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:
Let ai=seciA+taniA.
Notice that we can factor this. Let's attempt to pull out a factor of secA+tanA.
We need to find the terms ai by finding cosA using the information we are given about tanA.
cosAsinAsinAsinA=2cosA=2cos2A=2−2sin2A
Solving this equation and then solving for cosA we get that
cosA=4217−2.
This yields a0=2, a1=17+4, a2=17, ai=17+4(17−2) and then a4=9.
For this to be an integer, notice that for an integer ai that 4∣i.
Now, we have the recursion
an=9an−4−16an−8.
We now have to notice that these are all multiples of 4, so we make the substitution an=b4n and plugging this in directly you get
bw=9⋅4⋅bw−4−16⋅4⋅bw−8.
Notice that we have b0=2, b4=9. This yields
bi=9bi−4−16ai−8.
We notice that after i≥2, that bn≡−1(mod2) and we have to find when it's −1(mod5) in order for the units digit to be 9.
Notice that (mod5), we have
bn≡9bn−1−16bn−2≡−(bn−1+bn−2).
Trying out and seeing the cycle, we see that every 3, we get that 2 are −1(mod5) and then the first is 2(mod5).
In particular, the residues go 2,−1,−1,2,−1,−1,….
As long as w is not a multiple of 3, we are good, and notice that n=4w so the only condition for n is multiples of 4 that aren't multiples of 3, and there are 167 values.