If n is a positive integer, is 169 a factor of n?

If n is a positive integer, is 169 a factor of n?

(1) 52 is the greatest common divisor of 260 and n
260 = 52*5
Possible values of
n = 52 —> 169 is not a divisor of n
n = 52*13 —> 169 is a divisor of n

Insufficient

(2) 1,352 is the least common multiple of 104 and n

1352 = 169*8 = \(13^2\)*8
104 = 13*8
Possible values of
n = \(13^2\) —> 169 is a factor of n
n = \(13^2\)*2 —> 169 is a factor of n
n = \(13^2\)*4 —> 169 is a factor of n
n = \(13^2\)*8 —> 169 is a factor of n

Sufficient

IMO Option B

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(1) 52 is the greatest common divisor of 260 and n
not sufficient

(2) 1,352 is the least common multiple of 104 and n
divide number by 104 and 169 you can tell the number is divisible by 169
sufficient

If n is a positive integer, is 169 a factor of n?
Given: n > 0 & integer
n = 169*m ? where m is an integer from 1 to any m
=> n = 13^2*m?

(1) 52 is the greatest common divisor of 260 and n --> insufficient: 260 = 52*5, so GCD of 52*5 and n = 52, so n = 52*p=13*4*p, but p can or can't be multiple of 13
(2) 1,352 is the least common multiple of 104 and n --> sufficient: 104 = 13*2^3, so LCM of 13*2^3 & n = 1,352 = 13^2*2^3, so n must be multiple of 13^2

So the answer is B

If n is a positive integer, is 169 a factor of n?

(1) 52 is the greatest common divisor of 260 and n
52=2*2*13
n=\(2^2*13k\)
But 169=\(13^2\)
and n is not necessarily a multiple of 169
or 169 is not necessarily a factor of n
NOT SUFFICIENT

(2) 1,352 is the least common multiple of 104 and n
1352=13*104 = \(2^3*13^2\)
104=2*2*2*13 = \(2^3*13\)
Since 1352 is least common multiple of 104 and n
=>\(13^2\) is a factor of n
=> 169 is a factor of n
SUFFICIENT

IMO B

If n is a positive integer, is 169 a factor of n?

(1) 52 is the greatest common divisor of 260 and n
(2) 1,352 is the least common multiple of 104 and n

Question Stem Analysis:

According to the stem, we must find out whether n/169 , i.e \(\frac{n}{13^2}\) is an integer.

Statement One Alone:

Prime factorization of 52 is 13 X \(2^2\)
Prime factorization of 260 is 13 X 5 X \(2^2\)

The greatest common divisor (gcd), also known as the greatest common factor (gcf), or highest common factor (hcf), of two or more non-zero integers, is the largest positive integer that divides the numbers without a remainder.

For GCD, we pick the lowest power of common factors. we know that n consist a 13, but n can or cannot contain more than \(13^2\), because gcf of 260 and n is limited to 13 which we have in 260.

Hence this statement alone is not sufficient.

Statement Two Alone:

Prime factorization of 104 is 13 X\(2^3\)
Prime factorization of 1352 is 13 X 13 X \(2^3\)

The lowest common multiple or lowest common multiple (lcm) or smallest common multiple of two integers a and b is the smallest positive integer that is a multiple both of a and of b.

For LCM, we pick the highest power of common factors. we know that 104 just has \(13^1\), and that 1352 has \(13^2\) which definitively must have come from n. Hence we can say that n has\(13^2\) and that will make it divisible by 169.

Statement two alone is sufficient.

Hence the answer must be B

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