- Thread starter
- #1

#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

n is any odd positive integer

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

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

- Thread starter Albert
- Start date

- Thread starter
- #1

- Jan 25, 2013

- 1,225

n is any odd positive integer

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

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

- Mar 4, 2013

- 188

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...

Last edited:

- Thread starter
- #3

- Jan 25, 2013

- 1,225

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

- Mar 4, 2013

- 188

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

But can you give a mathematical proof?

- Jan 26, 2012

- 644

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?

- Mar 4, 2013

- 188

Yes,I got it...

\(\displaystyle 2^{2n}\) 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 \(\displaystyle 2^{2n+1} - 1\) is also always 7 with the second last digit any even number but it has to be double of the second last digit of \(\displaystyle 2^{2n}\).It turns out that all possible combinations give 28 as the last digit...

\(\displaystyle 2^{2n}\) 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 \(\displaystyle 2^{2n+1} - 1\) is also always 7 with the second last digit any even number but it has to be double of the second last digit of \(\displaystyle 2^{2n}\).It turns out that all possible combinations give 28 as the last digit...

Last edited by a moderator:

- Thread starter
- #7

- Jan 25, 2013

- 1,225

smart analysis !

- Mar 4, 2013

- 188

or A4 X (2A-10)7 = ....28

where A,4,(2A),(2a-1) and 7 are digits....

- Thread starter
- #9

- Jan 25, 2013

- 1,225

yes please go ahead

or A4 X (2A-10)7 = ....28

where A,4,(2A),(2a-1) and 7 are digits....

- Mar 4, 2013

- 188

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-372

850A 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

- Thread starter
- #11

- Jan 25, 2013

- 1,225

let n=2m+1 (m=0,1,2,3,4,5----)n is any odd positive integer

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

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

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

=$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

Last edited:

- Mar 4, 2013

- 188

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.