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Source: — Problem Solving |
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by pepeprepa » Fri Aug 01, 2008 2:15 am
2^1: unit digit is 2
2^2: unit digit is 4
2^3: unit digit is 8
2^4: unit digit is 6
(2^5: unit digit is 2, and you can see that the same set of 2-4-8-6 comes back all the time)

And 2^(8n+6) has the same digit as 2^2
because 2^(8*0+6) has a unit digit of 4, the same for 2^(8*1+6) ....

So we have unit digit 4 for 2^(8n+6) and we had 7 so it gives 1

I hope I am clear but as soon as you have the method it is fast
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by raunekk » Fri Aug 01, 2008 3:50 am
2^1: unit digit is 2
2^2: unit digit is 4
2^3: unit digit is 8
2^4: unit digit is 6
So the units digit repeats after every four

put n=1

thus 7+ 2^14..

( if we take the cyclic order 2^14 will have 4 as the unit place..)

thus 7+4=11

Thus units digit is 1..hence A
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by coffee5251 » Tue Jan 06, 2009 12:17 pm
put n=1

thus 7+ 2^14..

( if we take the cyclic order 2^14 will have 4 as the unit place..)
How do you know to use n=1? and then to use 2^14??

Thanks
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by sonu_thekool » Tue Jan 06, 2009 1:27 pm
coffee5251 wrote:
put n=1

thus 7+ 2^14..

( if we take the cyclic order 2^14 will have 4 as the unit place..)
How do you know to use n=1? and then to use 2^14??

Thanks
Since the question says n is a positive number, the first positive number is 1 and when n=1, the exponent is 8*1 + 6 = 14.

So, we try to find 7 + 2^14.

First, as one earlier post suggested, we need to find a sequence here...

2^1 = 2 (ends in 2)
2^2 = 4 (ends in 4)
2^3 = 8 (ends in 8)
2^4 = 16 (ends in 6)
2^5 = 32 (ends in 2)
2^6 = 64 (ends in 4)
...

If you see above, the sequence of digits in the units place is always 2, 4, 8, 6 and it repeats again with 2, 4, 8 and 6 as the exponent increases. So, the 14th term is 3 sets of this sequence (3 * 4 terms) plus second element, which is 4.

7 + 4 = 11 and so the units place of 11 is 1.

Hope this makes it clearer....thanks to all earlier posts...
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