A teacher has a bag of marbles to distribute to a class. If the marbles are divided equall… — ACT

A teacher has a bag of marbles to distribute to a class. If the marbles are divided equally among $8$ students, there are $6$ marbles left over. If the marbles are divided equally among $9$ students, there are $7$ marbles left over. What is the least possible number of marbles in the bag?

Answer: 70

Answer

The least possible number of marbles in the bag is 70.

The correct answer is 70. The problem describes a situation where the number of marbles,

N

, leaves a remainder of

6

when divided by

8

, and a remainder of

7

when divided by

9

. This means

N + 2

is a multiple of both

8

and

9

. The smallest positive integer that is a multiple of both

8

and

9

is their least common multiple, which is

72

. Therefore,

N + 2 = 72

, which gives

N = 70

Step-by-Step Solution

Analyze the relationship between the divisors and their respective remainders.

For divisor

8

, the remainder is

6

, which is

8 - 6 = 2

less than the divisor. For divisor

9

, the remainder is

7

, which is

9 - 7 = 2

less than the divisor. Therefore, adding

2

to the total number of marbles creates a number that is exactly divisible by both

8

and

9

This establishes that the target number plus

2

must be a common multiple of the two divisors.

Calculate the least common multiple (LCM) of the divisors

8

and

9

The prime factorization of

8

2^3

and of

9

3^2

. The LCM is the product of the highest powers of all prime factors involved:

2^3 \times 3^2 = 8 \times 9 = 72

To find the smallest positive common multiple of

8

and

9

Subtract

2

from the LCM to find the least possible number of marbles.

72 - 2 = 70

Since the bag is

2

marbles short of having a multiple of both

8

and

9

, we subtract

2

from the least common multiple.

Key Concept

Least Common Multiple (LCM) application with remainders

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic