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

Question

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

rajatchopra1994 · Accepted Answer

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
Answer is C

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rajatchopra1994 · Answer

Explanation: 14^22 + 6^22 divided by 5
=  ÷ 5
=  ÷ 5
=(1+1) ÷ 5 = 2/5
Remainder is 2.
Option 3

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Amlu95 · Answer

14 can be -> (2*7)^22
6 can be -> (2*3)^22

Using cyclicity, we arrive at (4*9)+(4*9) = 72 %5 = 2

KarishmaB · Answer

We can use binomial or cyclicity. Both approaches have been discussed above.

Samuel0709 · Answer

To find the remainder we need to know the sum of the units digits of both these expressions (sum)
Since both exponents are such high numbers it is not possible to multiply, instead lets try and observe if there is any pattern :

14^1= 14
14^2= 196
14^3= 2744
and so on, so  every even power of 14 has a unit digit of 6

6^1=6
6^2=36
6^3=216
and so on, every  even power of 6 also has a unit digit of 6

6+6=12 and 12/5 leaves a remainder 2. So option C=2 is our answer

BrushMyQuant · Answer

We know to find what is the remainder when  14^{22}+6^{22}  is divided by 5

Lets solve the problem using two methods:

Method 1: Binomial Theorem

We solve these problems by using Binomial Theorem, where we split the number into two parts, one part is a multiple of the divisor(5) and a big number, other part is a small number.

=>  14^{22}+6^{22}  =  (15 - 1)^{22}+(5 + 1)^{22}

Watch this video to   MASTER BINOMIAL Theorem

Now, when we expand this expression then all the terms except the last term will be a multiple of 5. 
=> All terms except the last term will give 0 as remainder then divided by 5
=> Problem is reduced to what is the remainder when the last term is divided by 5
=> What is the remainder when  22C22 * 15^0 * (-1)^ {22}  +  22C22 * 5^0 * (1)^ {22}  is divided by 5 = Remainder of 1 + 1 by 5 = 2

Method 2: Units' Digit Cycle

Theory: Remainder of a number by 5 is same as the Remainder of the unit's digit of the number by 5

( Watch this Video  to Learn  How to find Remainders of Numbers by 5 )

Units' digit of  14^{22} , will be same units' digit of  4^{22} 
 
We can do this by finding the pattern / cycle of unit's digit of power of 4 and then generalizing it.

Unit's digit of  4^1  = 4
Unit's digit of  4^2  = 6
Unit's digit of  4^3  = 4

So, unit's digit of power of 4 repeats after every  2^{nd}  number.
=> If power is odd then units' digit is 4
=> If power is even then units' digit is 6

=> Unit's digits of  14^{22}  = 6

Units' digit of positive integer power of 6 is ALWAYS 6
 => Unit's digits of  6^{22}  = 6

=> Unit's digits of  14^{22}  + Unit's digits of  6^{22}  = 6 + 6 = 12
=> Remainder = remainder of 12 by 5 = 2

So,  Answer will be C 
Hope it helps!

MASTER How to Find Remainders with 2, 3, 5, 9, 10 and Binomial Theorem

https://www.youtube.com/watch?v=h1HTEhmUb1w

bumpbot · Answer

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

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

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