Problem:
Let n be the number of ordered quadruples (x1​,x2​,x3​,x4​) of positive odd integers that satisfy ∑i=14​xi​=98. Find 100n​.
Solution:
Each xi​ can be replaced by 2yi​−1, where yi​ is a positive integer. Because
98=i=1∑4​xi​=i=1∑4​(2yi​−1)=2(i=1∑4​yi​)−4
it follows that 51=∑i=14​yi​. 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