To answer, we must recognize an important rule: the divisor must be greater than the remainder. Let's look at a few examples:
10 ÷ 4 = 2 remainder 2 (divisor 4 is greater than remainder 2)
23 ÷ 6 = 3 remainder 5 (divisor 6 is greater than remainder 5)
If the divisor weren't greater than the remainder, the divisor would be able to divide into the dividend at least one more time. Let's take an incorrect example to illustrate:
23 ÷ 6 = 2 remainder 11.
In this case, we've framed the operation such that the divisor is LESS than the remainder (6 is less than 11). The error is that 6 actually goes into 23 three times. The remainder is what is left over when the divisor has been divided into the dividend as many times as possible. Therefore, if a divided by b gives a remainder of 9, we can conclude that b is greater than 9: b ≥ 10.
Likewise, if c divided by d gives a remainder of 5, we can conclude that d is greater than 5: d ≥ 6. Therefore, we can determine a minimum for the sum: b + d ≥ 16.
Only 15 is too small.
The correct answer is E.
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Anaira Mitch