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M22-10

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M22-10  [#permalink]

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New post 16 Sep 2014, 01:16
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A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

76% (01:19) correct 24% (01:39) wrong based on 108 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  [#permalink]

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New post 10 Jul 2019, 09:03
Bunuel wrote:
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


I really didn't get this. So if A = 1543@2 and the sum of the digit is 15, then can be 0, or 3 (adding to 18), or 6 (adding to 21) or 9 (adding to 24). All these numbers make it divisible by 3 and hence insufficient.
As for statement 2, again the last 2 digits can be either 3 or 9 as 32 and 92 are divisible by 4. Again insufficient.
So shouldn't the answer be E?
Please explain where am I going wrong?
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Re: M22-10  [#permalink]

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New post 10 Jul 2019, 09:16
archimaitreya25 wrote:
Bunuel wrote:
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


I really didn't get this. So if A = 1543@2 and the sum of the digit is 15, then can be 0, or 3 (adding to 18), or 6 (adding to 21) or 9 (adding to 24). All these numbers make it divisible by 3 and hence insufficient.
As for statement 2, again the last 2 digits can be either 3 or 9 as 32 and 92 are divisible by 4. Again insufficient.
So shouldn't the answer be E?
Please explain where am I going wrong?


(1) says that A is divisible by 9, not by 3.
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Re: M22-10  [#permalink]

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New post 11 Jul 2019, 10:20
Bunuel wrote:
\(@\) represents the tens digit in integer \(A = 1543@2\). What is the value of \(@\) ?


(1) \(A\) is divisible by 9

(2) \(A\) is divisible by 4


#1
Has to be 3 sufficient
#2
can be 1,3, 4, so on insufficient
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Re: M22-10   [#permalink] 11 Jul 2019, 10:20
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