Question

find number of zeroes in product of 901*902*903............................999*1000

Answer 1

Question:

Find the no. of zeroes at the end of the product of 901 × 902 × 903 ×......× 999 × 1000.

Answer:

25

Step-by-step explanation:

⇒ First we have to find how many multiples of 5¹ = 5 are there between 901 and 1000.

• Least multiple of 5 after 901 = 905.
• Largest multiple of 5 before 1000 = 1000 itself.
• Let the multiples are in AP, such that, a = 905, a_n = 1000, d = 5.
• n = (a_n - a)/d + 1 = (1000 - 905)/5 + 1 = 95/5 + 1 = 19 + 1 = 20

Thus there are 20 multiples, so 20 zeroes are there.

⇒ Now we shall find how many multiples of 5² = 25 are there between 901 and 1000.

There are 4 multiples of 25, and those are 925, 950, 975 and 1000. Thus 4 more zeroes are there.

⇒ Now we have to find the no. of multiples of 5³ = 125 between 901 and 1000.

There is 1 multiple of 125, and i.e., 1000. So 1 more zero is there.

⇒ There is no multiples of 5⁴ = 625 between 901 and 1000. So we can conclude the explanation with the no. of multiples of 125.

⇒ Thus the no. of zeroes at the end of the product is 20 + 4 + 1 = 25.

Answer 2

Above is your answer.

@GENIUS