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Intern
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28 Sep 2005, 04:31
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Hi
antmavel mentioned this example in one of the previous discussions but he only mentioned it as an example to follow a certain approach to solve. i can`t find the way how to solve this one in a fast way. pls help. thx

47^10-45^8
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28 Sep 2005, 04:51
Good question !

47^10 - 45^8 = (47^10 - 45^10) + (45^10 - 45^8)

The best way to eat an elephant is bit by bit.

45^10 - 45^8 = 45^8 (45^2 - 1) = 45^8 * (2025-1)

Note special case of (x+y)(x-y)=x^2-y^2 when y=5
45^2 = 40 * 50 + 25 = 2025

47^10 - 45^10 = 45^10 (10^2 - 1) = 99 * 45^10
= 99 * 2025 *45^8
= 200,475 * 45^8

47^10 - 45^8 = (200,475 + 2,025 - 1) * 45^8
= 202,499 * 45^8

Not sure I plan to expand this any more.
I think it was a slightly flippant example by antmavel.
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28 Sep 2005, 09:52
richardj wrote:
Good question !

47^10 - 45^10 = 45^10 (10^2 - 1) = 99 * 45^10
= 99 * 2025 *45^8
= 200,475 * 45^8

can you explain bold part, please.
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28 Sep 2005, 14:25
Oops, I think I wrote complete rubbish !
Manager
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28 Sep 2005, 23:54
I think it would take some time to calculate that...

I also think that antmavel refered to calculating the last digit, not the total result.

For the last digit all we need is:
1 7^1=7
2 7*7=49
3 9*7=63
4 3*7=21
5 1*7=7
.... we have a cycle here so:

10mod4 = 2 thus, the last digit of 7^10=9

Last digit = 9-5=4
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