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# Is a 3-digit integer divisible by 3? (1) The hundreds' digit

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Intern
Joined: 10 Aug 2010
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Is a 3-digit integer divisible by 3? (1) The hundreds' digit [#permalink]

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10 Aug 2010, 05:48
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71% (01:30) correct 29% (01:01) wrong based on 7 sessions

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Is a 3-digit integer divisible by 3?
(1) The hundreds' digit of the number is equal to the tens' digit of the number.
(2) The units' digit of the number is 6 greater than the tens' digit of the number

OA
[Reveal] Spoiler:
C

A little confused

I'll start by naming the integers A B C

(1) if 100's digit is = 10's digit then A = B so it now becomes A A C unit place is not given so INSUFF

(2) if unit digit is 6 greater than the 10's digit the number will read as A B 6+B since 100's digit is not given it still is INSUFF (to me)

BOTH together then it will read something like this A A 6+A

so the numbers will be either 116 , 227 or 339 to me it is still INSUFF and i picked E

where am i going wrong ??

Sorry about posting in such a short time My exam is very very close!!!!
[Reveal] Spoiler: OA
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10 Aug 2010, 06:23
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Expert's post
amijags wrote:
Is a 3-digit integer divisible by 3?
(1) The hundreds' digit of the number is equal to the tens' digit of the number.
(2) The units' digit of the number is 6 greater than the tens' digit of the number

OA
[Reveal] Spoiler:
C

A little confused

I'll start by naming the integers A B C

(1) if 100's digit is = 10's digit then A = B so it now becomes A A C unit place is not given so INSUFF

(2) if unit digit is 6 greater than the 10's digit the number will read as A B 6+B since 100's digit is not given it still is INSUFF (to me)

BOTH together then it will read something like this A A 6+A

so the numbers will be either 116 , 227 or 339 to me it is still INSUFF and i picked E

where am i going wrong ??

Sorry about posting in such a short time My exam is very very close!!!!

First # should be 117, which is divisible by 3 and second number should be 228, which is also divisible by 3. So all 3 numbers possible: 117, 228 and 339 are divisible by 3.

Is a 3-digit integer divisible by 3?

Algebraic solution:

A number is divisible by 3 if the sum of its digits is divisible by 3. So 3-digit integer $$abc$$ will be divisible by 3 if $$a+b+c=3k$$, where k is integer or to make it simpler if $$a+b+c$$ is a multiple of 3.

(1) The hundreds' digit of the number is equal to the tens' digit of the number --> $$a=b$$ --> $$a+b+c=b+b+c=2b+c$$. Now, $$2b+c$$ may or may not be a multiple of 3, hence this statement is insufficient.

(2) The units' digit of the number is 6 greater than the tens' digit of the number --> $$c=b+6$$ --> $$a+b+c=a+b+b+6=a+2b+6$$. Now, $$a+2b+6$$ may or may not be a multiple of 3, hence this statement is insufficient.

(1)+(2) As from (1) $$a=b$$ and from (2) $$c=b+6$$ then $$a+b+c=b+b+b+6=3b+6=3(b+2)$$. As $$3(b+2)$$ has 3 as multiple hence it's divisible by 3. Sufficient.

 ! Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/

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10 Aug 2010, 06:36
OMG i cant believe i made this stupid mistake!!!!!!! shame on me
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10 Aug 2010, 08:21
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we have 100a+10b+c
from (1) we have a=b, thus
100a+10a+с
from (2) we have c=6+a, so
100a+10a+a+6 = 111a+6
plug any figure instead of a, you will see it divisible by 3.
(C)
Re: divisibility issues   [#permalink] 10 Aug 2010, 08:21
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