What is the remainder when 12^24 is divided by 5?

Question

A. 1
B. 2
C. 3
D. 4
E. 6

giobas · Answer

power of 2 ends on:
2 - 4 -8 - 6 and repeat

so 2^24 will repeat this endings 6 times and end on 6
and divided by 5 will result in reminder 1

answer A

RR88 · Answer

Option A

Remainder (  \frac{12^{24}}{5} ) = ?

Remainder (  \frac{12^{24}}{5} ) = Remainder (unit's digit of (12^{24})/5)

: ( 12^{24} ) = ( 2^{48} * 3^{24} ) 
: Unit's digit ( 2^{48} * 3^{24} ) = ( 6 * 1)   = 6

:Remainder ( \frac{6}{5} ) = 1

blackfish · Answer

12^24 can be written as (10+2)^24.

By Binomial expansion, we know that all the terms will have multiples of 10 except the last term which is 2^24. So, the remainder will depend upon 2^24.

Now, to find out the last digit of 2^24 :
2^1=2
2^2=4
2^3=8
2^4=16
2^5=32

As we can see the units digit start to repeat after 4, that means cyclicity is 4.
Dividing the power of 2, i.e, 24 by 4 we get 0. 
That means the last digit of 2^24 will be same as last digit of 2^4 which is 6. 
Hence, when divided by 5 the remainder will be 1.

Ans: A

Abhishek009 · Answer

12 = \frac{(10 + 2)}{5}  ( Remainder 2 )
 12^2 = \frac{(10 + 2)^2}{5}  ( Remainder 4 )
 12^3 = \frac{(10 + 2)^3}{5}  ( Remainder 3 )
 12^4 = \frac{(10 + 2)^4}{5}  ( Remainder 1 )

Now,  12^{24} = 12^4*6

Since,  12^4  will have a remainder of 1 ;  12^{24}  will also have a remainder of 1 when divided by 5

Thus, answer must be (A) 1

ScottTargetTestPrep · Answer

When solving this problem, we should recall the rule that we can determine the remainder when a number is divided by 5 by simply dividing the units digit of that number by 5. Thus, to determine the remainder when 12^24 is divided by 5, we need to first calculate the units digit of 12^24.
 
Since we only care about the units digit, we can evaluate the pattern of units digits for 2^n for positive integer values of n. That is, let’s look at the pattern of the units digits of powers of 2, which will result in the same pattern as 12^n. When writing out the pattern, notice that we are ONLY concerned with the units digit of 2 raised to each power.
 
2^1 = 2 
 
2^2 =4 
 
2^3 = 8 
 
2^4 = 6 
 
2^5 = 2
 
The pattern of the units digit of powers of 2 repeats every 4 exponents. The pattern is 2–4–8–6. In this pattern, all positive exponents that are multiples of 4 will produce a 6 as their units digit. Thus:
 
2^24 has a units digit of 6.
 
Finally, since the remainder is 1 when 6 is divided by 5, the remainder is also 1 when 12^24 is divided by 5.
 
Answer: A

Abhishek009 · Answer

\frac{12^{24}}{5}

=\frac{(10 + 2) ^{24}}{5}

\frac{10}{5}  will be completely divisible and leave no remainder.....

\frac{2^4}{5}  will leave remainder 1

Hence,  \frac{2^{24}}{5} = \frac{2^{4*6}}{5}  will leave remainder 1....

Hence, Answer must be (A) 1...

danbrown9445 · Answer

Alternate way:

12^{24}  
 = (12^{2)12}  
 = 144^{12}  
 = (145-1)^{12}  
 =(-1)^{12}  ......since 145/5 will not leave any remainder
 = 1

BrushMyQuant · Answer

We know to find what is the remainder when  12^{24}  is divided by 5

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

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

Using Above theory Remainder of  12^{24}  by 5 unit's digit of  2^{24}

Now, Let's find the unit's digit of  2^{24}  first.

We can do this by finding the pattern / cycle of unit's digit of power of 3 and then generalizing it.

Unit's digit of  2^1  = 2
Unit's digit of  2^2  = 4
Unit's digit of  2^3  = 8
Unit's digit of  2^4  = 6
Unit's digit of  2^5  = 2

So, unit's digit of power of 2 repeats after every  4^{th}  number.
=> We need to divided 24 by 4 and check what is the remainder
=> 24 divided by 4 gives 0 remainder

=>  2^{24}  will have the same unit's digit as  2^4  = 6
=> Unit's digits of  12^{24}  = 6

But remainder of  12^{24}  by 5 cannot be more than 5
=> Remainder = Remainder of 6 by 5 = 1

So,  Answer will be A 
Hope it helps!

Watch the following video to learn the Basics of Remainders

https://www.youtube.com/watch?v=P0S6j2uTaj4

findingmyself · Answer

12^1 =12 -->Remainer is 2
 12^2 =144---> Remainder is 4
 12^3 =Some number ending is 8, ---> Remainder is 3
 12^4 = Some number ending in 6, Remainer is 1
 12^5 = Some number ending in 12, Remainder is 2

Cyclicity of remainder= 2, 4, 3, 1, 2, 4....
 12^24 --->24/4= Perfectly divisible

12^24  will have remainder 1 ( 12^21  remainder is 2,  12^22- -remainder is 4, 12^23 remainder will be 3)

What is the remainder when 12^24 is divided by 5?

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What is the remainder when 12^24 is divided by 5?

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