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nislam
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Remainder when divided by 5:
\((-1)^{14} +(0)^{15} +(1)^{16} \)
\(= 1+0+1\)
\(=2\)
Option C

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Hi, here is my logic to solve this question;
14^14 would give you a number with the last digit being 6 (because 14^2 gives 6, 14^3 gives 4, 14^4 gives 6...14^14 giving you 6)
15^15 would give you a number with the last digit being 5
16^16 would give you a number with the last digit being 6 (last digits give you 6x6x6.. giving you always 6 as the last digit)

if you sum last digits and divide it 5;
6+5+6=17 divided by 5 gives 2 as the remainder
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gloomybison
Hi, here is my logic to solve this question;
14^14 would give you a number with the last digit being 6 (because 14^2 gives 6, 14^3 gives 4, 14^4 gives 6...14^14 giving you 6)
15^15 would give you a number with the last digit being 5
16^16 would give you a number with the last digit being 6 (last digits give you 6x6x6.. giving you always 6 as the last digit)

if you sum last digits and divide it 5;
6+5+6=17 divided by 5 gives 2 as the remainder

Hi gloomybison
The approach you are following is apt for the questions where you are supposed to find/work-with unit digits. But for such remainder problems, you can simply focus on the 'remainder raised to the even or odd power (and not the magnitude of power)'. If you are not aware of this method, visit https://www.expertsglobal.com/finding-r ... nt-on-gmat
(Source: ExpertsGlobal)

Best regards
:please:
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amritaarhea

nislam
What is the remainder when \(14^{14}+15^{15}+16^{16}\) is divided by 5?

A) 0

B) 1 ^

C) 2

D) 3

E) 4
I think the answer should be 5

Anything 4^4 will give a number that ends with 4. Hence, the remainder will be 4. Second number will give 0 remainder. And 6^6 will give a number a that ends with 6. Hence, the remainder will be 1. Thus, the total remainder becomes 5. Kindly let me know what am I missing here.
­4^x does not always give a number that ends in 4. 4^x pattern is 4, 6, 4, 6.... for all x>=1.­
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