Date: Sat, 27 Feb 1999 12:06:15 -0800 My name is Ali and I am in the 5th grade. I have a math question: What is the least positive integer meeting each of the following conditions: Dividing by 7 gives a remainder of 4 Dividing by 8 gives a remainder of 5 Dividing by 9 gives a remainder of 6 Help! Hi Ali Let me explain to you one method of solving these problems. Write down the numbers that dividing by 7 gives a remainder of 4: 4, 11, 18, 25, 32, 39, 46, 53, 70, 67, 74, 81, 88, 95, 102, 109, 116, 123, ... Do the same for dividing by 8 gives a remainder of 5 5, 13, 21, 29, 37, 45, 53, 61, 69, 77, 85, 93, 101, 109, ... and dividing by 9 gives a remainder of 6 6, 15, 24, 33, 42, 51, 60, 69, 78, 87, 96, 105, 114, 123, 132, ... It might take a long time to see where they overlap and what you want is a number in each sequence that is exactly the same. You can save some work and paper by first noticing that the first two sequences meet at 53. Now since the first sequnce goes up in steps of 7 and the second in steps of 8 they will meet again in 7x8 =56 later -- see they meet at 53 + 56 = 109. They will meet again at 109 + 56 = 165,and so on.     The problem now becomes one with just two sequences: What is the least positive integer meeting the following conditons: Dividing by 56 gives a remainder of 53 Dividing by 9 gives a remainder of 6 That is you need to look where, 53, 109, 165, 221, ... meets 6, 15, 24, 33, 42, 51, 60, 69, 78, 87, 96, 105, 114, 123, 132, ... There are general methods to solve such problems, one is called the Chinese Remainder Theorem, but they are quite complicated. Hope this helps! Denis

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