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2013 AMC 10B Problem 4

2013 AMC 10B Problem 4AMC 10 Step by Step

Name: Date: September 13, 2026

Number Properties · difficulty 1 of 5 · about 2 min · position P1-10

When counting from $3$ to $201$ , $53$ is the $51^\mathrm{st}$ number counted. When counting backwards from $201$ to $3$ , $53$ is the $n^\mathrm{th}$ number counted. What is $n$ ?

A)
$146$
B)
$147$
C)
$148$
D)
$149$
E)
$150$

Answer

Key insight

Counting backwards from 201, the number 53 is the (201 - 53 + 1)th, because inclusive counts need the extra 1.

Solution

Counting backwards from $201$ , the numbers said before reaching $53$ are $201, 200, \dots, 54$ , and then $53$ itself. The position of $53$ is the number of integers from $53$ to $201$ inclusive:
$201 - 53 + 1 = 149.$

The answer is $\boxed{\textbf{(D)}\ 149}$ .

Why this works

The number of integers from $a$ to $b$ inclusive is $b - a + 1$ , not $b - a$ . The forward count given in the problem ( $53 - 3 + 1 = 51$ ) is a hint to apply the same inclusive rule in reverse.

Alternative approach

The list $3, \dots, 201$ has $199$ numbers. A number in forward position $p$ has backward position $199 + 1 - p$ , so $n = 200 - 51 = 149$ .

The trap

Computing 201 - 53 = 148 and forgetting to add 1 for the inclusive count.

Common mistakes

Computing 201 - 53 = 148 and forgetting to add 1 for the inclusive count.
Using $199 - 51 = 148$ instead of $199 - 51 + 1$ when reversing the position.

Techniques

Set up the equation/formula and compute; no special trick needed