When the expression (7^1)(7^2)(7^3) ... (7^100) is written as an integer, what is the product of the tens digit and the ones digit?
productfor(n, 1, 100, 7^n) % 10^10 = 6696251249 - These are the last 10 digits.
Product =4 x 9 = 36
+397 Working to mod 100 throughout,
\(\displaystyle 7 \equiv 7, \\ 7^{2} \equiv 49\dots(1),\\ 7^{3} = 343 \equiv 43,\\ 7^{4}=2401 \equiv 1 \dots (2). \)
\(\displaystyle 7^{1}.7^{2}.7^{3}.\dots7^{100}=7^{5050}=7^{5048}.7^{2}=(7^{4})^{1262}.7^{2}.\)
Using (1) and (2) above,
\(\displaystyle (7^{4})^{1262}.7^{2} \equiv1^{1262}.49 \equiv 49.\)
So the final two digits are 49
