Question

If a = 1+2+3+4+ ... +n, and
b = 1+2^3+3^3+...+n^3
Find n, if gcf(a,b)+lcm(a,b)=4422

Answer 1

use sum of first n numbers, sum of cubes formulas

Answer 2

b has the curious property of being a squared. a is equal to \[(n^2+n)/2\]because there are n values averaging (n+1)/2, and n times that gives the equation above. b should therefore equal \[(n^4+2n^3+n^2)/4\]We can prove this by showing the relation for 1 and for n+1: it is clearly true for one, and for (n+1) we have: \[b=\sum_{1}^{n}n^3= ((n+1)^4+2(n+1)^3+(n+1)^2)/4=\]\[(n^4+4n^3+6n^2+4n+1+2(n^3+3n^2+3n+1)+n^2+2n+1)/4=\]\[(n^4+2n^3+n^2)/4+(4n^3+12n^2+12n+4)/4\]
Notice that our original term (n^3, expressed according to our first formula) remains the same while on the right we have \[n^3+3n^2+3n+1=(n+1)^3\]Therefore b=a^2
Q.E.D.
From that you have a very simple polynomial (think about what happens with the gcf and lcd of a and a^2) to solve.

Answer 3

Ok, now i know that b=a^2 and a = (n^2+n)/2
So that means that gcf(a,b) = a and lcm(a,b)=a^2
then
a+a^2 = 4422
a^2+a-4422=0
(a-66)(a+67)

so a = 66, because it's possitive and
(n^2+n)/2=66
n^2+n-132 = 0
(n-11)(n+12)=0

Therefore n = 11 !!!

Thanks.

Answer 4

Nice!

Answer 5

Oh, you should find the value of b as well, just for fun. I'd do it as a sum of cubes, but if you want to be boring you can take 66^2.

Answer 6

It's fun to watch b equaling a^2 all of the way up.

Answer 7

I better memorize that formulas , because the demonstration may be fun, but seems laborious.

Answer 8

There is also something about Pascal's triangle and sums of powers of n...