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.
cosxsinx=2cosA
sinx=2cos2A
sinx=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≡−1mod2 and we have to find when it's −1 mod 5 in order for the units digit to be 9.
Notice that mod 5, 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 mod 5 and then the first is 2 mod 5.
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.