Problem:
Which of the following is the correct order of the fractions 1115β, 1519β, 1317β, from least to greatest?
Answer Choices:
A. 1115β<1317β<1519β
B. 1115β<1519β<1317β
C. 1317β<1519β<1115β
D. 1519β<1115β<1317β
E. 1519β<1317β<1115β
Solution:
To determine the order of 1519β and 1317β, rewrite the fractions using a common
denominator: 15β
1319β
13β and 13β
1517β
15β. Because
19β
13=(16+3)(16β3)=162β32
17β
15=(16+1)(16β1)=162β12
and 162β32<162β12 , it follows that 1519β < 1317β.
Similarly, to determine the order of 1317β and 1115β, rewrite the fractions using a common denomi-nator:13β
1117β
11β and 11β
1315β
13β. Because
17β
11=(14+3)(14β3)=142β32,
15β
13=(14+1)(14β1)=142β12,
and 142β32<142β12 , it follows that 1317β < 1115β.
OR
Subtracting 1 from each fraction results in the fractions 114β,154β, and 134β. Because 154β < 134β < 114β it follows that 1519β < 1317β < 1115β.
OR
Given a fraction baβ where 0 < b < a, if n is a positive integer, then bn < an and so b(a+n) = ab + bn < ab + an = a(b + n). Thus b+na+nβ < baβ. Therefore 1519β< 1317β< 1115β.
Answer: Eβ.
The problems on this page are the property of the MAA's American Mathematics Competitions