M10 #36 : Retired Discussions [Locked]
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# M10 #36

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19 Sep 2010, 08:44
This question seems odd a little bit.

We have a range of figures expressed $$(10^n-1)^n$$, where we are given first three elements n=1, n=2, n=3.
Then I am asked what is the last digit of given range of figures.
in 2) it is mentioned that n is prime, I know that all primes are odd except for 2. But I see that n=2 is in range and a set $$(10^n-1)^n$$ has already n=1, n=2, n=3, so last figure can not contain $$(10^2-1)^2$$. Additional n=2 misleads.

Don't you think that for clarity we must add something like " n can be any number" or something like this.
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23 Sep 2010, 10:08
This question seems odd a little bit.

We have a range of figures expressed , where we are given first three elements n=1, n=2, n=3.
Then I am asked what is the last digit of given range of figures.
in 2) it is mentioned that n is prime, I know that all primes are odd except for 2. But I see that n=2 is in range and a set has already n=1, n=2, n=3, so last figure can not contain . Additional n=2 misleads.

Don't you think that for clarity we must add something like " n can be any number" or something like this

Here are my 2 cents-
A Range is given..and first 3 elements of the range also you can find out (given n1=1 n2=2 n3=3).So at least the range has 3 elements, otherwise if the range is bigger, for other elements of the range, n is a prime no and that is bigger than 3.

now $$(10^n-1= has 9 at the units place always$$. So $$9^(any odd prime digit)$$ will have 1 at its units place..hope i made sense
Re: M10 #36   [#permalink] 23 Sep 2010, 10:08
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# M10 #36

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