Question

find the number of zeroes at the end of the number
${2}^{2} \times {5}^{5} \times {7}^{10000} \times 11$
​

Answer 1

$\boxed{\bold{(i)\:For\:2^2:}}$

We know that,

The number with digit 0 must have prime factors as 2 and 5.

But, the prime factors of 2,

2 = 2 * 1

Therefore, 2^2 = 2 * 1 * 2 * 1 = 4

It means that,

If we insert any value insist of 2 in the power, the factors of that new number will be always 2 and 1.

For example, let's insert 3 in the place of 2,

Then, 2^3 = (2 * 1) (2 * 1) (2 * 1) = 8

As the number of exponent increases the number of pair of ( 2 * 1) will also increases.

Thus, we can see that the prime factors of 2^2 doesn't have 5 as a factor.

According to tje fundamental theorem of Arithmetic, every composite number has a unique solution.

Therefore, the number 2^2 will never end with zero.

$\boxed{\bold{(ii)\:For\:5^5:}}$

We know that,

The number with the digit 0 must have prime factors 2 and 5.

But the prime factors of 5 are,

5 = (5 * 1)

Therefore, 5^5 = (5 * 1) (5 * 1) (5 * 1) (5 * 1) (5 * 1) = 3,125

Thus, we can see that the prime factors of 5^5 doesn't have 2 as a factor.

According th the fundamental theorem of Arithmetic, every composite number has a unique factor.

Therefore, the number 5^5 never ends with 0.

$\boxed{\bold{(iii)\:For\:7^{10000}:}}$

We know that,

The number with digit 0 must have prime factor as 2 and 5.

As the number 10,000 ends with 0 as the prime factors of it will end with 2 and 5.

As the zero is repeated for 5 times, theerfore, number of zeros at the end of 7^10000 will be 5.

$\boxed{\bold{(iv)\:For\:11:}}$

We know that,

The number with the digit 0 must have prime factors 2 and 5.

But the prime factors of 11 are,

11 = 11 * 1

Thus, we can see that the prime factors of 11 doesn't have 5 and 2 as a factor.

According to the fundamental theorem of Arithmetic, every composite number has a unique factor.

Therefore, the number 11 never ends with 0.