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25 Jan 2021, 03:00
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Difficulty:
65%
(hard)
Question Stats:
55% (01:43) correct
45%
(02:13)
wrong
based on 260
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What is the remainder when \(((11^{12})^{13})^{14}\) is divided by9?
A. 1
B. 2
C. 3
D. 5
E. 7
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
A. 1
B. 2
C. 3
D. 5
E. 7
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
25 Jan 2021, 03:24
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Bunuel
Let us expand the term \(((11^{12})^{13})^{14}\)
\(((11^{12})^{13})^{14}\)= \(((9+2)^{12})^{13})^{14}\)
When we expand the term using binomial expansion, all terms except the last will be divisible by 9
\(((2^{12})^{13})^{14}\)
\(((2^3)^4)^{13})^{14}\)
\(((8)^4)^{13})^{14}\)
\(((9-1)^4)^{13})^{14}\)
Again only last term will not be divisible by 9, when the term is expanded
\(((-1)^4)^{13})^{14}\)
\((1^{13})^{14}=1\)
So remainder =1.
We can also find pattern
11 leaves a remainder 2
\(11^2.....2^2\)
\(11^3.....2^3=8\)
\(11^4......2^4=16=9+7\)
\(11^5......2^5=32=27+5\)
\(11^6......2^6=64=63+1\)
\(11^7......2^7=128=126+2\)
So pattern repeats after every 6 terms => 2,4,8,7,5,1,2,4,8....
Now the power 12 is multiple of 6, so \(11^{12}\) will leave 1 as the remainder.
A
General Discussion
07 Jun 2021, 02:44
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Bunuel
\(((11^{12})^{13})^{14}\)
\(((121^{6})^{13})^{14}\)
\((((117+4)^{6})^{13})^{14}\)
117 = 13*9 so it wont leave any remainder when divided by 9
\(((4^{6})^{13})^{14}\)
\(((64^{2})^{13})^{14}\)
\((((63+1)^{2})^{13})^{14}\)
\((((1)^{2})^{13})^{14}\)
\(=1\)
