2023 AMC 8 Problem 21

Question

Alina writes the numbers

1,

2,

\cdots,

9

on separate cards, one number per card. She wishes to divide the cards into

3

groups of

3

cards so that the sum of the numbers in each group will be the same. In how many ways can this be done?

0

1

2

3

4

Answer: C

Concepts:partitions and compositions casework

Difficulty rating: 1690

Small hint:

Each group must have sum

15

Big hint:

Look at the group containing

9

Video solution:

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Written solution:

The sum of all the numbers is

\dfrac{9 \cdot 10}{2} = 45.

This means that the sum of each group is

45 \div 3 = 15.

Consider the group with

9

in it. The other two numbers must add to

6.

Therefore, the other cards in this group are

1

and

5

or

2

and

4.

Case $1:$ One group is

1,

5,

and

9.

Consider the group with

8

in it. The other numbers must add to

7.

The only option is

3

and

4

with the remaining cards.

The other group is then

2,

6,

and

7.

This adds to

15,

so this case contributes one possibility.

Case $2:$ One group is

2,

4,

and

9.

Consider the group with

8

in it. As above, the other numbers have to add to

7.

The only option is

1

and

6.

The final group is

3,

5,

and

7,

which adds to

15.

This is another configuration.

We have gone through all the cases, which revealed that there are only

2

possible groupings.

Thus, C is the correct answer.

2023 AMC 8 Problem 21

Answer: C

Difficulty rating: 1690

Small hint:

Big hint:

Video solution:

Written solution:

Problem 21 in other years