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### Finding A Number Given Its Divisors and Remainders

```Date: 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.
```
Associated Topics:
High School Number Theory
High School Puzzles

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