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09 Oct 2010, 23:02
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
66% (02:05) correct
34%
(02:17)
wrong
based on 420
sessions
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If the infinite sequence, M, is defined as M1 = 6, M2 = 96, M3 = 996, … , Mk = 10^k – 4, is every term in this sequence divisible by q, if q is an even number?
(1) q is less than 45.
(2) At least 2 terms in the sequence are divisible by q.
(1) q is less than 45.
(2) At least 2 terms in the sequence are divisible by q.
10 Oct 2010, 00:02
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mrinal2100
I think you mean to write that M_k = 10^k - 4, and not 10k - 4.
When I first read the question, I ask - what even number q would divide every term in the sequence? The only possibilities are 2 and 6, since these are the only even divisors of the first term of the sequence, which is 6. Each other term in the sequence is clearly divisible by 2, and since the digits of each term are multiples of 3, each term will also be divisible by 3, and thus by 6. So the question is really asking if we can determine that q is either 2 or 6.
Statement 1 then is clearly insufficient. From Statement 2, q is an even number which divides at least two terms in the sequence. Certainly q might be 2 or 6, but if you notice that each term in the sequence besides the first term ends in 96, you can see that each term in the sequence besides the first must be a multiple of 4. So q could also be equal to 4, and even combined the two Statements are not sufficient, and the answer is E.
14 Oct 2010, 03:27
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With (1) q is less than 45...q can be 2 which can divide each term Mk=(10^k)-4=(2* something) but it can be 8 also which cant divide 6 .It means it is not sufficient to answer.
With (2) ...Here also it can be 2 or 12 ...2 can divide all whereas 12 cant divide 6.
So both are not suffictent .Hence answer is E.
Consider KUDOS if You like the solution.
With (2) ...Here also it can be 2 or 12 ...2 can divide all whereas 12 cant divide 6.
So both are not suffictent .Hence answer is E.
Consider KUDOS if You like the solution.
