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Divisibility and primes

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by apshara5 » Tue Oct 18, 2011 3:28 pm
If a and b are both single-digit positive integers, is a + b a multiple of 3?

(1) The two-digit number "ab" (where a is in the tens place and b is in the ones place) is a multiple of 3.

(2) a - 2b is a multiple of 3.
ans-d

i can tell why a is sufficient i don't know why b is sufficient?
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Source: — Data Sufficiency |
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by neelgandham » Tue Oct 18, 2011 3:50 pm
(1) The two-digit number "ab" (where a is in the tens place and b is in the ones place) is a multiple of 3.

=> 10a + b = multiple of 3
=> 9a + a + b = multiple of 3
=> a + b = multiple of 3 ( because 9a is already a multiple of 3), If you didn't understand this step, never mind, read on..

let 9a + a + b = 3*x (where x = positive number). Then divide both sides with 3

3a +(a+b)/3 = x => Integer + (a+b)/3 = Integer => (a+b)/3 = Integer => a+b is divisible by 3

Sufficient !

(2) a - 2b is a multiple of 3.

from the above we can tell that a > 2b

If b = 1, a = 5 or 8 (from the equation a-2b = 3 and a-2b = 6)
If b = 2, a = 7 (from the equation a-2b = 3)
If b = 3, a = 9 (from the equation a-2b = 3)

In all cases a+b(6,9,9,12) is a multiple of 3

Sufficient!

Hence, Option D.

Makes sense? Else let me know !
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by GMATGuruNY » Tue Oct 18, 2011 4:46 pm
apshara5 wrote:If a and b are both single-digit positive integers, is a + b a multiple of 3?

(1) The two-digit number "ab" (where a is in the tens place and b is in the ones place) is a multiple of 3.

(2) a - 2b is a multiple of 3.
ans-d

i can tell why a is sufficient i don't know why b is sufficient?
Statement 1: The two-digit number "ab" is a multiple of 3.
The sum of the digits of a multiple of 3 is a multiple of 3.
Thus, a+b is a multiple of 3.
SUFFICIENT.

Statement 2: a - 2b is a multiple of 3.
a+b = (a-2b) + 3b
a+b = (multiple of 3) + (multiple of 3)
Thus, a+b is a multiple of 3.
SUFFICIENT.

The correct answer is D.
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by apshara5 » Tue Oct 18, 2011 7:12 pm
Thanks guys! Both explanations were useful.
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by nandy1984 » Wed Oct 19, 2011 9:32 am
apshara5 wrote:If a and b are both single-digit positive integers, is a + b a multiple of 3?

(1) The two-digit number "ab" (where a is in the tens place and b is in the ones place) is a multiple of 3.

(2) a - 2b is a multiple of 3.
ans-d

i can tell why a is sufficient i don't know why b is sufficient?
I have a similar explanation as David said, but differently put up....

a-2b = a + b - b -2b = (a+b) -(3b)....If this expression is divisible by 3...
= ( (a+b) - (3b) ) / 3 = (a+b)/3 - b ..... if this expression is an integer then (a+b)/3 must also be an integer as b is an integer....So from this we can prove that statement (2) is sufficient...Hope u understand...Thanks...
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by Whitney Garner » Wed Oct 19, 2011 12:56 pm
apshara5 wrote:If a and b are both single-digit positive integers, is a + b a multiple of 3?

(1) The two-digit number "ab" (where a is in the tens place and b is in the ones place) is a multiple of 3.

(2) a - 2b is a multiple of 3.
ans-d

i can tell why a is sufficient i don't know why b is sufficient?
A fun way to look at Statement (2) is to do something I call "Conceptual Solving". Its sortof like math, but with some "notes" in there as well.

So Statement (2) tells us that a-2b is a multiple of 3, so I can say:

a-2b = (mult. of 3)

Now, I want to know if (a+b) is a multiple of 3, so I can take my little "conceptual equation" and use it to substitute in. So let's solve for a:

a = (mult. of 3) + 2b

So I can plug this in for a in the expression a+b:

a+b = [(mult. of 3) + 2b] + b = (mult. of 3) + 3b

So the sum (a+b) is the sum of a multiple of 3 and 3b. Since both terms are divisible by 3, then their sum must also be divisible by 3!

Sufficient!
:)
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by bpdulog » Fri Oct 21, 2011 8:59 am
GMATGuruNY wrote:
apshara5 wrote:If a and b are both single-digit positive integers, is a + b a multiple of 3?

(1) The two-digit number "ab" (where a is in the tens place and b is in the ones place) is a multiple of 3.

(2) a - 2b is a multiple of 3.
ans-d

i can tell why a is sufficient i don't know why b is sufficient?
Statement 1: The two-digit number "ab" is a multiple of 3.
The sum of the digits of a multiple of 3 is a multiple of 3.
Thus, a+b is a multiple of 3.
SUFFICIENT.

Statement 2: a - 2b is a multiple of 3.
a+b = (a-2b) + 3b
a+b = (multiple of 3) + (multiple of 3)
Thus, a+b is a multiple of 3.
SUFFICIENT.

The correct answer is D.
I don't understand your restatement of Statement 2, can you elaborate more on that? Specifically, where did the 3b come from?
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by GMATGuruNY » Fri Oct 21, 2011 9:25 am
bpdulog wrote:
GMATGuruNY wrote:
apshara5 wrote:If a and b are both single-digit positive integers, is a + b a multiple of 3?

(1) The two-digit number "ab" (where a is in the tens place and b is in the ones place) is a multiple of 3.

(2) a - 2b is a multiple of 3.
ans-d

i can tell why a is sufficient i don't know why b is sufficient?
Statement 1: The two-digit number "ab" is a multiple of 3.
The sum of the digits of a multiple of 3 is a multiple of 3.
Thus, a+b is a multiple of 3.
SUFFICIENT.

Statement 2: a - 2b is a multiple of 3.
a+b = (a-2b) + 3b
a+b = (multiple of 3) + (multiple of 3)
Thus, a+b is a multiple of 3.
SUFFICIENT.

The correct answer is D.
I don't understand your restatement of Statement 2, can you elaborate more on that? Specifically, where did the 3b come from?
3b is the difference between the expression in the question stem (a+b) and the expression in Statement 2 (a-2b):

(a+b) - (a-2b) = 3b.
Thus:
(a+b) = (a-2b) + 3b.
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by bpdulog » Fri Oct 21, 2011 9:38 am
GMATGuruNY wrote:
bpdulog wrote:
GMATGuruNY wrote:
apshara5 wrote:If a and b are both single-digit positive integers, is a + b a multiple of 3?

(1) The two-digit number "ab" (where a is in the tens place and b is in the ones place) is a multiple of 3.

(2) a - 2b is a multiple of 3.
ans-d

i can tell why a is sufficient i don't know why b is sufficient?
Statement 1: The two-digit number "ab" is a multiple of 3.
The sum of the digits of a multiple of 3 is a multiple of 3.
Thus, a+b is a multiple of 3.
SUFFICIENT.

Statement 2: a - 2b is a multiple of 3.
a+b = (a-2b) + 3b
a+b = (multiple of 3) + (multiple of 3)
Thus, a+b is a multiple of 3.
SUFFICIENT.

The correct answer is D.
I don't understand your restatement of Statement 2, can you elaborate more on that? Specifically, where did the 3b come from?
3b is the difference between the expression in the question stem (a+b) and the expression in Statement 2 (a-2b):

(a+b) - (a-2b) = 3b.
Thus:
(a+b) = (a-2b) + 3b.
Thanks for clarifying. Why are we taking the difference between these 2 statements?
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