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

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Manager
Joined: 22 Sep 2008
Posts: 117
a, b, c are positive integers. If a, b and c are assembled  [#permalink]

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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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VP
Joined: 17 Jun 2008
Posts: 1462

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29 Sep 2008, 23:33
1
1
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.
VP
Joined: 18 May 2008
Posts: 1161

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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.
Manager
Joined: 22 Sep 2008
Posts: 117

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

it is divisible by 13

it was tricky one.. specially when you encounter in exam.
Thanks
VP
Joined: 17 Jun 2008
Posts: 1462

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

it is divisible by 13

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

My silly mistake....got trapped in basic arithmatic.
Director
Joined: 23 May 2008
Posts: 730

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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
VP
Joined: 05 Jul 2008
Posts: 1317

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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.
Manager
Joined: 30 Sep 2008
Posts: 111

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

it is divisible by 13

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
Intern
Joined: 16 Feb 2006
Posts: 29
Location: ZURICH

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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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VP
Joined: 17 Jun 2008
Posts: 1265

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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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Joined: 09 Sep 2013
Posts: 8179
Re: a, b, c are positive integers. If a, b and c are assembled  [#permalink]

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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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