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

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?

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.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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


In the original condition, there are 2 variables(p,q), which should match with the number of equations. So you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. When 1) & 2), they become p=5(10^5)k(k is 2 or 5, which are not prime factors but integer). So, only q=5 is possible, which is unique and sufficient.
For 1), just like p=5(2^5), 5^2(2^5), it is not unique and not sufficient.
For 2), just like p=2(5^6), 2^2(5^6), it is not unique and not sufficient.
Therefore, the answer is C.


--> For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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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.

P/(10)ˆq=integer
or
P/(5*2)ˆq=Integer

1) P/2ˆ5=Integer and P/2ˆ6= not an integer but we have no idea about the power of 5 hence not sufficient
2) P/5ˆ6=Integer and P/5ˆ7=Not an integer but we have no idea about the power of 2 hence not sufficient
Combined we know that p is divisible by 2ˆ5 and also by 5ˆ5(divisible by 5ˆ6 means that it is divisible by 5ˆ1 or 2 or 3... upto 6)
We need a pair of 2 and 5 to make zero. Hence here the limiting factor is 2 since it has lower power of 2.
also p is not divisible by q+1
combined statement is sufficient.

The answer is C.

S1 - Says about 2-s no information about 5-s.
S2 - Says about 5-s no information about 2-s.

S1+S2 = Sufficient as total 5 trailing zeros are possible.
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