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2010 AMC 10B Problem 4

2010 AMC 10B Problem 4AMC 10 Step by Step

Name: Date: September 13, 2026

Functions · difficulty 1 of 5 · about 1 min · position P1-10

For a real number $x$ , define $\heartsuit(x)$ to be the average of $x$ and $x^2$ . What is $\heartsuit(1)+\heartsuit(2)+\heartsuit(3)$ ?

A)
$3$
B)
$6$
C)
$10$
D)
$12$
E)
$20$

Answer

Key insight

The heart of x is (x + x^2)/2, so the three values are 1, 3 and 6, the first three triangular numbers.

Solution

By definition $\heartsuit(x) = \dfrac{x + x^2}{2}$ . Evaluate at the three inputs:
$\heartsuit(1) = \frac{1+1}{2} = 1,\qquad \heartsuit(2) = \frac{2+4}{2} = 3,\qquad \heartsuit(3) = \frac{3+9}{2} = 6.$
Their sum is $1 + 3 + 6 = 10$ .

The answer is $\boxed{\textbf{(C)}\ 10}$ .

Why this works

A made-up symbol is just a function with a stated rule; write the rule as a formula once, then substitute. Note that $\frac{x+x^2}{2} = \frac{x(x+1)}{2}$ is the $x$ -th triangular number, which explains the values $1, 3, 6$ .

The trap

Averaging x with 2x or with x^2 incorrectly, or adding x + x^2 without dividing by 2 (giving 20, choice E).

Common mistakes

Averaging x with 2x or with x^2 incorrectly, or adding x + x^2 without dividing by 2 (giving 20, choice E).
Computing $\heartsuit(1+2+3) = \heartsuit(6) = 21$ instead of summing three separate values.

Techniques

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