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[ Information | Background | The CyberTeaser | Quantum News | Miscellaneous Stuff ] Solution to CyberTeaser B146Any number in question can be written as A × 104 + 1995. Clearly, it is exactly divisible by A if and only if A is a divisor of 1995. Since 1995 = 3 × 5 × 7 × 19, any divisor of 1995 is a product of some of its four prime factors taken no more than once each. So the number of divisors is equal to the number of subsets of a four-element set (including an empty subset, which corresponds to the divisor 1). So the answer is 24 = 16. To put it slightly differently, all the divisors are formed by choosing or not choosing the factor 3, choosing or not choosing the factor 5, and so on, and then multiplying all the chosen factors. The number of choices is 2 × 2 × 2 × 2. CyberTeaser Archive
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