Find the last non zero digit of 96!a)2b)4c)8d)6Correct answer is option 'D'. Can you explain this answer?

Question

Answer

The provided answer is wrong..The correct answer should be 6.
The correct procedure of solving this involves:
Suppose we have to calculate the factorial of 'N'
so the formula for last non-zero unit's digit is:
2^a * a!*b! where a is the quotient obtained when 'N' is divided by 5 and b is the remainder when 'N' is divided by 5.
So following the above formula we have-
96!= 2^19 * 19!* 1! which gives unit digit as 6.

{Note- Here you can use the above mentioned formula for calculating 19! and 2^19 can be calculated with the help of cyclicity which comes out as 2^3 i.e 8}

HAPPY TO HELP!

Answer

Main formula=2 ki power A multiply A! multiply B!

96!= 96/5= quotient A=19, Remainder B=1
Then, 2 ki power 19= unit digit is 8
A!= 19! = unit digit is 2
B!= 1!= unit digit is 1
hence,
8×2×1= 16 then unit digit is 6 answer.

Answer

It should be (c) .

Answer

Solution:

To find the last non zero digit of 96!, we need to find the last non zero digit of its factorials. We can use the concept of prime factorization of factorial to find the same.

Prime Factorization of 96!:

- We can write 96! as product of prime factors.
- Prime factorization of a number is the expression of that number as a product of prime factors.
- To find the prime factorization of a number, we keep dividing it by the smallest prime number until the quotient is no longer divisible by that prime number.
- Then we continue this process with the next smallest prime number until the quotient is 1.
- The prime factorization of a number is the product of all the prime factors.

Prime factorization of 96! can be found using the following steps:

- Divide 96 by the smallest prime number, which is 2. We get 48 as quotient and 2 as factor.
- Divide 48 by 2. We get 24 as quotient and 2 as factor.
- Divide 24 by 2. We get 12 as quotient and 2 as factor.
- Divide 12 by 2. We get 6 as quotient and 2 as factor.
- Divide 6 by 2. We get 3 as quotient and 2 as factor.
- Divide 3 by the smallest prime number, which is 3. We get 1 as quotient and 3 as factor.
- Prime factorization of 96! is 2^92 x 3^38.

Last Non Zero Digit of 96!:

- To find the last non zero digit of 96!, we need to find the last non zero digit of 2^92 x 3^38.
- The last non zero digit of a number is the same as the last non zero digit of its prime factorization.
- For 2, the last non zero digit is 2. For 3, the last non zero digit is 3.
- To find the last non zero digit of 2^92 x 3^38, we need to find the last non zero digit of 2^92 and 3^38 separately.
- The last non zero digit of 2^92 can be found by looking at the pattern of last digits of powers of 2.

- The last digit of 2^1 is 2.
- The last digit of 2^2 is 4.
- The last digit of 2^3 is 8.
- The last digit of 2^4 is 6.
- The pattern repeats after every 4 powers of 2.

- Therefore, the last non zero digit of 2^92 will be the same as the last non zero digit of 2^4, which is 6.

- The last non zero digit of 3^38 can be found by looking at the pattern of last digits of powers of 3.

- The last digit of 3^1 is 3.
- The last digit of 3^2 is 9.
- The last digit of 3^3 is 7.
- The last digit of 3^4 is 1.
- The pattern repeats after every 4 powers of 3.

- Therefore, the last non zero digit of 3^38 will be the same as the last non zero digit of 3^2

Answer

Computing 96!
then
4*z(96/5)*z(6)
=4*z(19)*8
=8*(4*z(19/5)*z(9))
=8*(4*z(3)*8)
=8*(4*6*8)
=1536 i.e last digit is 6

Answer

Answer-D

Answer

Correct answer is option 'D''

Similar Quant Doubts

Let S be a two-digit number such that both S and S2 end with the same digit and none of the digits in S equals zero. When the digits of S are written in the reverse order, the square of the new number so obtained has the last digit as 6 and is less than 3000. How many values of S are possible?

Find the remainder when 4^96 is divided by 6.

The sum of the digits of a two-digit number is 6. If the digits are reversed, the number is decreased by 36. Find the number?

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