Remainders Problem Exam Pack 1
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Translate the question.sukhman wrote:If k is a positive integer and n = k(k + 7), is n divisible by 6?
(1) k is odd
(2) When k is divided by 3 the remainder is 2.
Does n have among its prime factors at least one 2 and one 3?
Statement 1:
If k is odd, then k + 7 is even. So if k + 7 is a factor of n, then n is even, meaning that n has 2 among its prime factors.
However k could be 3 or another positive integer.
If k = 3, then n has 3 among its prime factors.
If, for instance, k = 19, then k(k + 7) = 19(26) and n does not have 3 among its prime factors.
So there is not enough information for determining whether n has 3 among its prime factors.
Insufficient.
Statement 2:
This means that k = 3x + 2. So k + 7 = 3x + 9. So 3 is a factor of k + 7 and of n.
If k is even, then n has 2 among its prime factors. If k is odd, then k + 7 is even and n has 2 among its prime factors.
So Statement 2 makes clear that n has 2 and 3 among its prime factors and is divisible by 6.
Sufficient.
The correct answer is B.
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Statement 1: k is oddIf k is a positive integer and n = k(k + 7), is n divisible by 6?
(1) k is odd.
(2) When k is divided by 3, the remainder is 2.
If k=1, then n = 1(1 + 7) = 8, which is NOT divisible by 6.
If k=3, then n = 3(3 + 7) = 30, which IS divisible by 6.
INSUFFICIENT.
Statement 2: When k is divided by 3, the remainder is 2
In other words, k is 2 more than a multiple of 3.
In math terms:
k = 3a + 2, where a is a nonnegative integer.
Options for k:
k = 2, 5, 8, 11...
If k=2, then n = 2(2 + 7) = 18, which is divisible by 6.
If k=5, then n = 5(5 + 7) = 60, which is divisible by 6.
If k=8, then n = 8(8 + 7) = 120, which is divisible by 6.
In every case, n is divisible by 6.
SUFFICIENT.
The correct answer is B.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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