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Consider a collection of pennies with the following constraints: When the pennies are put in groups of 2 there is one penny left over. When they are put in groups of three, five and six there is 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?

User Calin Don
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2 Answers

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The number of pennies that could be in the collection is 91.

To find the number of pennies that could be in the collection, we need to consider the given constraints.

1. When the pennies are put in groups of 2, there is one penny left over.

2. When they are put in groups of 3, 5, and 6, there is also one penny left over.

3. But when they are put in groups of 7, there are no pennies left over.

To find a solution, we can start by considering the concept of remainders. Let's work step-by-step to determine the number of pennies.

1. Start with the first constraint: When the pennies are put in groups of 2, there is one penny left over. This means the total number of pennies is odd.

2. Moving on to the second constraint: When the pennies are put in groups of 3, 5, and 6, there is one penny left over. Again, this indicates that the total number of pennies is odd.

3. Next, consider the third constraint: When the pennies are put in groups of 7, there are no pennies left over. This means the total number of pennies is divisible evenly by 7.

4. Combining the information from the first two constraints, we can conclude that the total number of pennies must be odd and divisible by 7.

5. Let's consider the smallest odd number that is divisible by 7, which is 7 itself. If we divide 7 by 7, there are no pennies left over, satisfying the third constraint. However, the first two constraints are not met.

6. Now, let's try the next odd number, which is 7 + 2 = 9. If we divide 9 by 7, there is a remainder of 2, not satisfying the third constraint. We can continue this process until we find a number that satisfies all the given constraints.

7. Continuing with this process, we find that the number 91 satisfies all the given constraints. If we divide 91 by 7, there are no pennies left over. Similarly, when divided by 2, 3, 5, and 6, there is always one penny left over.

Therefore, the number of pennies that could be in the collection is 91.

User Tute
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3 votes

Answer:

91 pennies.

Explanation:

It is given that 5 types of groups of pennies: group of 2, group of 3, group of 5, group of 6, and group of 7. If the pennies are arranged in the groups of 7, no pennies are left over. This means that the number of pennies has to be a multiple of 7 in order to satisfy this constraint. In the rest of the groups, there will always be 1 penny remaining without any remaining. This means that the number of pennies will yield the remainder of 1 if it is divided by 2, 3, 5, and 6. Possible multiples of 7:

0, 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98...

The number which satisfies the above conditions is 91. Since 91 divided by 7 is 13 and if 91 is divided by other numbers, the remainder will always be 1. Therefore, there are 91 pennies!!!

User Daniel Khoroshko
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