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If a and b are positive integers, is 10^a + b divisible by 3

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If a and b are positive integers, is 10^a + b divisible by 3 [#permalink] New post 06 Nov 2008, 09:53
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A
B
C
D
E

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Question Stats:

58% (02:18) correct 42% (02:08) wrong based on 69 sessions
If a and b are positive integers, is 10^a + b divisible by 3?

(1) b/2 is an odd integer.

(2) The remainder of b/10 is b.
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Oct 2013, 08:15, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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Re: No. Properties [#permalink] New post 06 Nov 2008, 10:08
for a number to be divisible by 3, sum of the digits have to be divisible by 3.
So I'd try looking to get at the sum of digits from \(10^a + b\)
\(10^a\) gives us just 1+how many ever 0s. so the options for sum of the digits of b ought to be 2, 4, 8 etc..

From (1) we are told b/2 is odd.. Take for example b = 6, so b/2 = 3. But then sum of digits yields 1+3, which is not divisible by 3. Alternatively, take b=10, b/2 = 5. Then sum of digits yields 5+1 = 6, which is divisible by 3. Since we have conflicting results from (1), it is not sufficient by itself. Cross out A and D.

(2) says remainder of b/10 is b. That means, b < 10. We'll get conflicting results if we were to chose b=5 or b = 4. So (2) alone is not sufficient enough. Cross out B.

(1) and (2) together, says b/2 is odd and b < 10, it gives only choice for b as 6. that is conclusive enough. \(10^a + 6\) is not divisible by 3. So my pick for the answer is C
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Re: No. Properties [#permalink] New post 06 Nov 2008, 10:36
[quote="study"]If a and b are positive integers, is 10^a + b divisible by 3?

1. b/2 is an odd integer
2. the remainder of b/10 is b

when 10^a is devided by 3 the remainder is always 1 thus the question asks is 1+b devisable by 3
b can be 2,5,8,11,14..etc

from 1

b is even .....insuff

from 2

b is is less than 10 (0,1,2................9).........insuff

both

2,4,6,8 are all even and less than 10.........ONLY 6 WHEN DEVIDED BY 2 GIVES AN ODD

SUFF

C
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Re: No. Properties [#permalink] New post 06 Nov 2008, 10:58
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agree with C..

stmnt 2 says B<10

stmnt 1 says B/2=odd

together we know b=6
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Re: No. Properties [#permalink] New post 06 Nov 2008, 11:04
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study wrote:
If \(a\) and \(b\) are positive integers, is \(10^a + b\) divisible by 3?

1. \(\frac{b}{2}\) is an odd integer
2. the remainder of \(\frac{b}{10}\) is \(b\)

what positive integer divided by itself gives a remainder of itself?


It should be E because b could be 2 or 6.

If b is either 2, 10^a + b is divisible by 3. But if b is 6, then 10^a + b is not divisible by 3.
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Re: No. Properties [#permalink] New post 06 Nov 2008, 11:14
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yup ur right..i overlooked 2/2=1
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Re: No. Properties [#permalink] New post 06 Nov 2008, 13:38
thanks for pointing out a silly mistake :oops:

stand corrected. Somehow on DS, I gravitate towards C. Need to keep that in check on the real test.
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Re: No. Properties [#permalink] New post 06 Nov 2008, 19:49
Yeah its E

b/2 is odd integer means b = 2 X odd means even

b= {0,1,..9}

Together, b can be {0,2,4,6,8}. We can clearly see yes and no. Insuff
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Re: If a and b are positive integers, is 10^a + b divisible by [#permalink] New post 12 Oct 2013, 06:36
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study wrote:
If \(a\) and \(b\) are positive integers, is \(10^a + b\) divisible by 3?

1. \(\frac{b}{2}\) is an odd integer
2. the remainder of \(\frac{b}{10}\) is \(b\)

what positive integer divided by 10 gives a remainder of itself?


Hi all, let me try to explain this one.

So we have 10^a + b /3 what is the remainder?

First of all note that you are told that a,b are positive integers. THEY ALWAYS TELL YOU THIS FOR SOME REASON, DON'T IGNORE IT

So we have that 10^a = (10,100 etc...)

Now, what is the divisibility rule for multiples of 3? That's right the digits of the number must add to a multiple of 3
We already have the 1 on the 10^a. Let's see what we can get out of 'b'

Statement 1: b/2 is odd integer. So b could be (2,6,10,14 etc...)
If b is 2 then remainder is 0
If b is 6 then remainder is 1
If b is 10 then remainder is 2
Not good enough

Statement 2: b/10 remainder is b. Now wait a second. What this is telling us is that b<10. But that could be (1,2,3,4,5 etc...)
This is stil not enough

Statement (1) and (2) - Now going back to statement 1. We still have 2 or 6 as possible answer giving different remainders

Hence (E) is the correct answer

Kudos if you like! :)
Re: If a and b are positive integers, is 10^a + b divisible by   [#permalink] 12 Oct 2013, 06:36
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