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

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Senior Manager
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10^25 тАУ 7 is divisible by (1) 2 (2) 3 [#permalink] New post 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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 [#permalink] New post 07 Nov 2003, 06:20
If answer 3 is correct, mustn't answer 2 be correct, also?

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.
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 [#permalink] New post 07 Nov 2003, 06:46
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.
One question for your practise:
What is the remainder if 7^384 is divided by 5.

a) 0
b) 1
c) 2
d) 3
e) 4
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 [#permalink] New post 07 Nov 2003, 07:19
"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?
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 [#permalink] New post 07 Nov 2003, 08:03
Vicky...give me your solution...
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 [#permalink] New post 08 Nov 2003, 16:23
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..
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 [#permalink] New post 08 Nov 2003, 18:46
stoofli your approach is perfect... good..
i wud say this may be one of the tough questions u will face at the test...
thanks
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 [#permalink] New post 11 Nov 2003, 00:36
the number will be 9999...99993
it is clearly divisible by 3 but not by 9.
  [#permalink] 11 Nov 2003, 00:36
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10^25 тАУ 7 is divisible by (1) 2 (2) 3

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