+0

# number theory

0
53
3

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?

Oct 27, 2020

#1
0

productfor(n, 1, 100, 7^n) % 10^10 = 6696251249 - These are the last 10 digits.

Product =4  x  9 = 36

Oct 27, 2020
#2
+146
+1

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

Oct 28, 2020
#3
+111600
0

That is a really cool technique.  Thanks Tiggsy.

Nov 10, 2020