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 _________________