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

benjamingu22  Aug 21, 2017
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4+0 Answers

 #1
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It is 8

Guest Aug 21, 2017
 #2
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+1

 

Note that (21) (22) (23) (24)......(299)(2100)   can be rearranged as

 

(21)(2100) * ( 22)(299)  * ( 23)(298)  =

 

(2101)50   =  

 

(250)101 =

 

( [(2^10)]^5)^101

 

Note that  2^10  ends in 24

 

And 24^(2n + 1)  ends in 24 for n = any whole number......so we can write

 

[( ......24)^5]^101  =

 

[ (......24)^(2(2) + 1) ]^101  =

 

[(.....24) ] ^(2(50) + 1)  =

 

24

 

And the product of the tens digit and  ones digit  = 8

 

 

cool cool cool

CPhill  Aug 21, 2017
edited by CPhill  Aug 21, 2017
 #3
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These exponents (2^1) (2^2) (2^3) (2^4)......(2^99)(2^100) can be written thus:

 

2^(100*101/2) = 2^5050. This, in turn, can be split into: 2^5045 x 2^5. And since 2 to any exponent ending in 5 will result ending in 32. Therefore, 2^5045 x 2^5 =.......32 x 32 =1,024, or......24. And their product = 2 x 4 = 8.

Guest Aug 22, 2017
 #4
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No, not true !!!.  2^15 =32,768 !!!.

2^1 =2

2^2 =4

2^3 =8

2^4 =16

2^5 =32

2^6 =64

2^7 =128

2^8 =256

2^9 =512

2^10 =1024 .........etc. Do you see any pattern??

Guest Aug 22, 2017

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