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a, b, c are positive integers. If a, b and c are assembled

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Manager
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a, b, c are positive integers. If a, b and c are assembled  [#permalink]

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New post 29 Sep 2008, 23:16
1
3
a, b, c are positive integers. If a, b and c are assembled into the six-digit number abcabc, which one of the following must be a factor of abcabc?
16
13
None of these
5
3

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Re: Factor of abcabc  [#permalink]

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New post 29 Sep 2008, 23:33
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agree with bigtreez.

My approach.

abcabc = 100000a + 10000b + 1000c + 100a + 10b + c
= 100100a + 10010b + 1001c
=1001(100a + 10b + c)

The above expression has 3 as a factor (1001 is divisible by 3).

Hence, 3 should be the answer.
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Re: Factor of abcabc  [#permalink]

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New post 29 Sep 2008, 23:44
1001 is not divisible by 3?
scthakur wrote:
agree with bigtreez.

My approach.

abcabc = 100000a + 10000b + 1000c + 100a + 10b + c
= 100100a + 10010b + 1001c
=1001(100a + 10b + c)

The above expression has 3 as a factor (1001 is divisible by 3).

Hence, 3 should be the answer.
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Re: Factor of abcabc  [#permalink]

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New post 30 Sep 2008, 00:00
how could 1001 is divisible by 3?

it is divisible by 13

answer is 13 : OA

it was tricky one.. specially when you encounter in exam.
Thanks
VP
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Re: Factor of abcabc  [#permalink]

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New post 30 Sep 2008, 02:40
vr4indian wrote:
how could 1001 is divisible by 3?

it is divisible by 13

answer is 13 : OA

it was tricky one.. specially when you encounter in exam.
Thanks



My silly mistake....got trapped in basic arithmatic.
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Re: Factor of abcabc  [#permalink]

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New post 30 Sep 2008, 10:38
scthakur wrote:
agree with bigtreez.

My approach.

abcabc = 100000a + 10000b + 1000c + 100a + 10b + c
= 100100a + 10010b + 1001c
=1001(100a + 10b + c)

The above expression has 3 as a factor (1001 is divisible by 3).

Hence, 3 should be the answer.



nice approach and nice question
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Re: Factor of abcabc  [#permalink]

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New post 30 Sep 2008, 21:03
scthakur wrote:
agree with bigtreez.

My approach.

abcabc = 100000a + 10000b + 1000c + 100a + 10b + c
= 100100a + 10010b + 1001c
=1001(100a + 10b + c)

The above expression has 3 as a factor (1001 is divisible by 3).

Hence, 3 should be the answer.


Great Approach! I did trial and error and narrowed down to 13 and none. But I picked 3 numbers and all of them were divisible by 13.
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Re: Factor of abcabc  [#permalink]

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New post 01 Oct 2008, 01:37
vr4indian wrote:
how could 1001 is divisible by 3?

it is divisible by 13

answer is 13 : OA

it was tricky one.. specially when you encounter in exam.
Thanks


Agree with you

abcabc = abc + 1000abc = 1001abc

Then see which number can be divided
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Re: Factor of abcabc  [#permalink]

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New post 01 Oct 2008, 03:12
2
Remember that
any three digit number when multiplied by 1001 , the resultant is the same three digit number side by side
abc X 1001 = abcabc

or, abc X 13 X 11 X 7 = abcabc

which is surely divisible by 13, 11 and 7

answer is 13 as per the choices
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Re: Factor of abcabc  [#permalink]

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New post 01 Oct 2008, 22:25
vr4indian wrote:
a, b, c are positive integers. If a, b and c are assembled into the six-digit number abcabc, which one of the following must be a factor of abcabc?
16
13
None of these
5
3


IMO B
abcabc = c+10b+100a+1000c+10000b+100000a = 1001c+10010b+100100a => this is div by 1001 or 13 hence IMO B
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Re: a, b, c are positive integers. If a, b and c are assembled  [#permalink]

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New post 27 Jun 2018, 10:28
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Re: a, b, c are positive integers. If a, b and c are assembled &nbs [#permalink] 27 Jun 2018, 10:28
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