p and q are integers. If p is divisible by 10^q and cannot

From condition 1 and condition 2 it we got \(q\leq{5}\) and \(q\leq{6}\) so possible cases can be \(q\leq{5}\) so q can be anything 5,4,3,2,...... So both together aren't sufficient right?

Yeah right on, what the question is really asking is what is the maximum number of trailing zeroes that P can have

With both statements together we get that the maximum number must be 5

Hence C is the correct answer

Hope it helps
Cheers!
J

How do we know the question is asking us to use the concept of trailing zeroes. I understand the concept but how do we know that we have to apply trailing zeroes concept.

Hi, I did not get part where you said it is asking for maximum number of trailing zeros. Is it because of statement 1) ? thanks in advance.

What is the meaning of trailing zeroes?

102000 has 3 trailing zeroes i.e. 3 zeroes at the end.
1920040 has only 1 trailing zero.
Trailing zeroes means the number of zeroes at the end - after a non zero number.

Now think, 19200 = 192*100
So 19200 is divisible by 100 but not by 1000 or 10000 etc.
Since 19200 has 2 trailing zeroes, it will be divisible by 10^2 but not by 10^3 or 10^4 or 10^5 etc.

The question stem tells you that p is divisible by 10^q but not by 10^(q+1); this means that p has exactly q trailing zeroes.
For every 0 at the end, you need the number to be a multiple of 10 i.e. a 2 and a 5. Stmt 1 tells you that p has 5 2s and stmnt 2 tells you that p has 6 5s. Together, you can make only 5 10s. So q must be 5.

from 1

we can see 2^5 can divide p but not 2^6. so can conclude that q is at most 5 or less(q<=5) depending upon no. of 5's available.(5*2=10)
so not sufficient.

from 2

we can see 5^6 can divide p but not 5^7. so can conclude that q is at most 6 or less(q<=6) depending upon no. of 2's available.(5*2=10)
so not sufficient.

Now taking both (1) & (2)
maximum no. of 2's available is 5 and maximum no. of 5's available is 6. so we can take five 2's and 5's i.e (2*5) five times.
hence sufficient.

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.

Yes ! Thank you that was very clear. I got indeed mistaken because I assumed p was a factorial and generalized that 5 would be the limiting factor. But as you said in this case 2 is the limiting factor and thus there are 5 trailing zeros (q = 5).