## 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?

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July 7th, 2012 at 5:44 pm

Assign a through d ascending values

Suppose that ab is the missing product,

1) ac=2

2) ad=3

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

2) ad=3

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

July 7th, 2012 at 6:47 pm

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)

July 7th, 2012 at 8:23 pm

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.

July 7th, 2012 at 8:28 pm

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)

July 7th, 2012 at 8:35 pm

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

July 7th, 2012 at 9:01 pm

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.

July 8th, 2012 at 4:13 am

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.

July 8th, 2012 at 2:49 pm

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