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# For what value of 'n' will the remainder of 351^n and 352^n

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CEO
Joined: 15 Aug 2003
Posts: 3469
Followers: 61

Kudos [?]: 701 [0], given: 781

For what value of 'n' will the remainder of 351^n and 352^n [#permalink]  11 Oct 2003, 04:52
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For what value of 'n' will the remainder of 351^n and 352^n be the same when divided by 7?

a) 2
b) 3
c) 6
d) 4
e ) Both 3 and 6

EDIT : 5th option added ..mistake in the problem

Last edited by Praetorian on 13 Oct 2003, 00:39, edited 1 time in total.
SVP
Joined: 03 Feb 2003
Posts: 1609
Followers: 6

Kudos [?]: 75 [0], given: 0

351^n=7p+R1
352^n=7q+R2

351=350+1
352=350+2

n=2) (350+1)*(350+1); R1=1 so, R2=4
n=3) R1=1 R2=1 OK

n=6) R1=1 R2=1 OK
n=4) R1=1 R2=2

CEO
Joined: 15 Aug 2003
Posts: 3469
Followers: 61

Kudos [?]: 701 [0], given: 781

For what value of 'n' will the remainder of 351^n and 352^n be the same when divided by 7?

When 351 is divided by 7, the remainder is 1.
When 352 is divided by 7, the remainder is 2.

Let us look at answer choice (1), n = 2

When 351^2 is divided by 7, the remainder will be 1^2 = 1.
When 352^2 is divided by 7, the remainder will be 2^2 = 4.

So when n = 2, the remainders are different.

When n = 3,

When 351^3 is divided by 7, the remainder will be 1^3 = 1.
When 352^3 is divided by 7, the remainder will be 2^3 = 8.

As 8 is greater than 7, divide 8 again by 7, the new remainder is 1.
So when n = 3, both 351^n and 352^n will have the same remainder when divided by 7.

thanks
praetorian

Last edited by Praetorian on 13 Oct 2003, 00:34, edited 1 time in total.
SVP
Joined: 03 Feb 2003
Posts: 1609
Followers: 6

Kudos [?]: 75 [0], given: 0

351^N leaves 1 with any positive N, when divided by 7
(350+1)*...*(350+1) all the multiplied members are divisible by 7, save for the last 1.

check out 352=350+2

N=2) (350+2)*(350+2) leaves 4 =7a+4
N=3) (7a+4)*(350+2) leaves 8, or actually 1, =7b+1
N=4) (7b+1)*(350+2) leaves 2, or =7c+2
N=5) (7c+2)*(350+2) leaves 4, or =7d+4
N=6) (7d+4)*(350+2) leaves 8, or actually 1

remainders (from N=2) go in the order of 4, 1, 2, 4, 1, 2, ...

so N=3 and N=6 both match
CEO
Joined: 15 Aug 2003
Posts: 3469
Followers: 61

Kudos [?]: 701 [0], given: 781

stolyar wrote:

351^N leaves 1 with any positive N, when divided by 7
(350+1)*...*(350+1) all the multiplied members are divisible by 7, save for the last 1.

check out 352=350+2

N=2) (350+2)*(350+2) leaves 4 =7a+4
N=3) (7a+4)*(350+2) leaves 8, or actually 1, =7b+1
N=4) (7b+1)*(350+2) leaves 2, or =7c+2
N=5) (7c+2)*(350+2) leaves 4, or =7d+4
N=6) (7d+4)*(350+2) leaves 8, or actually 1

remainders (from N=2) go in the order of 4, 1, 2, 4, 1, 2, ...

so N=3 and N=6 both match

i agree

When 351^6 is divided by 7, the remainder will be 1^6 = 1.
When 352^6 is divided by 7, the remainder will be 2^6 = 64.

But 64 > 7 ,so divide 64/ 7 , remainder is 1.

Both 3 and 6 work...

let me make necessary changes to the problem

thanks
praetorian
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