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

 

Product =4  x  9 = 36

 Oct 27, 2020
 #2
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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
avatar+111600 
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That is a really cool technique.  Thanks Tiggsy.

 Nov 10, 2020

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