Units Digit

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by goelmohit2002 » Sun Aug 23, 2009 5:43 am
DanaJ wrote:Actually, you should consider that there are some situations where number picking is the only way. This is one situation. There are countless others where you can solve algebraically, but just breaking it down into cases according to x and y is waaay too tedious and not worth it (at least in my head).

Normally, for this type of exercise, you'd use one of the classical rules of units digit:

A number with units digit 1, 5 or 6 will keep its units digit throughout all its >=1 powers.

It's important to remember that rule, so you'll have to check for some <1 powers (as you did with 0.5).

Extra: @wccotton: don't use negative powers when establishing the units digit (unless specifically asked). Most of the time, when we talk about units digits, we only use positive powers. Units digits for non-integers (i.e. 0.5) are usually not discussed.
Thanks Danaj,

Can you please tell....how to quickly find out which approach we should take...number substitution or algebraic....as in Test time most likely we will not have that much liberty to try first method...if that does not work then try second....IMO it will be better if we somehow have fair bit of idea of how to attack a question before hand.....

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by DanaJ » Sun Aug 23, 2009 7:15 am
My guess is that you should save picking numbers for cases when you can offer counterexamples (as you did above). There is no clear cut method of telling when you should go for one way or the other, but I suppose that if the algebraic method doesn't work/seems too difficult, you should go for number picking.

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Re: Units Digit

by umaa » Sun Aug 23, 2009 7:29 am
goelmohit2002 wrote:What is the units' digit of (p^(xy)) ?
1) The units' digit of p^x is 1.
2) The units' digit of p^y is 1.
You can rewrite P^(xy) as (p^x)^y or (p^y)^x

Statement 1:

If (p^x) is 1 then (p^x)^y's unit digit is 1 for any value of y.

eg, (121)^5

Statement 2:

If (p^y) is 1 then (p^y)^x's unit digit is 1 for any value of x.

(81)^3

So, the answer should be D.
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