Maths Olympiad Prep

Library / /451 of 860

Algebra Difficulty 5.2 AIME, harder Find the answer

Find the number of sequences a1,a2,,a10a_{1}, a_{2}, \ldots, a_{10} of positive integers with the property that an+2=an+1+ana_{n+2}=a_{n+1}+a_{n} for n=1,2,,8n=1,2, \ldots, 8, and a10=2002a_{10}=2002.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

3 Let a1=a,a2=ba_{1}=a, a_{2}=b; we successively compute a3=a+b;a4=a+a_{3}=a+b ; \quad a_{4}=a+ 2b;;a10=21a+34b2 b ; \quad \ldots ; \quad a_{10}=21 a+34 b. The equation 2002=21a+34b2002=21 a+34 b has three positive integer solutions, namely (84,7),(50,28),(16,49)(84,7),(50,28),(16,49), and each of these gives a unique sequence.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.