A) 2
B) 3
C) No value
D) 5
Correct Answer: C
Solution :
| If any number ends with the digit 0 or 5, it is always divisible by 5. If \[{{12}^{n}}\]ends with Adie digit zero it must be divisible by 5. This is possible only if prime factorization of \[{{12}^{n}}\]contains the prime number 5. Now, |
| \[12=2\times 2\times 3={{2}^{2}}\times 3\] |
| \[\Rightarrow \,\,{{12}^{n}}={{\left( {{2}^{2}}\times 3 \right)}^{n}}={{2}^{2n}}\times {{3}^{n}}\] |
| [since, there is no term contains 5] Hence, there is no value of n e N for which \[{{12}^{n}}\]ends with digit zero or five. |
You need to login to perform this action.
You will be redirected in
3 sec
