Discussion
How many natural numbers are there which give a remainder of 28 when 1458 is divided by the numbers?
Answer: B
There are 8 such numbers.
A divisor d leaves remainder 28 on 1458 exactly when d divides \(1458 - 28 = 1430\) and d is greater than 28 (a remainder must be smaller than the divisor).
\[ 1430 = 2 \times 5 \times 11 \times 13 \]
1430 has 16 divisors: 1, 2, 5, 10, 11, 13, 22, 26, 55, 65, 110, 130, 143, 286, 715 and 1430. Those greater than 28 are 55, 65, 110, 130, 143, 286, 715 and 1430, which is 8 numbers. Answers of 9 or 10 come from including divisors that are not greater than 28, such as 26 or 22, which actually leave remainders of 2 and 6.
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