The sum of three prime numbers is 112. If one number exceeds the other by 36, what is the product of these numbers?
5402
Since the sum of three primes (112) is even, one of the primes must be 2 (the only even prime); the other two primes p and q satisfy p + q = 110 and p - q = 36, giving p = 73 and q = 37 (both prime).
The product is \(2 \times 73 \times 37 = 5{,}402\).
Hence, the product of the three prime numbers is 5,402.
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