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p and q are integers. If p is divisible by 10^q and cannot Updated on: 29 Oct 2013, 06:22
Originally posted by banksy on 10 Feb 2011, 16:05.
Last edited by Bunuel on 29 Oct 2013, 06:22, edited 1 time in total.
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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
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p and q are integers. If p is divisible by 10^q and cannot 10 Feb 2011, 16:43
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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?

p being divisible by 10^q and not by 10^(q + 1) implies that the number of trailing zeros in p is exactly q (p ends with exactly q zeros).

(1) p is divisible by 2^5, but is not divisible by 2^6.

This implies that, the number of trailing zeros, q, is less than or equal to 5: \(q\leq{5}\) (since each trailing zero in p requires one 2 and one 5 in its prime factorization, this statement confirms five 2s, but the number of 5s remains unknown). Not sufficient.

(2) p is divisible by 5^6, but is not divisible by 5^7

This implies that, the number of trailing zeros, q, is less than or equal to 6: \(q\leq{6}\) (since each trailing zero in p requires one 2 and one 5 in its prime factorization, this statement confirms six 5s, but the number of 2s remains unknown). Not sufficient.

(1)+(2) We have five 2s and six 5s, which give exactly 5 trailing zeros: \(q=5\) (as the number of 2s is the limiting factor here). Sufficient.

Answer: C.
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p and q are integers. If p is divisible by 10^q and cannot 14 Jan 2014, 22:34
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Vidhi1


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.
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Isn't that the core challenge of GMAT? The concepts tested are quite basic. Why then, does everyone not get Q50? Because you really need to understand them very well to be able to see which particular concept is used in a particular question.
Here, when you see

"If p is divisible by 10^q and cannot be divisible by 10^(q + 1)"
you should understand that p has q trailing 0s (that's how it is will be divisible by 10^q) but it does not have q+1 trailing 0s. This means it has exactly q trailing 0s.

If p has q trailing 0s, it must have at least q 2s and at least q 5s (but both 2s and 5s cannot be more than q)
Statement 1 tells you about the 2s but not about the 5s. Statement 2 tells you about the 5s but not about the 2s. Both statements together tell you that you can make five 0s. and hence q must be 5.
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p and q are integers. If p is divisible by 10^q and cannot 15 Nov 2012, 12:28
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Bunuel
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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.
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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?
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p and q are integers. If p is divisible by 10^q and cannot 05 Jan 2014, 06:58
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banksy
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
Show more


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 :)
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p and q are integers. If p is divisible by 10^q and cannot 14 Jan 2014, 22:08
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Bunuel
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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.
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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.
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p and q are integers. If p is divisible by 10^q and cannot 11 Dec 2014, 06:45
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jlgdr
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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


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 :)
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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.
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p and q are integers. If p is divisible by 10^q and cannot 14 Dec 2014, 04:09
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jlgdr
banksy
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


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 :)

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.
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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.
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p and q are integers. If p is divisible by 10^q and cannot 28 Feb 2017, 12:52
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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.
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p and q are integers. If p is divisible by 10^q and cannot 11 Nov 2023, 11:27
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Bunuel
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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.
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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 !
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p and q are integers. If p is divisible by 10^q and cannot 12 Nov 2023, 08:25
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Bunuel
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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 !
Show more

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.
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p and q are integers. If p is divisible by 10^q and cannot 13 Nov 2023, 12:07
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Bunuel
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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.
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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).
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p and q are integers. If p is divisible by 10^q and cannot 19 Jul 2026, 01:38
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How can we have a definite answer for trailing 0's even from I and II combined. This gives us that q less than equal to 5 i.e. it can 1,2,3,4 or 5.
Bunuel
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?

p being divisible by 10^q and not by 10^(q + 1) implies that the number of trailing zeros in p is exactly q (p ends with exactly q zeros).

(1) p is divisible by 2^5, but is not divisible by 2^6.

