Problem: Which of the following is satisfied by all numbers x of the form
x=3a1ββ+32a2ββ+β¦+325a25ββ
where a1β is 0 or 2,a2β is 0 or 2,β¦,a25β is 0 or 2?
Answer Choices:
A. 0β©½x<1/3
B. 1/3β©½x<2/3
C. 2/3β©½x<1
D. 0β©½x<1/3 or 2/3β©½x<1
E. 1/2β©½xβ©½3/4
Solution:
Using the formula for the sum of a geometric series
0β€x<n=1βββ(32β)n=1β31β32ββ=1
If a1β=0, then
0β€x<n=2βββ(32β)n=1β31β92ββ=31β
If Ξ±1β=2, then
32β+0+0+β―β€x<1
So either 0β€x<31β or 32ββ€x<1.