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# When the expression $(2^1)(2^2)(2^3)\cdots (2^{99})(2^{100})$ is written as an integer, what is the product of the tens digit and the ones d

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When the expression $$(2^1)(2^2)(2^3)\cdots (2^{99})(2^{100})$$ is written as an integer, what is the product of the tens digit and the ones digit?

Feb 3, 2018

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2^1 x 2^2 x2^3 x.........x 2^99 X 2^100 =2^(1+2+3+.......+99+100) =2^(100 x 101) / 2 =2^5050.

Any power of 2 ending in: 10, 30, 50, 70, 90 will have the last 2 digits as =.....24.

Feb 3, 2018
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When the expression  $$(2^1)(2^2)(2^3)\cdots (2^{99})(2^{100})$$    is written as an integer,

what is the product of the tens digit and the ones digit?

1+2+3+.....100 = 100/2(1+100)= 50*101 = 5050

so this is

$$2^{5050}=(2^{25*202})\\ =((2^{25})^{202})\\ \text{now only considering the last two digits}\\ \equiv (32)^{202}\\ \equiv (32)^{101*2}\\ \equiv (32^2)^{101}\\ \equiv (24)^{101}\\ \equiv 24*(24)^{100}\\ \equiv 24*((24)^{5})^{20}\\ \equiv 24*(24)^{5*4}\\ \equiv 24*24^4\\ \equiv 24$$

This is probably not the best way to do it though :/

Feb 3, 2018
edited by Melody  Feb 3, 2018
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(2^1) (2^2) (2^3)......(2^99) (2^100)  =

2^(1 + 2 + 3 + ...+  99 + 100)  =

2^(5050)  =

(2^10)^505

Note that 2^10    ends in 24

And note the pattern of 24

24^5 ends in 24

24^10  ends in 76

24^15  ends in 24  ....etc....

So......  (....24)^[5 (2n - 1) ]    ends   in 24....when n is a positive integer

So

(2^10)^505    =

(.....24)^505  =

(  ...24 ) ^[ 5 (2 * 51 - 1) ]   .......ends in 24

Feb 3, 2018
edited by CPhill  Feb 3, 2018