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

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

\(12 = \frac{(10 + 2)}{5}\) ( 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

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

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

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

\(\frac{10}{5}\) 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...

Alternate way:

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

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

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