A positive integer, x, is divisible by 15, 25, and 30. Which of the following is the smallest possible value of x?
a) 75 b) 90 c) 120 d) 150 e) 450 From an old ACT Exam. SOLUTION The answer is choice d. The smallest number that is divisible by 15, 25, and 30 is the Least Common Multiple (LCM) of all the numbers. First factor all the given numbers. 15 = 3 * 5 25 = 5 * 5 30 = 2 * 3 * 5 Then multiply each factor the greatest number of times it occurs in either number. If the same factor occurs more than once in both numbers, you multiply the factor the greatest number of times it occurs. 2: 1 occurrence 3: 1 occurrence 5: 2 occurrences LCM = 2 * 3 * 5 * 5 = 150 Therefore, the answer is choice d) 150.
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Math Problem of the Day ANSWERS PART II
Here are the solutions to The Math Problems of The Day. The problems and solutions shown here come
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