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

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p and q are integers. If p is divisible by 10^q and cannot [#permalink]

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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 29 Oct 2013, 06:22, edited 1 time in total.
Added the OA.

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Re: p and q are integers. If p is divisible by 10q and canno [#permalink]

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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.
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Re: p and q are integers. If p is divisible by 10q and canno [#permalink]

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New post 15 Nov 2012, 12:28
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.

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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Re: p and q are integers. If p is divisible by 10^q and cannot [#permalink]

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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



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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Re: p and q are integers. If p is divisible by 10q and canno [#permalink]

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New post 14 Jan 2014, 22:08
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.


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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Re: p and q are integers. If p is divisible by 10q and canno [#permalink]

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Vidhi1 wrote:

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.


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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Re: p and q are integers. If p is divisible by 10^q and cannot [#permalink]

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New post 11 Dec 2014, 06:45
jlgdr 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



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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Re: p and q are integers. If p is divisible by 10^q and cannot [#permalink]

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sunilp wrote:
jlgdr 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



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.


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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Re: p and q are integers. If p is divisible by 10^q and cannot [#permalink]

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New post 27 Jan 2016, 18:51
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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Re: p and q are integers. If p is divisible by 10^q and cannot [#permalink]

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New post 16 Mar 2016, 06:18
Excellent Question here we need to work on the values of q
clearly combination statement is sufficient as q=5 else q=no specific value in individual statements
hence C
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Re: p and q are integers. If p is divisible by 10^q and cannot [#permalink]

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New post 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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Re: p and q are integers. If p is divisible by 10^q and cannot [#permalink]

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New post 04 Aug 2017, 06:53
MathRevolution wrote:
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.



Hi ,
Why cant we consider the two conditions, which tell us the divisibility as equations ? :?
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p and q are integers. If p is divisible by 10^q and cannot [#permalink]

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New post 08 Aug 2017, 22:15
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.

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p and q are integers. If p is divisible by 10^q and cannot [#permalink]

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New post 22 Aug 2017, 00:55
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



what a fancy way of asking a very easy question..hihihihi :grin: ... Actually this is what the test makers do to make a question hard...
very interesting and didactic question it is, indeed...thanks ...

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p and q are integers. If p is divisible by 10^q and cannot   [#permalink] 22 Aug 2017, 00:55
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p and q are integers. If p is divisible by 10^q and cannot

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