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Math Expert V
Joined: 02 Sep 2009
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Difficulty:   15% (low)

Question Stats: 76% (01:19) correct 24% (01:39) wrong based on 108 sessions

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$$@$$ 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

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Math Expert V
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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.

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+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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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.

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?
Math Expert V
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Posts: 58335

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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.

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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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
IMOA
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