Finding units digit

acc100jt · #1 $Old$ November 6th, 2009, 09:08 PM

Hi all,

I got one question here.

Find the units digit in the sum $1^{3}-2^{3}+3^{3}-4^{3}+...+39^{3}$

Thanks!

Bacterius · #2 $Old$ November 6th, 2009, 09:17 PM

Can't you just evaluate it and read the last digit ?

Or it is specifically stated you must do it the analytic way ?
By the way it is not a sum ... well not completely

acc100jt · #3 $Old$ November 6th, 2009, 09:24 PM

I am supposed to find the units digit without evaluating the sum.

There must be hidden pattern, i can't figure it out yet.

Bacterius · #4 $Old$ November 6th, 2009, 09:28 PM

Difference of two cubes, maybe ? You could look at your expression like this :

$(1^3 - 2^3) + (3^3 - 4^3) + ... + (37^3 - 38^3) + 39^3$

Does that make sense ?

Or maybe by considering modular arithmetic modulo $10$ , which would conservate only the last digit ...

acc100jt · #5 $Old$ November 6th, 2009, 09:38 PM

I tried grouping the terms, still couldn't see any pattern.

I used calculator to find the sum, which is 30800. So the units digit is zero.
But how to do this without evaluating the sum?

Debsta · #6 $Old$ November 6th, 2009, 09:41 PM

I'd split it into 1^3 + 3^3 + 5^3 +... 39^3
and 2^3 + 4^3 +6^3 +... 38^3 and then consider the difference.
Write down the units digits for cubes of odd numbers and then for even numbers. You'll see a pattern!

Bacterius · #7 $Old$ November 6th, 2009, 09:46 PM

All good then ... is it possible to user-delete newly irrelevant posts of ours ?

acc100jt · #8 $Old$ November 6th, 2009, 09:50 PM

GREAT! there is a pattern. ThankS

acc100jt · #9 $Old$ November 6th, 2009, 09:53 PM

[quote=Bacterius;399101] $(1^3 - 2^3) = (1 - 2)(1^2 + 2 \times 1 \times 2 + 2^2)$
$= (1 - 2)(1^2 + 4 + 2^2) = (1 - 2)(1 + 4 + 4) = -1 \times 9 = -9$

But $1^{3}-2^{3}=-7$ , not -9.

Bacterius · #10 $Old$ November 6th, 2009, 09:57 PM

Yes, I made a small (but fatal) error, sorry.