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# What is the remainder when 14^(22)+6^(22) is divided by 5

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Math Expert
Joined: 02 Aug 2009
Posts: 8284
What is the remainder when 14^(22)+6^(22) is divided by 5  [#permalink]

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23 Nov 2019, 21:42
00:00

Difficulty:

35% (medium)

Question Stats:

63% (01:01) correct 38% (01:27) wrong based on 56 sessions

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What is the remainder when $$14^{22}+6^{22}$$ is divided by 5?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

Chetan’s question

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Manager
Joined: 16 Feb 2015
Posts: 124
Location: United States
Concentration: Finance, Operations
Re: What is the remainder when 14^(22)+6^(22) is divided by 5  [#permalink]

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23 Nov 2019, 21:55
2
2
Explanation: 14^22 + 6^22 divided by 5
=[(15-1)^22 + (5+1)^22] ÷ 5
=[(-1)^22 + (1)^22] ÷ 5
=(1+1) ÷ 5 = 2/5
Remainder is 2.
Option 3

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Manager
Joined: 16 Feb 2015
Posts: 124
Location: United States
Concentration: Finance, Operations
Re: What is the remainder when 14^(22)+6^(22) is divided by 5  [#permalink]

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23 Nov 2019, 23:25
2
There is more method. That is cyclicity.
14^22 + 6^22 divided by 5
Last digit cycle.
For 4: last digit cyclicity 4,6,4,6
For 6:similarly it will be 6,6,6,6
Therefore, we can write 14^22 + 6^22 as below
= 4^22 + 6^22 (as our focus is at last digit only)
Using cyclicity, we know last digit is
= 6+6
=12
12÷5 = 2 which is remainder

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Re: What is the remainder when 14^(22)+6^(22) is divided by 5   [#permalink] 23 Nov 2019, 23:25
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