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## Product problem

Posted by Chris on July 7, 2012 – 8:02 am

I have four positive numbers, a, b, c and d (not necessarily integers).

Obviously, there are six ways to multiply pairs of them, yielding the products ab, ac, ad, bc, bd and cd.

I tell you what five of these products are (but not which product is what):

2, 3, 4, 5 and 6

What’s the sixth product?

This post is under “MathsChallenge” and has 8 respond so far.

### 8 Responds so far- Add one»

1. 1. AnonymousCam Said：

Assign a through d ascending values

Suppose that ab is the missing product,

1) ac=2
3) bc=4
4) bd=5
5) cd=6
from 2 and 1 we have ad/ac=d/c=3/2
and from 4 and 3 we have bd/bc=d/c= 5/4

inconsistent so ab is not the missing product

Suppose that ac is the missing product,

1) ab=2
3) bc=4
4) bd=5
5) cd=6
from 2 and 1 we have ad/ab=d/b=3/2
and from 5 and 3 we have cd/bc=d/b= 6/4 = 3/2 (good so far!)
from d/b=3/2 and 4 we have b*(3/2*b)=5
or b^2=10/3
or b=sqrt(10/3) or approx. 1.83
sub into 1 gives us a= 2/( sqrt(10/3) ) or approx. 1.10
sub into 3 gives us c= 4/( sqrt(10/3) ) or approx. 2.19

( d= 3/2* (sqrt(10/3)) or approx. 2.74 )

a through d are ascending so we are consistent

so ac= 2/( sqrt(10/3) ) * 4/( sqrt(10/3) ) = 8/(10/3)=24/10=12/5

ac=12/5 or 2.4

The sixth product is
12/5 or 2.4

2. 2. Wizard of Oz Said：

Let ab be the missing product.
Multiply all the products together to get (abcd)^3 = ab*720.
Now 720 = 2^4 * 3*2 * 5
So multiply by (2^2 * 3 * 5^2) to bring this up to a cube.
So (abcd)^3 = 216000 = 60^3
So ab (the missing product) = 300 (and cd = 1/5)

3. 3. Wizard of Oz Said：

We could generalise the above post 2 by saying that ab could be any number which when multiplied by 720 forms a cube.
A smaller solution would be ab = 75/2 leading to a cube of 27000, so that abcd would be 30.
cd would then be 4/5.
Obviously there is (are) an infinity of solutions for ab.

4. 4. Carc Said：

wiz, from the 2nd line to the 5th, you only replaced a*b by 300.
besides, how could c*d be 1/5? all other products are given
however, if you multiply all known products, you get (ab)^2*(cd)^3=720, or (ab)^2=720/(cd)^3
so by inserting 2,3,4,5 and 6 for cd, and take the square root from 720/(cd)^3, only for cd=5 ab becomes a rational number, which (i think) is needed to make it possible to get the other products.
therefore ab has to be 2.4
(besides, anonymouscam, your calculating achtually doesn’t works 100% safe, as you assume that it is given, which product equals which number)

5. 5. Carc Said：

wiz, i might misunderstand you, but cd HAS to be either 2,3,4,5 or 6, therefore there is a Maximum of 5 possible situations

6. 6. Wizard of Oz Said：

Absolutely right, Carc. I’d overlooked the blindingly obvious fact that cd had to be one of the integers 2 to 6.
Disregard posts 2 & 3. I’ll go back into my corner.

7. 7. Chris Said：

There’s no need to say it, but that won’t stop me, Cam has the correct answer. However, there’s an almost elegant way to do it.

Later: Wiz was kinda going in the right direction.

8. 8. Chris Said：

abcd = ab cd = ac bd = ad bc = const. Note that 2*6 = 3*4 = 12.
So the remaining pair must satisfy 5x = 12 => x = 12/5 = 2.4

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