AMC 10 Step by Step

2003 AMC 10A Problem 15

P11-15~2 minPrint

What is the probability that an integer in the set {1,2,3,...,100}\{1,2,3,...,100\} is divisible by 22 and not divisible by 33 ?

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Key insight

Count the 50 even numbers, then take away the even numbers that are also multiples of 3, which are the 16 multiples of 6. That leaves 34.

Solution

Start with all the even numbers, then remove the ones that are also multiples of 33.

1. Even numbers from 11 to 100100: 2,4,6,…,1002, 4, 6, \ldots, 100. There are 100÷2=50100 \div 2 = 50 of them.
2. An even number that is also a multiple of 33 is a multiple of 66. Those are 6,12,18,…6, 12, 18, \ldots. Since 6×16=966 \times 16 = 96 and 6×17=1026 \times 17 = 102 is too big, there are 1616 of them.
3. Even numbers that are not multiples of 33: 50−16=3450 - 16 = 34.
4. The probability is
34100=1750.\frac{34}{100} = \frac{17}{50}.

The answer is (C) 1750\boxed{\textbf{(C)}\ \frac{17}{50}}.

Why this works

"Divisible by 22 but not by 33" means: take the multiples of 22, then throw out the ones that are multiples of both, which are the multiples of 66. The count of multiples of 66 up to 100100 must be a whole number, 1616, so the answer is not exactly one third of the evens. Whenever the upper limit is not a multiple of the step, count carefully instead of using a fraction.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Multiples of 22 up to 100100: ⌊100/2⌋=50\lfloor 100/2 \rfloor = 50.

Among these, the ones also divisible by 33 are the multiples of lcm⁡(2,3)=6\operatorname{lcm}(2,3) = 6: ⌊100/6⌋=16\lfloor 100/6 \rfloor = 16 (namely 6,12,…,966, 12, \dots, 96).

Even numbers that are not multiples of 33: 50−16=3450 - 16 = 34. The probability is 34100=1750\dfrac{34}{100} = \dfrac{17}{50}.

The answer is (C) 1750\boxed{\textbf{(C)}\ \frac{17}{50}}.

Review in the AoPS books

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