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Find the number of integers x with 0 ≤ x < 18000 satisfying x^2 ≡ 1…
Question
Find the number of integers x with 0 ≤ x < 18000 satisfying x^2 ≡ 1 (mod 18000).
✓ Verified answer: 16checked by our engine — not a guess
Step-by-step solution
Factor 18000 = 2^4 · 3^2 · 5^3 = 16·9·125.
By CRT, the number of solutions of x^2≡1 modulo a number is the product of the counts modulo each prime power.
For odd prime powers p^k there are exactly 2 solutions (x≡±1).
For the modulus 16 = 2^4: x^2≡1 (mod 2^k) has 4 solutions when k≥3.
So counts: mod 16 → 4, mod 9 → 2, mod 125 → 2.
Total = 4·2·2 = 16.
Final answer16
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