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m15 q29

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Senior Manager
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Status: Happy to join ROSS!
Joined: 29 Sep 2010
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Kudos [?]: 132 [0], given: 48

Concentration: General Management, Strategy
Schools: Ross '14 (M)
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m15 q29 [#permalink]

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New post 05 Dec 2010, 13:11
If in a six-digit integer N , F(k) is the value of the k- th digit, is N divisible by 7 (For example, F(4) is the value of the hundreds digit of N )?

1. F(1)= F(4); F(2) = F(5); F(3)=F(6)
2. F(1)= F(2)= F(3) =.. = F(6)

OA: D
Explanation
Statements (1) and (2) by themselves are sufficient. S1 tells us that the last three digits of are the same as the first three digits: N=abcabc . Note that abc*1000+abc = abc*1001 . As 1001 is divisible by 7, is also divisible by 7.

==
I did not get it. Why do we talk about 1001? and abc*1001 if we only have 6-digit integer N?
And what about Statement 2 - where is not a word about its sufficiency.
Thanks!

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Re: m15 q29 [#permalink]

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New post 05 Dec 2010, 20:37
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Let f(1) = f(4) = A ; f(2) = f(5) = B ; f(3) = f(6) = C then,

A. (10^5 * C ) + ( 10^4 *B) + (10^3 *A ) + ( 10^2 * C ) + (10 * B) + A

Reducing the terms we have

( 10^2 * C ) * [ 10^3 + 1 ] + (10 * B) * [ 10^3 + 1 ] + A * [ 10^3 + 1 ]

Taking [ 10^3 + 1 ] common term outside

[ 10^3 + 1 ] * {(10^2 * C) + (10 * B) + A }

[ 10^3 + 1 ] = 1001 is divisible by 7. Hence A is sufficient.


Let f(1) = f(2) = ...... = f(6) = a ; then

B. (10^5 * a ) +(10^4 * a)+ ( 10^3 * a )+ (10^2 * a )+ (10 * a ) + a

taking a the common term outside ,

(10^5 + 10^4 + 10^3 + 10^2 + 10 + 1 ) a

111111 * a , here 111111 is divisible by 7. Hence B is sufficient. So, OA D.

I believe this helps.
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Kudos [?]: 286 [1], given: 10

Re: m15 q29   [#permalink] 05 Dec 2010, 20:37
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