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# A positive integer X has 'n' total factors, another positive integer Y

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Joined: 21 Aug 2013
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A positive integer X has 'n' total factors, another positive integer Y  [#permalink]

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21 May 2018, 20:34
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65% (hard)

Question Stats:

61% (02:40) correct 39% (02:25) wrong based on 76 sessions

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A positive integer X has 'n' total factors, another positive integer Y has 'm' total factors. Is X+Y odd?

(1) m = 2*n.

(2) 2*X has 2*n total factors.
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Re: A positive integer X has 'n' total factors, another positive integer Y  [#permalink]

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29 May 2018, 11:35
amanvermagmat wrote:
A positive integer X has 'n' total factors, another positive integer Y has 'm' total factors. Is X+Y odd?

(1) m = 2*n.

(2) 2*X has 2*n total factors.

1) m is even. Therefore y has an even number of factors. Therefore y can be any number as long as it's not a perfect square.

y = 15 ----> 4 factors
x = 3 ------> 2 factors
-----------------------> x + y = 18 = even

y = 22 ---> 4 factors
x = 3 -----> 2 factors
-----------------------> x + y = 25 = odd

-------------------------------------------------> INSUFFICIENT

2) The only way to double the number of factors when you double an integer is if that integer is odd. Therefore x is odd. We know nothing about y though.
-------> INSUFFICIENT

1 & 2) Combining the statements we know x is odd and y has double the factors of x. Both of our examples from statement 1 satisfy this requirement therefore x + y can still be even or odd.
-------------------------> INSUFFICIENT

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Re: A positive integer X has 'n' total factors, another positive integer Y  [#permalink]

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31 May 2018, 05:57
amanvermagmat wrote:
A positive integer X has 'n' total factors, another positive integer Y has 'm' total factors. Is X+Y odd?

(1) m = 2*n.

(2) 2*X has 2*n total factors.

The sum (X + Y) will be odd if one of X or Y is odd and the other is even. The sum will be even if both are odd or both are even.
Given: X has n factors. Y has m factors.

Statement 1: m = 2*n
Y has twice as many factors as X. It is possible with one being odd and the even or otherwise.

Example: X = 2. Y = 6
X has 2 factors and Y has 4 factors. X+Y is even.
Counter example: X = 3. Y = 6
X has 2 factors. Y has 4 factors. X + Y is odd.

Statement 1 alone is NOT sufficient.

Statement 2: 2*X has 2*n
If 2X has 2n factors and X has n factors, it is possible only when 2 is NOT a prime factor of X. So, we can infer that X is odd.
But we do not know anything about Y.
If Y is also odd, (X + Y) will be even. If Y is even, (X + Y) will be odd.

Statement 2 alone is NOT sufficient.

Combining the statements: m = 2*n and 2*X has 2*n
From statement 2, X is odd.

Example: X = 5 and Y = 6.
X has 2 factors. 2X = 10 has 4 factors. So, satisfies statement 2.
X has 2 factors and Y has 4 factors. So, satisfies statement 1.
Sum (X + Y) is odd.

Example: X = 5 and Y = 15.
X has 2 factors. 2X = 10 has 4 factors. So, satisfies statement 2.
X has 2 factors and Y has 4 factors. So, satisfies statement 1.
Sum (X + Y) is even.

Statements together NOT sufficient. Choice E.

Theory behind number of factors: If a number N can be prime factorised as a^p * b^q, number of factors of N will be (p + 1)(q + 1).

If N is even, then one of a or b will be 2. Let us say a = 2. p > 0.
2N will take power p to (p + 1).
Therefore, number of factors of 2N = (p + 2)(q + 1)

Number of factors of 2N/Number of factors of N = (p + 2)/(p + 1)

If this value has to be 2, it is possible only when p = 0. If that is the case, then 2 could not have been a factor of N.
Inference: If multiplying a prime c to a number N doubles the number of factors of N, then c is not a factor of N.
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Re: A positive integer X has 'n' total factors, another positive integer Y &nbs [#permalink] 31 May 2018, 05:57
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