2023 AMC 10B Problem 23

Problem 23 of 25HarderAlgebra Number Theory

An arithmetic sequence of positive integers has

n \ge 3

terms, initial term

a,

and common difference

d \gt 1.

Carl wrote down all the terms in this sequence correctly except for one term, which was off by

1.

The sum of the terms he wrote down was

222.

What is

a + d + n?

Show solution

Solution

The true sum is

S = na + \frac{n(n-1)}{2}d.

Since one term is off by

1,

the written total satisfies

222 = S \pm 1,

S = 221

223.

Also

2S = n\bigl(2a + (n-1)d\bigr),

n

divides

2S.

Since

a \ge 1

and

d \ge 2,

we have

2S \ge 2n^2,

hence

n^2 \le S.

For

S = 223,

no divisor of

446

lies between

3

and

\sqrt{223}.

For

S = 221 = 13 \cdot 17,

the only divisor of

442

in this range is

n = 13.

Thus

2a + 12d = 34,

a + 6d = 17.

Since

a

and

d

are positive integers with

d \gt 1,

we get

a = 5,

d = 2.

Then

a + d + n = 5 + 2 + 13 = 20.

Thus, B is the correct answer.