Count Divisible Power Sums

For every exponent triple (x, y, z) with 0 <= x <= maxExponent2, 0 <= y <= maxExponent3, and 0 <= z <= maxExponent5, form 2^x + 3^y + 5^z.
Return the number of exponent triples whose formed number is divisible by divisor.
Different exponent triples are counted separately, even if they produce the same numeric sum.
Implement countDivisiblePowerSums with integer parameters maxExponent2, maxExponent3, maxExponent5, and divisor. Return the number of valid triples as a long.

Function
countDivisiblePowerSums(maxExponent2: int, maxExponent3: int, maxExponent5: int, divisor: int) → long

Examples
Example 1
maxExponent2 = 0
maxExponent3 = 0
maxExponent5 = 0
divisor = 3
return = 1
The only triple is (0, 0, 0), producing 1 + 1 + 1 = 3, which is divisible by 3.
Example 2
maxExponent2 = 1
maxExponent3 = 1
maxExponent5 = 0
divisor = 2
return = 2
The triples (1, 0, 0) and (1, 1, 0) produce 4 and 6. The other two sums are odd.
Example 3
maxExponent2 = 2
maxExponent3 = 2
maxExponent5 = 2
divisor = 5
return = 7
Seven of the 27 exponent triples have a power sum congruent to 0 modulo 5.

Constraints
0 <= maxExponent2, maxExponent3, maxExponent5 <= 2000
1 <= divisor <= 2000
The answer fits in a signed 64-bit integer.

Compute every power mod divisor, not the real power. Build three arrays of residues: 2^x mod d, 3^y mod d, 5^z mod d, each built by multiplying the previous residue and reducing. Then put the 5^z residues into a count map (or an array of size divisor). Loop over every (x, y) pair, about 4 million at most, compute need = (d - (a + b) % d) % d, and add count[need] to the answer. That's roughly 4 million lookups plus building the arrays. The common pitfalls are overflow from computing real powers, forgetting the outer mod when the sum is already 0 mod d, and using int for the answer. Use a long accumulator. Triples are counted separately even when sums match, so don't dedupe anything. When divisor is 1, every triple counts. StealthCoder is the hedge if the live OA clock gets to you and the two-sum-style counting idea won't come.

Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.