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2008 AMC 10B Problem 8

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2008 AMC 10B Problem 8AMC 10 Step by Step
Name:                             Date: September 13, 2026
Diophantine Equations · difficulty 2 of 5 · about 2 min · position P1-10

A class collects $50\$50 to buy flowers for a classmate who is in the hospital. Roses cost $3\$3 each, and carnations cost $2\$2 each. No other flowers are to be used. How many different bouquets could be purchased for exactly $50\$50 ?

  1. A)

    11

  2. B)

    77

  3. C)

    99

  4. D)

    1616

  5. E)

    1717

Answer

C

Key insight

In 3r + 2c = 50 the term 3r must be even, so r is even; r can be 0, 2, ..., 16, which is 9 values.

Solution

Let rr be the number of roses and cc the number of carnations, with r,c≥0r, c \ge 0 integers and
3r+2c=50.3r + 2c = 50.
Since 5050 and 2c2c are even, 3r3r must be even, so rr is even. Also 3r≤503r \le 50 gives r≤16r \le 16. Thus r∈{0,2,4,…,16}r \in \{0, 2, 4, \dots, 16\}, and each such rr determines c=(50−3r)/2c = (50 - 3r)/2 as a nonnegative integer.

That list has 16/2+1=916/2 + 1 = 9 values, so there are 99 bouquets.

The answer is (C) 9\boxed{\textbf{(C)}\ 9}.

Why this works

A two-variable linear equation in nonnegative integers is solved by isolating a divisibility condition on one variable (here, parity of rr) and then bounding it. Once the constrained variable is pinned to an arithmetic progression, the count is just the length of that progression; do not forget the endpoints.

The trap

Forgetting the all-carnation bouquet (r = 0) and answering 8, or including r = 18 even though 3(18) exceeds 50.

Common mistakes

  • Forgetting the all-carnation bouquet (r = 0) and answering 8, or including r = 18 even though 3(18) exceeds 50.
  • Counting cc over 00 to 2525 without checking that 50−2c50 - 2c is divisible by 33, which gives 1717 or 1616.

Techniques

Use an invariant, parity, or coloring argument