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General concept

by ketkoag » Tue May 19, 2009 4:01 am
What is the remainder when 7^625 is divided by 15?
[spoiler]Now, i got that by power cycle the unit's digit will be 7, but how the answer to the question above is 7?[/spoiler]
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Re: General concept

by Pranay » Tue May 19, 2009 4:33 am
ketkoag wrote:What is the remainder when 7^625 is divided by 15?
[spoiler]Now, i got that by power cycle the unit's digit will be 7, but how the answer to the question above is 7?[/spoiler]
Hi,

When 7^8 is divided by 15 the remainder left is 1.

Thus,

7^625 = [(7^8)^78].7

Since, when 7^8 divided by 15 leaves a remainder of 1

=> [(1)^78].7 leads a remainder of 7.

Hope it helps.

Please let me know if there is any other approach to solve the problem.

regards,

Pranay
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by sagiroz » Tue May 19, 2009 6:06 am
Hi,
I found out that it is easier to use 7^4=2401. It is now easy to see that 2400 is divided by 15, hence the reminder is 1.

It is easier to see that 625 when dived with 4 gives a reminder of 1.

Given that, 7^625 = ( (7^4)^x)*7
Let x be 624/4.

We know that 7^4 / 15 leaves a reminder of 1. Hence (7^4)^x) /15 leaves a reminder of 1 also.
Multiply it by 7 and you get a reminder of 7.

Regards
Sagi.
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by odod » Wed May 20, 2009 6:40 am
Hello...

I've been reviewing the last couple posts for the last day and its not sinking in. I guess my questiosn are

1) Given that, 7^625 = ( (7^4)^x)*7

Why are you multiplying it by 7????


2) If I get this type of question on the GMAt, I wouldn't even know where to start. For Example, if it was 6^350 and I want to see if it is divisible by 15, where would I start.

I noticed that the first post picked 7^4, the second post picked 7^8. No idea how. Any help would be appreciated .Thanks!
ODOD
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