Finding A Number Given Its Divisors and RemaindersDate: 10/22/2003 at 17:22:20 From: Kaytlyn Subject: complex algebra Find the smallest whole number that when divided by 5, 7, 9, and 11 gives remainders of 1, 2, 3, and 4 respectively. I think there is a formula out there, or a way to do it with relationships, but I can't figure it out. Date: 10/22/2003 at 20:20:56 From: Doctor Greenie Subject: Re: complex algebra Hi, Kaytlyn - I don't think there is a formula, but there is a general strategy for solving problems like this. We are looking for a whole number "x" such that x/5 leaves remainder 1 x/7 leaves remainder 2 x/9 leaves remainder 3 x/11 leaves remainder 4 What we need to do is change this problem into one where some simple expression involving "x" leaves zero remainder when divided by any of the numbers 5, 7, 9, or 11; then the problem can be worked as a least common multiple problem. We do this by comparing the divisors and the remainders we are given: For the number "x" which we are looking for, the divisors and remainders are divisor remainder ------------------ 5 1 7 2 9 3 11 4 When the divisors increase by 2, the remainders increase by 1. We want to find an expression involving x for which the remainders will increase by the same amount as the divisors (you will see why shortly). Since the remainders with the number "x" are increasing half as fast as the divisors, let's look at the remainders when we divide "2x" by these divisors. Each remainder will simply double. For example, if 7/6 has remainder one, then (2*7)/6 = 14/6 has remainder 2. So, for the number "2x", the divisors and remainders are divisor remainder ------------------ 5 2 7 4 9 6 11 8 The reason we want the remainders to increase at the same rate as the divisors is that now we can see that each remainder is 3 less than the corresponding divisor. This tells us that the number "2x + 3" will leave 0 remainder when divided by any of these divisors. So we are looking for the smallest whole number "x" for which the number "2x + 3" is divisible by 5, 7, 9, and 11. Because these divisors are relatively prime, the smallest number divisible by 5, 7, 9, and 11 (their least common multiple) is 5*7*9*11 = 3465 And so we have 2x + 3 = 3465 2x = 3462 x = 1731 This is the number we are looking for. I hope this helps. Please write back if you have any further questions about any of this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ Date: 10/23/2003 at 22:13:07 From: Kaytlyn Subject: Thank you (complex algebra) Thank you so much for helping me. It's nice to know that if I need more help I can trust the Dr. Math services. |
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