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### Number * Sum of Remaining Four Numbers

```Date: 04/03/2003 at 20:33:29
From: John
Subject: Adding/multiplying groups of numbers

Find 5 numbers such that when each number is multiplied by the sum of
the remaining 4 numbers, the following values will result: 152, 245,
297, 320, 360.

Thank you!
```

```
Date: 04/04/2003 at 00:11:42
From: Doctor Greenie
Subject: Re: Adding/multiplying groups of numbers

Hi, John -

Cool problem! Lots of good mental exercise and mathematical detective
work here.

Let's call the 5 numbers a, b, c, d, and e.  Then we have

(1) a(b+c+d+e)=152
(2) b(a+c+d+e)=245
(3) c(a+b+d+e)=297
(4) d(a+b+c+e)=320
(5) e(a+b+c+d)=360

It seems that with the equations written this way, the numbers a, b,
c, d, and e should be in numerical order. This can in fact be
demonstrated quite easily by comparing any two consecutive equations.
For example, distributing in equations (2) and (3) and subtracting
equation (2) from equation (3), we have

ca+cb+cd+ce = 297
ba+bc+bd+be = 245
-------------------
(c-b)(a+d+e) = 52

Since this product is positive, (c-b) must be positive, which means c
is greater than b. Using other pairs of "consecutive" equations, we
can similarly show that b is greater than a, d is greater than c, and
e is greater than d.

So we know

a < b < c < d < e

From equations (2) and (3), since the products are odd, we know that
b and c are odd integers. Looking at those two equations from a
different perspective, we see that, since the same 5 numbers occur on
the left side of both equations, it is necessary to find number p, q,
r, and s such that

pq = 245
rs = 297

with the restriction that

p+q = r+s

We have the following possible factorizations of 245 and 297:

245 = 5*49
= 7*35

297 = 3*99
= 9*33
= 11*27

The only factorizations of these two numbers that satisfy the
requirements are

245 = 7*35;  297 = 9*33

So we know

(*) b = 7
(*) c = 9

Using these two values in either equation (2) or (3) above, we can
also determine that

a+d+e = 26

And from this we know that

a+b+c+d+e = 42

Now let's look at equations (1), (4), and (5) again, knowing the
values of b and c. We have

a(16+d+e) = 152
d(16+a+e) = 320
e(16+a+d) = 360

Because a+d+e=26, we need to find a factorization of 152 into the
product of two integers whose sum is 42; and likewise factorizations
of 320 and 360 each into the product of two integers whose sum is 42.
With some trial and error we find

152 = 4*38
320 = 10*32
360 = 12*30

And so we conclude

(*) a = 4
(*) d = 10
(*) e = 12

So we have

a = 4; b = 7; c = 9; d = 10; e = 12

Checking our results in the original equations:

4(7+9+10+12) = 4(38) = 152
7(4+9+10+12) = 7(35) = 245
9(4+7+10+12) = 9(33) = 297
10(4+7+9+12) = 10(32) = 320
12(4+7+9+10) = 12(30) = 360

Looking back at this solution, it appears there could have been many
different paths to the answer using the same types of logical
analysis....

I hope this helps. Please write back if you still have any questions
about any of this. And thanks for providing me with some good mental
exercise!

- Doctor Greenie, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra
Middle School Algebra
Middle School Word Problems

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