## Divisible by 3

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### Divisible by 3

by akhilsuhag » Tue Nov 18, 2014 1:10 pm
Is the tens digit of a three-digit positive integer p divisible by 3?

(1) p-7 is a multiple of 3

(2) p-13 is a multiple of 3
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by GMATGuruNY » Tue Nov 18, 2014 1:28 pm
akhilsuhag wrote:Is the tens digit of a three-digit positive integer p divisible by 3?

(1) p-7 is a multiple of 3

(2) p-13 is a multiple of 3
Note:
0 is considered divisible by EVERY POSITIVE INTEGER.

Statement 1: p-7 is a multiple of 3
Make a list of options for p-7:
p-7 = 93, 96, 99, 102, 105...

Adding 7 to each value in the list, we get the following options for p:
p = 100, 103, 106, 109, 112...

If p=100, then its tens digit is 0, which is divisible by 3.
If p=112, then its tens digit is 1, which is NOT divisible by 3.
INSUFFICIENT.

Statement 2: p-13 is a multiple of 3
Make a list of options for p-13:
p-13 = 87, 90, 93, 96, 99...
Adding 13 to each value in the list, we get the following options for p:
p = 100, 103, 106, 109, 112...

Both statements yield the same list of options for p.
Implication:
Even when the two statements are combined, it's possible that p=100 (in which case its tens digit is divisible by 3) or that p=112 (in which case its tens digit is NOT divisible by 3).
Thus, the the two statements combined are INSUFFICIENT.

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by [email protected] » Tue Nov 18, 2014 1:29 pm
akhilsuhag wrote:Is the tens digit of a three-digit positive integer p divisible by 3?

(1) p-7 is a multiple of 3
(2) p-13 is a multiple of 3
Target question: Is the tens digit of a three-digit positive integer p divisible by 3?

IMPORTANT: When I scan the two statements, I see that they both tell me the SAME THING.
There's a rule that says: If N and K are both divisible by d, then N+K and N-K are also divisible by d.

OBSERVE: p-13 is 6 less than p-7. In other words, (p-7) - 6 = p-13

Statement 1 tells us that p-7 is divisible by 3, and we know that 6 is divisible by 3. So, by the above rule, p-13 must be divisible by 3.

When two statements provide the SAME information, we can conclude that the correct answer will be either D or E.

Statement 1: p-7 is a multiple of 3
There are several values of p that satisfy this condition. Here are two:
Case a: p = 139, in which case p-7 = 132, and 132 is divisible by 3. In this case, the tens digit of p (3) IS divisible by 3
Case b: p = 118, in which case p-7 = 111, and 111 is divisible by 3. In this case, the tens digit of p (1) is NOT divisible by 3
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: p-13 is a multiple of 3
We already learned that statements 1 and 2 provide the SAME information. So, if statement 1 is NOT SUFFICIENT, we know that statement 2 is NOT SUFFICIENT

If you're not convinced, check out these two conflicting cases:
Case a: p = 139, in which case p-13 = 126, and 126 is divisible by 3. In this case, the tens digit of p (3) IS divisible by 3
Case b: p = 118, in which case p-13 = 105, and 105 is divisible by 3. In this case, the tens digit of p (1) is NOT divisible by 3
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Since both statements are not sufficient, and since both statements provide the SAME information, the combined statements are NOT SUFFICIENT

Cheers,
Brent

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by Mathsbuddy » Wed Nov 19, 2014 7:47 am
Consider the value p = ABC where A,B,C are the three digits.
Working backwards, firstly, for B to be a multiple of 3, it can only be 3, 6 or 9.

So, these are possibilities:
p = A3C, A6C, A9C

If we subtract 7 (= -10 + 3)we get:
p = A2C+3, A5C+3, A8C+3
For p to be a multiple of 3, then the sum of all digits must be a multiple of 3
As A and C are both undefined, then Statement A is NOT SUFFICIENT

Instead if we subtract 13 (= -10 - 3)we get:
p = A2C-3, A5C-3, A8C-3
For p to be a multiple of 3, then the sum of all digits must be a multiple of 3
As A and C are both undefined, then Statement A is NOT SUFFICIENT

Combined statements will still leave A and C undefined. NOT SUFFICIENT>

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by [email protected] » Wed Nov 19, 2014 8:52 am
Mathsbuddy wrote: Working backwards, firstly, for B to be a multiple of 3, it can only be 3, 6 or 9.
Hey Mathsbuddy,

You're absolutely right about the answer being E, but be careful. If B is a multiple of 3, then B can equal 0, 3, 6 or 9

Zero is a multiple of all integers. Likewise, zero is divisible by all integers.

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by Mathsbuddy » Wed Nov 19, 2014 10:05 am
[email protected] wrote:
Mathsbuddy wrote: Working backwards, firstly, for B to be a multiple of 3, it can only be 3, 6 or 9.
Hey Mathsbuddy,

You're absolutely right about the answer being E, but be careful. If B is a multiple of 3, then B can equal 0, 3, 6 or 9

Zero is a multiple of all integers. Likewise, zero is divisible by all integers.

Cheers,
Brent
Thanks Brent, I did wonder if I should have included zero!
However, as it is not sufficient for the other values, it's sufficiency for zero becomes obsolete.
Nonetheless, I haven't totally convinced myself if 'working backwards' is 100% valid!

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