Question

When the pennies are put in groups of two, there is one penny left over. When they are put in groups of 3, 5, and six, there's also one penny left over, but when they are put in groups of seven, there are no pennies left over. How many pennies could there be?

A. 41
B. 49
C. 71
D. 97

Answer 1

The number of pennies that fits the given criteria in the problem is 49, as it is the only number divisible by 7 with no remainder and leaves a remainder of 1 when divided by 2, 3, 5, and 6.

Step-by-step explanation:

The student's question is about finding the number of pennies which meet specific divisibility criteria. To do this, we look for a number that when divided by 2, 3, 5, and 6 leaves a remainder of 1 each time, but is divisible evenly by 7 with no remainder. Among the choices given, we will apply these conditions to each option.

• Option A: 41 – Not divisible by 7.
• Option B: 49 – Divisible by 7 and also meets the other conditions since 49 − 1 = 48, which is divisible by 2, 3, 6; and 48 + 1 is divisible by 5.
• Option C: 71 – Not divisible by 7.
• Option D: 97 – Not divisible by 7.

Therefore, the number of pennies is likely to be 49, which is option B.