# The last two digits

#### Albert

##### Well-known member
n is any odd positive integer

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

#### mathmaniac

##### Active member
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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#### Albert

##### Well-known member
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

#### mathmaniac

##### Active member
Re: the last two digits

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

But can you give a mathematical proof?

#### Bacterius

##### Well-known member
MHB Math Helper
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?

#### mathmaniac

##### Active member
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...

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smart analysis !

#### mathmaniac

##### Active member
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....

#### Albert

##### Well-known member
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

##### Active member
(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-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

#### Albert

##### Well-known member
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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#### mathmaniac

##### Active member
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.