Official Solution:


II. \(a_1\) is the only integer in the sequence. If \(a_1=1\), then all other terms will be non-integers, because in this case we would have \(a_1=1=2a_2=3a_3=...\), which leads to \(a_2=\frac{1}{2}\), \(a_3=\frac{1}{3}\), \(a_4=\frac{1}{4}\), and so on. Hence this option could be true.

III. The sequence does not contain negative numbers. Since given that \(a_1=\text{positive integer}=n*a_n\), then \(a_n=\frac{\text{positive integer}}{n}=\text{positive number}\), hence this option is always true.

Official Solution:


The sequence \(a_1\), \(a_2\), \(a_3\), ..., \(a_n\), ... is such that \(i*a_i=j*a_j\) for any pair of positive integers \((i, j)\). If \(a_1\) is a positive integer, which of the following could be true?

I. \(2*a_{100}=a_{99}+a_{98}\)

II. \(a_1\) is the only integer in the sequence

III. The sequence does not contain negative numbers


A. I only
B. II only
C. I and III only
D. II and III only
E. I, II, and III


Given that the sequence of numbers \(a_1\), \(a_2\), \(a_3\), ... have the following properties: \(i*a_i=j*a_j\) and \(a_1=\text{positive integer}\), so \(1*a_1=2*a_2=3*a_3=4*a_4=5*a_5=...=\text{positive integer}\).

We should determine whether the options given below can occur (notice that the question is which of the following COULD be true, not MUS be true).

I. \(2a_{100}=a_{99}+a_{98}\). Since \(100a_{100}=99a_{99}=98a_{98}\), then \(2a_{100}=\frac{100}{99}a_{100}+\frac{100}{98}a_{100}\). Reduce by \(a_{100}\): \(2=\frac{100}{99}+\frac{100}{98}\) which is not true. Hence this option could NOT be true.

II. \(a_1\) is the only integer in the sequence. If \(a_1=1\), then all other terms will be non-integers, because in this case we would have \(a_1=1=2a_2=3a_3=...\), which leads to \(a_2=\frac{1}{2}\), \(a_3=\frac{1}{3}\), \(a_4=\frac{1}{4}\), and so on. Hence this option could be true.

III. The sequence does not contain negative numbers. Since given that \(a_1=\text{positive integer}=n*a_n\), then \(a_n=\frac{\text{positive integer}}{n}=\text{positive number}\), hence this option is always true.


Answer: D

From \(100a_{100}=99a_{99}\) --> \(a_{99}=\frac{100}{99}a_{100}\);

From \(100a_{100}=98a_{98}\) --> \(a_{98}=\frac{100}{98}a_{100}\);

So, option I. \(2a_{100}=a_{99}+a_{98}\) becomes: \(2a_{100}=\frac{100}{99}a_{100}+\frac{100}{98}a_{100}\).

Hope it's clear.