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August 26th, 2012, 10:26 PM   #1
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The last two digits







Ans:28
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August 27th, 2012, 09:42 AM   #2
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Re: The last two digits

I just saw this problem, have not thought about it yet, but I will give it try right here:
if we just replace n with m, we get:
(2^(4m+2))(2^(4m+3)-1)
and we know m is a multiple of 0.5 that is greater or equal to 0 because that is the only way n will be a natural number
now if we set 4m+2=2n=x, therefore 4m+3=x+1 we have
2^x(2^(x+1)-1)
=(2^x)(2^x*2-1)

since n is a natural number, x must be a natural number that is even
so:
2^2*(2^3-1)=28
2^4*(2^5-1)=496
2^6*(2^7-1)=8128
2^8(2^9-1)=130816
2^10(2^11-1)=...28

I have a question now, when n=2, the last two digits are 96, could you please check your problem? Or did I do something wrong?
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August 27th, 2012, 05:01 PM   #3
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Re: The last two digits

wuzhe :



so n is an odd number, not even numbers
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August 27th, 2012, 06:15 PM   #4
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Re: The last two digits

But I thought m could be 0.5 or 1.5, did I miss something basic?
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August 27th, 2012, 06:58 PM   #5
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Re: The last two digits

Quote:
wuzhe wrote :But I thought m could be 0.5 or 1.5, did I miss something basic?
Sorry! I did not make it clear , where m is a non-negative integer

I have edited the original post
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August 27th, 2012, 07:02 PM   #6
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Re: The last two digits

If m is to be a non-negative integer, then you could use or .

is used to denote the integers.

is used to denote the natural numbers.

is used to denote the natural numbers and includes zero as well, i.e., the whole numbers.
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August 27th, 2012, 07:06 PM   #7
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Re: The last two digits

MarkFL : Thanks
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August 27th, 2012, 07:30 PM   #8
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Re: The last two digits

Another off-topic suggestion:

To use to display text, use the text command:

The code \text{ your text here } produces

This way, your text is not italicized like variables.
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August 27th, 2012, 07:39 PM   #9
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Re: The last two digits

Hint :to prove this, one of the possible solutions

you may use Mathematical Induction (MI), please try it

I will post using another method later.
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