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M22-10

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M22-10  [#permalink]

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New post 16 Sep 2014, 01:16
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

73% (00:52) correct 27% (01:18) wrong based on 125 sessions

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Re M22-10  [#permalink]

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New post 16 Sep 2014, 01:16
Official Solution:


(1) \(A\) is divisible by 9. In order an integer to be divisible by 9, the sum of its digit must be divisible by 9. Hence \(1+5+4+3+@+2=15+@\) must be divisible by 9, and since \(@\) is a digit then it can only equal to 3. Sufficient.

(2) \(A\) is divisible by 4. In order an integer to be divisible by 4, its last two digits must be divisible by 4. Hence \(@2\) must be divisible by 4, which is possible if \(@=1\), \(@=3\), \(@=5\), \(@=7\), or \(@=9\). Not sufficient.


Answer: A
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Re: M22-10  [#permalink]

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New post 05 Jan 2018, 07:18
+1 for A. A number is divisible by 9 if sum of individual digits is divisible by 9. The sum of digits in this case is 15+@.

Statement 1 : The max sum possible is 24. From 15 to 25, only 18 is divisible by 9. Hence we get a unique value for @.
Statement 2 : The number is divisible by 4 if @2 is divisible by 4. @ could be 1,3, or 5. Ie no unique value !!!

Hence the answer is option A.
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Re: M22-10 &nbs [#permalink] 05 Jan 2018, 07:18
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