Problem:
Let n be the number of ordered quadruples (x1,x2,x3,x4) of positive odd integers that satisfy ∑i=14xi=98. Find 100n.
Solution:
Each xi can be replaced by 2yi−1, where yi is a positive integer. Because
98=i=1∑4xi=i=1∑4(2yi−1)=2(i=1∑4yi)−4
it follows that 51=∑i=14yi. Each such quadruple (y1;y2,y3,y4) corresponds in a oneto-one fashion to a row of 51 ones that has been separated into four groups by the insertion of three zeros. For example, (17,5,11,18) corresponds to
11111111111111110111110111111111110111111111111111111.
There are (350)=19600 ways to insert three zeros into the fifty spaces between adjacent ones. Thus there are n=19600 of the requested sums, and 100n=196.
The problems on this page are the property of the MAA's American Mathematics Competitions