How many natural numbers divide 3920, 5880 and 7350 leaving the same remainder in each case?
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3
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6
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12
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18
The numbers must divide the differences: 5880-3920=1960, 7350-5880=1470, 7350-3920=3430. The GCD of 1960, 1470, 3430 is 70. The number of divisors of 70 (2*5*7) is (1+1)(1+1)(1+1) = 8. However, the question asks for natural numbers leaving the same remainder, which involves divisors of the GCD.
Let the common remainder be r and the divisor be d. Then d divides the differences 5880 - 3920 = 1960, 7350 - 5880 = 1470, and 7350 - 3920 = 3430. The number of such divisors is the number of common factors of 1960, 1470, and 3430, which equals the number of factors of their greatest common divisor. The greatest common divisor is 490, whose prime factorization is 2 * 5 * 7 * 7, so its factors are found by multiplying the choices from each prime: 2 * 2 * 3 = 12.