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# 10^25 тАУ 7 is divisible by (1) 2 (2) 3

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Senior Manager
Joined: 30 Aug 2003
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Location: dallas , tx
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10^25 тАУ 7 is divisible by (1) 2 (2) 3 [#permalink]  06 Nov 2003, 22:12
10^25 тАУ 7 is divisible by

(1) 2 (2) 3
(3) 9 (4) Both (2) and (3)

any shortcut and understandable way to solve this?
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shubhangi

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Joined: 28 Oct 2003
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And I'm pretty sure the number is not divisible by 9!

10^25 is a bunch of nines with a three on the end, and according to:

http://www.counton.org/explorer/primes/ ... y911.shtml

it's not divisible.
Senior Manager
Joined: 21 Aug 2003
Posts: 258
Location: Bangalore
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i think what jaydi meant was that her/his answer is: 3... )

Its correct answer choice is 2. (i.e number 3)

shubh, just try few simple choices and u will know that number 3 is answer.
10 - 7 = 3, divisible by 3
100 - 7 = 93, divisible by 3
1000 -7 = 993, divisible by 3
what u realize by now is that no matter what the number of zero's the number subtracted from 7 will always yield some 9's followed by 3.
Thus u know it will be divisble by 3 always...
In such questions just start with siimple numbers and try to see the pattern.
What is the remainder if 7^384 is divided by 5.

a) 0
b) 1
c) 2
d) 3
e) 4
Director
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"What is the remainder if 7^384 is divided by 5."

7^2=49-----4
7^3=a number ending in 3
7^4=a number ending in 1
7^5=a number ending in 7
7^6=a number ending in 9
7^7=a number ending in 3
7^8=a number ending in 1
7^9=a number ending in 7
7^10=a number ending in 9

when the power is divisible evenly by 4, the answer is 1
when the power, divided by 4, yields a remainder of 1, the answer is 2.
when the power, divided by 4, yields a remainder of 2, the answer is 4.
when the power, divided by 4, yields a remainder of 3, the answer is 3.

When we divide 384 by 4 and get no remainder, the answer is 1

Am I right, Vicky? And would you say that this is a difficult question?
Senior Manager
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shubhangi

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stoolfi wrote:
"What is the remainder if 7^384 is divided by 5."

7^2=49-----4
7^3=a number ending in 3
7^4=a number ending in 1
7^5=a number ending in 7
7^6=a number ending in 9
7^7=a number ending in 3
7^8=a number ending in 1
7^9=a number ending in 7
7^10=a number ending in 9

when the power is divisible evenly by 4, the answer is 1
when the power, divided by 4, yields a remainder of 1, the answer is 2.
when the power, divided by 4, yields a remainder of 2, the answer is 4.
when the power, divided by 4, yields a remainder of 3, the answer is 3.

When we divide 384 by 4 and get no remainder, the answer is 1

Am I right, Vicky? And would you say that this is a difficult question?

Looks like correct approach..
Senior Manager
Joined: 21 Aug 2003
Posts: 258
Location: Bangalore
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stoofli your approach is perfect... good..
i wud say this may be one of the tough questions u will face at the test...
thanks
SVP
Joined: 03 Feb 2003
Posts: 1609
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the number will be 9999...99993
it is clearly divisible by 3 but not by 9.
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