Question

the least value of natural number n for wich the number (2☆2☆2☆3☆75)^n ends with the digit 0 is 1) 1 2) 3 3) such value of n can't be found 4) 2 ​

Answer 1

Given: the number $(2\times 2\times 2\times 3\times 75)^{n}$

To find: the least value of natural number $n$ for which the given number ends with the digit $0$

Solution:

• Given number is $(2\times 2\times 2\times 3\times 75)^{n}$

• $=1800^{n}$

• We see that the number inside the exponent already ends with the digit $0$ and so we can say that when $n$ has the least value $n=1,$ the number given surely will end with $0$ since $1800^{1}=1800,$ ends with $0.$

Answer: Option 1) 1 is correct.