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If M=P*Q, where P and Q are different positive integers,

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If M=P*Q, where P and Q are different positive integers,  [#permalink]

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Updated on: 08 Aug 2013, 11:13
2
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Difficulty:

45% (medium)

Question Stats:

67% (01:38) correct 33% (01:35) wrong based on 242 sessions

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Hi Guys,

If M=P*Q, where P and Q are different positive integers, including 1 and M, how many factors does M have?
(1) Including 1 and itself, P has 3 factors.
(2) Including 1 and itself, Q has 3 factors.

I got the answer to this, my question is - if you had to calculate, how many factors does M have in total and which ones would they be?

Source: Not Sure

Thanks !

Originally posted by yaj on 08 Aug 2013, 11:08.
Last edited by Zarrolou on 08 Aug 2013, 11:13, edited 1 time in total.
Edited the question, renamed the topic.
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Re: If M=P*Q, where P and Q are different positive integers,  [#permalink]

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08 Aug 2013, 11:19
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1
Theory:
the total number of factors,including 1 and n, of a number $$n=a^x*b^y*c^z*...$$ (where $$a,b,c,...$$ are prime numbers) is given by the formula $$(x+1)(y+1)(z+1)...$$

If M=P*Q, where P and Q are different positive integers, including 1 and M, how many factors does M have?

(1) Including 1 and itself, P has 3 factors.
No info about Q, not sufficient.
(2) Including 1 and itself, Q has 3 factors.
No info about P, not sufficient.

(1+2)From statement 1 we get that $$P=a^2$$ (so its factors are 2+1=3) and that $$Q=b^2$$, and a,b are different prime numbers.
Total number of factors of $$M=a^2*b^2$$ is $$3*3=9$$.
Sufficient
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Re: If M=P*Q, where P and Q are different positive integers,  [#permalink]

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08 Aug 2013, 11:27
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yaj wrote:
Hi Guys,

If M=P*Q, where P and Q are different positive integers, including 1 and M, how many factors does M have?
(1) Including 1 and itself, P has 3 factors.
(2) Including 1 and itself, Q has 3 factors.

I got the answer to this, my question is - if you had to calculate, how many factors does M have in total and which ones would they be?

Source: Not Sure

Thanks !

Adding on to what Zarrolou said, refer here for more theory : math-number-theory-88376.html
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Re: If M=P*Q, where P and Q are different positive integers,  [#permalink]

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18 Nov 2015, 04:24
I solved it by assuming some values. Obviously Statement 1 and 2 alone are not enough. However, using the information given in each of them, I was able to assume some values - 4 (1, 2, 4), 9 (1, 3, 9) and 25 (1, 5, 25). Using these, I got three numbers - 36, 100 and 225. Each of these numbers has 9 factors.
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Re: If M=P*Q, where P and Q are different positive integers,  [#permalink]

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21 Nov 2015, 09:55
2
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.

If M=P*Q, where P and Q are different positive integers, including 1 and M, how many factors does M have?
(1) Including 1 and itself, P has 3 factors.
(2) Including 1 and itself, Q has 3 factors.

There are 3 variables (m,p,q) and one equation (m=pq) in the original condition, 2 equations in the given conditions, so there is high chance (C) will be our answer.
Looking at the conditions together,
p=k^2(k is a prime) and q=t^2(t is a prime different k), so m=(k^2)(t^2), and the number of factors becomes (2+1)(2+1)=9.

For cases where we need 2 more equation, 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: If M=P*Q, where P and Q are different positive integers,  [#permalink]

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25 Apr 2019, 12:35
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Re: If M=P*Q, where P and Q are different positive integers,   [#permalink] 25 Apr 2019, 12:35
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