This implies that, the number of trailing zeros, q, is less than or equal to 5: \(q\leq{5}\) (since each trailing zero in p requires one 2 and one 5 in its prime factorization, this statement confirms five 2s, but the number of 5s remains unknown). Not sufficient.

(2) p is divisible by 5^6, but is not divisible by 5^7

This implies that, the number of trailing zeros, q, is less than or equal to 6: \(q\leq{6}\) (since each trailing zero in p requires one 2 and one 5 in its prime factorization, this statement confirms six 5s, but the number of 2s remains unknown). Not sufficient.

(1)+(2) We have five 2s and six 5s, which give exactly 5 trailing zeros: \(q=5\) (as the number of 2s is the limiting factor here). Sufficient.

Answer: C.
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p and q are integers. If p is divisible by 10^q and cannot 19 Jul 2026, 04:23
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minininini
How can we have a definite answer for trailing 0's even from I and II combined. This gives us that q less than equal to 5 i.e. it can 1,2,3,4 or 5.

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The combined statements do not merely give q ≤ 5. Statement (1) tells us that p contains exactly five factors of 2, while statement (2) tells us that p contains exactly six factors of 5.

Since each trailing zero requires one factor of 2 and one factor of 5, exactly five pairs can be formed. Therefore, q must be 5.

For more please go back and study the discussion above.
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p and q are integers. If p is divisible by 10^q and cannot 19 Jul 2026, 05:57
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Isn't the first and second statement in itself sufficient , since prime factor of 10 is 2 and 5 , so if there are 2^5 , that means 5 twos are there , so to make 10 , there would 5 5s as well . Since p and q are both integers
Bunuel
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?

p being divisible by 10^q and not by 10^(q + 1) implies that the number of trailing zeros in p is exactly q (p ends with exactly q zeros).

(1) p is divisible by 2^5, but is not divisible by 2^6.

This implies that, the number of trailing zeros, q, is less than or equal to 5: \(q\leq{5}\) (since each trailing zero in p requires one 2 and one 5 in its prime factorization, this statement confirms five 2s, but the number of 5s remains unknown). Not sufficient.

(2) p is divisible by 5^6, but is not divisible by 5^7

This implies that, the number of trailing zeros, q, is less than or equal to 6: \(q\leq{6}\) (since each trailing zero in p requires one 2 and one 5 in its prime factorization, this statement confirms six 5s, but the number of 2s remains unknown). Not sufficient.

(1)+(2) We have five 2s and six 5s, which give exactly 5 trailing zeros: \(q=5\) (as the number of 2s is the limiting factor here). Sufficient.

Answer: C.
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p and q are integers. If p is divisible by 10^q and cannot 19 Jul 2026, 07:56
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kaustubhkohli12
Isn't the first and second statement in itself sufficient , since prime factor of 10 is 2 and 5 , so if there are 2^5 , that means 5 twos are there , so to make 10 , there would 5 5s as well . Since p and q are both integers

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For example, p = 160 = 2^5 * 5. Then p is divisible by 10^1 but not by 10^2, so q = 1, not 5.


P.S. Pure algebraic questions are no longer a part of the DS syllabus of the GMAT.

DS questions in GMAT Focus encompass various types of word problems, such as:

  • Word Problems
  • Work Problems
  • Distance Problems
  • Mixture Problems
  • Percent and Interest Problems
  • Overlapping Sets Problems
  • Statistics Problems
  • Combination and Probability Problems

While these questions may involve or necessitate knowledge of algebra, arithmetic, inequalities, etc., they will always be presented in the form of word problems. You won’t encounter pure "algebra" questions like, "Is x > y?" or "A positive integer n has two prime factors..."

Check GMAT Syllabus for Focus Edition

You can also visit the Data Sufficiency forum and filter questions by OG 2024-2025, GMAT Prep (Focus), and Data Insights Review 2024-2025 sources to see the types of questions currently tested on the GMAT.

So, you can ignore this question.

Hope it helps.
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