Question

Find the last two digits of the number $(17{)}^{10}$.

• A. 49
• B. 43
• C. 83
• D. 81

Answer 1

The correct option is A

49

We want to find the last two digits of the number $(17{)}^{10}$. If we can express it in terms of a number which is a multiple of 10 or 100, it will be easy to find the last two digits. We know $(17{)}^2$ = 289 = 290-1. We can start with this

$(17{)}^{10}={289}^5=(290-1{)}^5$

$(290-1{)}^5$ = $\left({}^5{C}_0\right(290{)}^5-{}^5{C}_1(290{)}^4+{}^5{C}_2(290{)}^3............+{}^5{C}_4(290{)}^1-{}^5{C}_5)$

$(290-1{)}^5$ = $\left({}^5{C}_0\right(290{)}^5-{}^5{C}_1(290{)}^4+{}^5{C}_2(290{)}^3............+{}^5{C}_4(290{)}^1-1)$

$(290-1{)}^5$ = (A multiple of 1000) + $5(290{)}^1-1$

$(290-1{)}^5$ = (A multiple of 1000) + 1449

$\Rightarrow$ The last three digits are 449