Question

Find the ${202}^{nd}$ digit from the right in the product of $4!\times 5!\times 6!...............71!?$

• A. 1
• B. 2
• C. 3
• D. 0
• E. 4

Answer 1

The correct option is D 0

Note that 71! has 16 zeroes at its end.(You can find this out by finding the highest power of 5 in 71!)

So does 70!.

66! To 69! Have 15 zeroes at its end. Multiplication of these itself amounts to $>202$

Hence the ${202}^{nd}$ digit is 0

To find the highest power of a number in a factorial

a)Highest power of a prime number in a factorial:

To find the highest power of a prime number (x) in a factorial (N!), continuously divide N by x and add all the quotients.

Eg) Find the highest power of 100!

Solution:

$\frac{100}{5}=20;\frac{20}{5}=4;$

Adding the quotients, its 20+4=24. So highest power of 5 in 100! = 24

b)Highest number of a composite number in factorial

1)Factorize the number into primes.

2)Find the highest power of all the prime numbers in that factorial using the previous method.

3)Take the least power.