As promised to a certain few (i.e. <<1):
The numbers p, q, r, s, and t are consecutive positive integers, arranged in increasing order. If p + q + r + s + t is a perfect cube and q + r + s is a perfect square, find the smallest possible value for r.
The sum of five positive integers x, y, z, u and v is equal to their product. If x ≤ y ≤ z ≤ u ≤ v, the how many distinct solutions (x,y,z,u,v) are there?
There you go Mr Full Marks For Common :O
EDITL: zomfgbbq how did you get it that fast? o.O I still don't get these Diaphantine equations...
I don't know if this is the smallest but for the first one, r = 675 works. Dunno about second.
ReplyDeletewell to explain JM's hyperspeed calculation it goes like this:
ReplyDelete1st of all p+q+r+s+t is a perfect cube. Hence the other numerical factors must come in a cube. Whilst having a factor of 5.
2nd q+r+s is a perfect square. Hence factors must come in threes. Also the other numerical factores must come in squares
Hence: 3^3x5^2 = 675