Problem:
Let A be an acute angle such that tanA=2cosA. Find the number of positive integers n less than or equal to 1000 such that secnA+tannA is a positive integer whose units digit is 9.
Solution:
Answer (167):
Let an=secnA+tannA and let s and c denote sinA and cosA, respectively. Because angle A is acute, it follows that s,c>0. Then cs=2c, so s=2(1−s2) and s=417−1. Thus tanA=2c=217−1 and tanA⋅secA=2. It follows that
an=secnA+tannA=(217+1)2n+(217−1)2n.
There are three cases to consider.
If n is an even integer not divisible by 4 , then n=4k+2 for some nonnegative integer k and
an=22k+1(17+1)2k+1+(17−1)2k+1
When the two powers in the numerator of this fraction are expanded using the Binomial Theorem, the terms with even powers of 17 will cancel, leaving a positive integer multiple of 17. Therefore an is not a rational number when n≡2(mod4).
If n is an odd integer, it follows immediately from the previous case and the following identity that an is not a rational number:
an2=(secnA+tannA)2=a2n+2n+1
If n is a multiple of 4 , then consider the sequence bk=a4k. Then
bk=a4k=(29+17)k+(29−17)k.
Note that bk satisfies the simple recursion bk=9bk−1−16bk−2, with b0=2 and b1=9. Hence bk is an integer for all k and
bk≡4bk−2−bk−1(mod10)
Computing several more terms modulo 10 yields b2≡4⋅2−9≡9,b3≡4⋅9−9≡7,b4≡4⋅9−7≡9, and b5≡4⋅7−9≡9. Hence for k>0,bk is the repeating sequence 9,9,7, with period 3. There are 250 such bk consisting of 83 cycles and one 9 left over for a total of 2⋅83+1=167 values of n satisfying the required conditions.