2018 AMC 8 Problem 13

Question

Laila took five math tests, each worth a maximum of

100

points. Laila’s score on each test was an integer between

0

and

100,

inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was

82.

How many values are possible for Laila’s score on the last test?

4

5

9

10

18

Answer: A

Concepts:mean modular arithmetic

Difficulty rating: 1250

Small hint:

Let the repeated score be

f

and the last test score be

l

.

Big hint:

Use

4f+l=410

and the fact that

l

is larger than the average.

Video solution:

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Written solution:

Since the average score on the five tests is

82,

the total score of those five tests must be

5\cdot82 = 410 .

Now, let

f

be the score on the first

4

tests and let

l

be the score for the last test.

We know that

f < l \leq 100

and

4f + l = 410.

And as

410 = 4f + l < 5l ,

we know

\frac{410}{5} = 82 < l .

Also, since

4f + l = 410 ,

and dividing

410

by

4

gives us a remainder of

2,

we know that dividing

l

by

4

must leave a remainder of

2

as

4f

will leave no remainder when divided by

4.

Equivalently:

l \equiv 2 \mod 4 .

Since

82 < l \leq 100

and

l \equiv 2 \mod 4,

the only options for

l

are

86,90,94,98.

This yields four distinct solutions as follows:

\begin{align*} (f,l) =& (81,86);\\ &(80,90);\\ &(79,94);\\ &(78,98) \end{align*}

Therefore, there are

4

solutions, and A is the correct answer.