is less than or equal to 5: \(q\leq{5}\) (as for each trailing zero we need one 2 and one 5 in prime factorization of p then this statement says that there are enough 2-s for 5 zeros but we don't know how many 5-s are there). Not sufficient.
(2) p is divisible by 5^6, but is not divisible by 5^7 --> # of trailing zeros, q, is less than or equal to 6: \(q\leq{6}\) (there are enough 5-s for 6 zeros but we don't know how many 2-s are there). Not sufficient.
(1)+(2) 2-s and 5-s are enough for 5 trailing zeros: \(q=5\) (# of 2-s are limiting factor). Sufficient.
Answer: C.
Hi
Bunuel, I understand this reasoning but I don't understand how the # of 2s could be the limiting factor ? Indeed I thought that 5s were always the limiting factor considering that there were always enough 2s. Therefore at first I said that statement 2 ALONE was sufficient. Thank you for your explanation !
Given that p is divisible by 10^q but not by 10^(q+1), the number of trailing zeros in p, which will be q, is determined by the lesser of the count of 2s and 5s in its prime factorization, since 10 = 2*5. In many cases, like when calculating the number of trailing zeros in p!, 2s are more common and thus 5s become the limiting factor. However, in this specific problem, we are dealing with p, not p!.
Statement (1) tells us that p is divisible by 2^5 but not by 2^6, meaning there are exactly five 2s in p's prime factorization. However, we don't know the number of 5's
Statement (2) says that p is divisible by 5^6 but not by 5^7, indicating there are exactly six 5s in p's prime factorization. However, we don't know the number of 2's.
When considering the statements together, the 2s are the limiting factor for the number of trailing zeros in p. Therefore, q equals 5, as there aren’t enough 2s to pair with all the 5s for more than five zeros.
Hope it's clear.