Last 2 Digits of $2^{2n}(2^{2n+1}-1)$

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Discussion Overview

The discussion revolves around determining the last two digits of the expression $2^{2n}(2^{2n+1}-1)$, specifically for odd positive integers n. Participants explore various mathematical approaches and reasoning to arrive at a solution.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests that the last digit could be either 8 or 4, raising questions about the second last digit's possibilities.
  • Another participant calculates the result for n=1 as 28 and proposes that this might hold for all odd positive integers.
  • A participant challenges others to provide a mathematical proof for the conjecture that the last two digits are consistently 28.
  • One participant analyzes the last digits of $2^{2n}$ and $2^{2n+1}-1$, concluding that they lead to a consistent pattern yielding 28.
  • Further mathematical manipulations are presented, suggesting that the expression can be factored in a way that confirms the last two digits are 28.
  • Another participant reflects on the importance of perseverance in problem-solving, emphasizing the value of sticking to a method despite challenges.

Areas of Agreement / Disagreement

While some participants express confidence in the result being 28, the discussion includes various approaches and reasoning, indicating that multiple viewpoints and methods are being explored without a definitive consensus on the proof's validity.

Contextual Notes

The discussion involves various mathematical assumptions and manipulations that may not be fully resolved, particularly regarding the conditions under which the last two digits are determined.

Albert1
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n is any odd positive integer

please find the last two digits of :$2^{2n}(2^{2n+1}-1)$
 
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Re: the last two digits

Albert,I got the last digit to be 8 or 4 .But are you sure the second last also has one or two possibilities only?

I think if the last one is 2,then the second last is one among (1,3,5,7,9) and if the last is 4 then second last is one among (0,4,2,4,6,8)...

Correct me if I am wrong...
 
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Re: the last two digits

if n=1 the result is 28

since it holds for all "odd" positive integers

we may guess the last two digits will be 28

now you have to prove that for n=3,5,7,9----

this is also true
 
Re: the last two digits

OOPS!I thought about 2n and 2n+1 separately...

But can you give a mathematical proof?
 
Re: the last two digits

It's a challenge, we're supposed to figure it out ourselves. Mathmaniac, I suggest expanding the product into a difference of powers of 2, and looking at the sequence of last two digits for each term separately and notice how you always get 28. Can you see a pattern, and prove it?
 
Yes,I got it...
$$2^{2n}$$ for all n has the last digit 4 always and the second last any even number(otherwise it will not be divisible by 4).The last digit of $$2^{2n+1} - 1$$ is also always 7 with the second last digit any even number but it has to be double of the second last digit of $$2^{2n}$$.It turns out that all possible combinations give 28 as the last digit...
 
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smart analysis !
 
In the end we would be proving A4 X (2A)7
or A4 X (2A-10)7 = ...28

where A,4,(2A),(2a-1) and 7 are digits...
 
mathmaniac said:
In the end we would be proving A4 X (2A)7
or A4 X (2A-10)7 = ...28

where A,4,(2A),(2a-1) and 7 are digits...
yes please go ahead
 
  • #10
(10A+4) X (20A+7)=

200A^2+150A+28

Surely 200A^2 cannot go into the last two digits,but 150 can go for 1,3,5,7,9 but we are concerned with even numbers only.So the last two digits are 28...

Now (10A+4)(20A-100+7)=(10A+4)(20A-93)
=200A^2-850A-372850A will give a number with 2 zeros (for even A) so will 200A^2.So it becomes ...00 - 372
which will surely end in 28
 
  • #11
Albert said:
n is any odd positive integer
please find the last two digits of :$2^{2n}(2^{2n+1}-1)$

let n=2m+1 (m=0,1,2,3,4,5----)

$S=2^{2n}(2^{2n+1}-1)-28$

$=2^{4m+2}(2^{4m+3}-1)-28$

=$4($$2^{8m+3}-2^{4m}-7)$

=$4[8$($2^{4m})^2-2^{4m}-7]$

=$4(2^{4m}-1)(8\times2^{4m}+7)$

=$4(16^m-1)[8\times(16^m-1)+15]$

S is a multiple of 100

so the proof is done ,the last two digits is 28
 
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  • #12
Perfect! I never thought it could be done this way.

I realize that in some problems of proving something all you need to have is faith to stick with your method no matter how long it goes.I have abandoned many problems after things felt slipping.
 

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