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Remainder

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Director
Joined: 23 May 2008
Posts: 839
Followers: 3

Kudos [?]: 39 [0], given: 0

Re: Remainder [#permalink]  15 May 2009, 08:37
00:00

Difficulty:

(N/A)

Question Stats:

100% (01:00) correct 0% (00:00) wrong based on 0 sessions
gmatee wrote:
What is the units digit of 6^15 - 7^4 - 9^3?
A:8
B:7
C:6
D:5
E:4

6^15 ends in 6
7^4 ends in 1
9^3 ends in 9

6-1=5

5-9 or 15 -9 after carrying 10 =6

C
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Joined: 29 Aug 2007
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Re: Remainder [#permalink]  15 May 2009, 11:12
bigtreezl wrote:
gmatee wrote:
What is the units digit of 6^15 - 7^4 - 9^3?
A:8
B:7
C:6
D:5
E:4

6^15 ends in 6
7^4 ends in 1
9^3 ends in 9

6-1=5

5-9 or 15 -9 after carrying 10 =6

C

Correct. This is how it is supposed to be solved.

gmatee wrote:
Why do u carry 10?
it may ends in 4 or 6.
Consider this:
155 - 99 = 56
15 - 99 = -84

Reminder can never be in -ve. 15 - 99 = -84 is not correct.

In this case, 6^15 is way larger than (7^4 + 9^3). So there is no question of -ve reminder.
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Re: Remainder [#permalink]  16 May 2009, 12:36
gmatee wrote:
thanks tiger, but the question now is changed to : is 6^15 > 7^4 + 9^3 ?it is tough for me too

Thats much easier: Its true that 6^15 > 7^4 + 9^3 because 6^15 is way larger than (7^4 + 9^3).
It is always true that n^(n+1) > (n+1)^n for all n>2.

Moreover, the issue here is not of addition rather multiplication. In that case, 6^15 is way greater than (7^4 * 9^3).
Therefore, 6^15 is undoubtdly even greater than (7^4 + 9^3).
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Joined: 15 May 2009
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Re: Remainder [#permalink]  16 May 2009, 15:22
gmatee wrote:
What is the units digit of 6^15 - 7^4 - 9^3?
A:8
B:7
C:6
D:5
E:4

Units digit of 6^15 is 6;
for 7^4 it'll be 1;
for 9^3 it'll be 9;

XXXX6-XXXX1-XXXX9 = XXXX6; Thus answer is (C).
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Re: Remainder [#permalink]  17 May 2009, 07:28
gmatee wrote:
ok,i think 6^15 - 7^4 - 9^3 > 0 in the first place (guessing).
btw, how do u know n^(n+1) > (n+1)^n for all n>1, what if n=2, then 8>9?

Thanks for the good catch.

That should be n>2. I updated my earlier post.
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Re: Remainder   [#permalink] 17 May 2009, 07:28
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