Question

will the number
${6}^{m} \times {25}^{n}$
,where m<n , have zeroes at the end. if yes,how much​

Answer 1

$\boxed{\bold{For\:6^m:}}$

We know that,

The number which ends with 0 must have its prime factorisation with number 2 and 5.

But the prime factors of 6 are,

6 = 2 * 3

$\therefore 6^m = (2 \times 3)^m$

It means that,

If we insert any number instead of m the same numbers 2 and 3 will repeat.

As we take m = 2

Then, $6^m = (2 \times 3)^m$

=> 6^2 = 2 * 3 * 2 * 3 = 36

If we take m = 3

Then, 6^3 = 2 * 3 * 2 * 3 * 2 * 3 = 216

Here, the same numbers will repeat as we increase the value of m.

Thus, we can see that the prime factors of $6^m$ doesn't have 5 as a factor.

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

Therefore, $6^m$ never ends with the digit 0.

$\boxed{\bold{Now,\:for\:25^n:}}$

We know that,

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

But, the prime factors of 25 are,

$\therefore 25^n = (5 \times 5)^n$

Thus, we can see that the prime factors of 25ⁿ doesn't have 2 has a factor.

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

Therefore, 25ⁿ never ends with 0.

As both the numbers doesn't ends with 0, there product too will not end with 0.

Therefore, the number $\bold{6^m}$ and 25ⁿ never ends with 0.