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27 Mar 2020, 11:12
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
44% (02:00) correct
56%
(02:05)
wrong
based on 174
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Not Attempted Yet
What is the remainder when \(9^{100} + 9^{101} + 9^{102} + . . . . . + 9^{110}\) is divided by 6?
A. 0
B. 1
C. 2
D. 3
E. 4
Posted from my mobile device
A. 0
B. 1
C. 2
D. 3
E. 4
Posted from my mobile device
27 Mar 2020, 17:26
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Dillesh4096
The first thing I noticed is that, since all of the terms in the sum are multiples of 9, they are all also multiples of 3. Therefore, the whole sum is a multiple of 3.
If a number is a multiple of 3 and is also even, then it's a multiple of 2*3, or 6. So, if you divide an even multiple of 3 by 6, you get a remainder of 0.
If you divide an odd multiple of 3 by 6, you get a remainder of 3.
The question is really asking whether this sum is even or odd (which will tell us whether the right answer is A or D).
Every number in the sum is odd (for instance, 9^101 is odd). There are also an odd number of these terms in total (specifically, there are 11 terms). Therefore, the sum is odd, and the remainder is 3. The correct answer is D.
General Discussion
27 Mar 2020, 12:54
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Each of these 11 terms is a positive odd number and is a multiple of 3. Hence, their sum is also a positive odd number that is multiple of 3.
We can write any odd number as 2k+1
Sum = 3*(2k+1)= 6k+3
We will get 3 as a remainder when we divide their sum by 6.
We can write any odd number as 2k+1
Sum = 3*(2k+1)= 6k+3
We will get 3 as a remainder when we divide their sum by 6.
Dillesh4096
