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unit digit problem

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by nafiul9090 » Thu Mar 31, 2011 10:02 am
If s^n= 4^n+ 5^n+l+ 3, what is the units digit of s^100?

this is mgmat problem....but the explanation doesnt satisfy me. there must be a short cut.....help needed
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by Spike142 » Thu Mar 31, 2011 11:29 am
Working the problem out right now, but cannot seem to comprehend what you have written here, is it:
1)s^n = 4^n + 5^(n+L)+3?
2)s^n = 4^n + 5^(n+1) + 3?
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by emalby » Thu Mar 31, 2011 12:31 pm
If the question is s^n = 4^n + 5^(n+1) + 3?

Then:

- 4^n gives a sequence 4, 16, 64, ... so jumps with units from 4 to 6 to 4 to 6 etc... with 6 when exponent (n) is even and 4 when n is odd. So in our case since n is 100 (even) it is equal to 6.
- 5^(n+1) gives always a unit digit equal to 5 since 25, 125, 625, etc...
- 3 is equal to 3.

So the sum is 6+5+3 = 14. So the unit digit is 4.

OK?
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by force5 » Thu Mar 31, 2011 1:41 pm
If s^n= 4^n+ 5^n+l+ 3, what is the units digit of s^100?

this is mgmat problem....but the explanation doesnt satisfy me. there must be a short cut.....help needed
if i consider s^n= 4^n+5^(n+1)+3

when n= even (as 100 is even) = 6+5+3 (4^even will have unit digit 6)
hence unit digit of s^100 = 4
please edit your question if there is a change..........
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