2023 AMC 8 Problem 25

Question

Fifteen integers

a_1,

a_2,

a_3,

\cdots,

a_{15}

are arranged in order on a number line. The integers are equally spaced and have the property that

1 \leq a_1 \leq 10,

13 \leq a_2 \leq 20,

and

241 \leq a_{15} \leq 250.

What is the sum of the digits of

a_{14}?

8

9

10

11

12

Answer: A

Concepts:arithmetic sequence bounding to limit cases digits

Difficulty rating: 1950

Small hint:

Let

d

be the common difference

Big hint:

Use the widest possible bounds for

a_1

and

a_{15}

to force

d

Video solution:

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Written solution:

Let

d

be the common difference. Using the largest possible value

10

for

a_1

and the smallest possible value

241

for

a_{15},

we have

d \geq \dfrac{241 - 10}{14} = 16.5.

Since all the numbers are integers,

d

must be at least

17.

Using the smallest possible value

1

for

a_1

and the largest possible value

250

for

a_{15},

we have

d \leq \dfrac{250 - 1}{14} \approx 17.8.

Since all the numbers are integers,

d

is at most

17.

Therefore,

d=17.

Note that

17 \cdot 14 = 238.

Since

a_{15}

is at least

241,

a_1

must be at least

3.

On the other hand, if

a_1

were greater than

3,

then

a_2=a_1+17

would be greater than

20,

which is not allowed.

Now we know that

a_1 = 3

and

d = 17.

This tells us that

a_{14} = 3 + 13 \cdot 17 = 224.

Therefore, sum of the digits is

2 + 2 + 4 = 8.

Thus, A is the correct answer.