Problem:
For how many ordered triples (x,y,z) of nonnegative integers less than 20 are there exactly two distinct elements in the set {ix,(1+i)y,z}, where i=β1β ?
Answer Choices:
A. 149
B. 205
C. 215
D. 225
E. 235 Solution:
There are three cases to consider.
First, suppose that ix=(1+i)yξ =z. Note that β£ixβ£=1 for all x, and β£(1+i)yβ£β₯β£1+iβ£=2β>1 for yβ₯1. If y=0, then (1+i)y=1=ix if x is a multiple of 4. The ordered triples that satisfy this condition are (4k,0,z) for 0β€kβ€4 and 0β€zβ€19,zξ =1. There are 5β 19=95 such triples.
Next, suppose that ix=zξ =(1+i)y. The only nonnegative integer value of ix is 1 , which is assumed when x=4k for 0β€kβ€4. In this case ix=1 and yξ =0. The ordered triples that satisfy this condition are (4k,y,1) for 0β€kβ€4 and 1β€yβ€19. There are 5β 19=95 such triples.
Finally, suppose that (1+i)y=zξ =ix. Note that (1+i)2=2i, so (1+i)y is a positive integer only when y is a multiple of 8 . Because (1+i)0=1, (1+i)8=(2i)4=16, and (1+i)16=162=256, the only possible ordered\
triples are (x,0,1) with xξ =4k for 0β€kβ€4 and (x,8,16) for any x. There are 15+20=35 such triples.
The total number of ordered triples that satisfy the given conditions is 95+95+35=225β.