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Unformatted text preview: , 376 ? Solution: Since both numbers are even, they are at least divisible by 2. The sum of the digits in the numerator is 24. The sum of the digits in the denominator is 30. Since these sums are both divisible by 3, each number is divisible by 3. Since these numbers meet the divisibility tests for 2 and 3, they are each divisible by 6. Example: Simplify to simplest form: 52, 832 Solution: Since both numbers are even, they are at least divisible by 2. However, to save time, we would like to divide by a larger number. The sum of the digits in the numerator is 22, so it is not divisible by 3. The number formed by the last two digits of each number is divisible by 4, making the entire number divisible by 4. The numbers formed by the last three digits of each number is divisible by 8. Therefore, each number is divisible by 8. Dividing by 8, we have
5459 . Since these numbers are no 6604
43, 672 135, 492 Then its last digit is 0, 2, 4, 6, or 8 the sum of the digits is divisible by 3 the number formed by the last 2 digits is divisible by 4 the last digit is 5 or 0 the number meets the tests for divisibility by 2 and 3 the number formed by the last 3 digits is divisible by 8 the sum of the digits is divisible by 9 longer even and divisibility by 3 was ruled out earlier, there is no longer a single digit factor common to numerator and denominator. It is unlikely, at the level of this examination, that you will be called on to divide by a twodigit number. www.petersons.com 26 Chapter 2 Exercise 3
Work out each problem. Circle the letter that appears before your answer. 1. Which of the following numbers is divisible by 5 and 9? (A) 42,235 (B) 34,325 (C) 46,505 (D) 37,845 (E) 53,290 Given the number 83,21p, in order for this number to be divisible by 3, 6, and 9, p must be (A) 4 (B) 5 (C) 6 (D) 0 (E) 9 If n! means n(n  1)(n  2) ... (4)(3)(2)(1), so that 4! = (4)(3)(2)(1) = 24, then 19! is divisible by I. 17 II. 54 III. 100 IV. 39 (A) I and II only (B) I only (C) I and IV only (D) I, II, III, and IV (E) no...
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This note was uploaded on 08/15/2010 for the course MATH a4d4 taught by Professor Colon during the Spring '10 term at EmbryRiddle FL/AZ.
 Spring '10
 Colon
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