Fools problem of the day:

Find the last two digits of

a) $22^3+24^3+26^3+28^3 $
b) $ 23^3+24^3+25^3+26^3+27^3 $

PS:This may be too easy.

Question

Fools problem of the day:

Find the last two digits of

a) $22^3+24^3+26^3+28^3 $
b) $ 23^3+24^3+25^3+26^3+27^3 $

PS:This  may be too easy.

zarkon · Answer

it is ;)

across · Answer

Gotta love modular arithmetic. ;)

turingtest · Answer

I can find all the last digit easy enough, but the other one...
still processing the stuff I tried to learn yesterday

anonymous · Answer

If this is just too easy: http://openstudy.com/study#/updates/4f5c5fc7e4b0602be43841dc But this ain't hard either.

anonymous · Answer

@across: I think we can solve these using just algebraic manipulations.

anonymous · Answer

Everyone just said easy, too easy and expressed their love for math but none solved it. Now someone please post the solution.

anonymous · Answer

$$ 22^3+24^3+26^3+28^3 = (22^3+28^3)+(24^3+26^3) $$
$$ = 50 \times \left( ( 22^2+28^2+22\times 28) + (24^2+26^2+24\times 26) \right) $$

It's easy to notice that the above is divisible by 100. Hence the last two digits are 00.

Can you do the second one?