Question

The product of three prime numbers is seven times their sum. Find the three prime numbers.

Answer 1

If we let the numbers as: x, y and z

Then :

xyz = 7(x + y + z)

And @hartnn is right..

Ha haha..

Answer 2

shortest way is to do trials,
i started with 2,3,5
then 3,5,7 and i got it :P

Answer 3

xyz = 7(x + y + z)
xyz - 7z = 7(y+x)
z(xy -7) = 7(y+x)
----------------- i don't have a complete logic here
z=7,
problem reduces to
xy -7 = y+x

Answer 4

And after that you will use Hit and Trial Method again??

Answer 5

only physicist are allowed to do that.

Answer 6

So, what will you use if you are not a physicist ??

Answer 7

I'm half physicist .. :P

Answer 8

Then I think @hartnn is full physicist..

Ha ha ha..

Answer 9

xy = y+x +7
y(x-1) = x+7
y = (x+7)/(x-1) <--- this should be an integer (prime ) maybe we could impose this constrain here.

Answer 10

mukushla is perfect for this kind of problem ... where did he go?

Answer 11

Yes I am looking for him too..

Answer 12

He is offline but we should mention his name @mukushla

Answer 13

it is easy to see that 2,3 will satisfy this criteria
(x+7)/(x-1) = 2+7 = 9 <-- not a prime
(x+7)/(x-1) = 10/2 = 5

Answer 14

There if x is 3 then we get y = 5

And they both are prime and anyhow according to your logic z = 7 too.

Answer 15

this isn't the problem ... you have to prove either this is infinite solutions (of some form) /// some finite solution /// or just unique solution.

Answer 16

Yeah may be we have other prime numbers which will satisfy the given condition..

Or may not be..

Answer 17

maybe the word 'consecutive' prime numbers, in the question can help simplify the solution....

Answer 18

also there is hidden logic here (probably not nice)

z(xy -7) = 7(y+x)
----------------- i don't have a complete logic here
z=7,

Answer 19

But there isn't any word named "Consecutive" there..

Answer 20

The hidden logic:
xyz = 7 (x+y+z)
Since a, b and c are prime, so, one of a, b, c should be 7

Answer 21

Oh wow hidden but not complete logic..

Answer 22

x, y and z will be multiple of 7 if one out of them is multiple of 7 or 7 itself..

Answer 23

It can't be multiple of 7 since it is a PRIME!

Answer 24

of course except 7 :P

Answer 25

And x and y and z are prime too..

I am saying according to the equation @RolyPoly

Answer 26

According to this equation:

xyz = 7(x + y + z)

Product of x, y and z will be multiple of 7 so, one out of x, y and z should be 7..

Answer 27

Put x = 7, then yz = 7 +y +z
So, z = (y+7) / (y-1)

We must use trial and error here? Or is there any other better way?

Answer 28

If we use the word consecutive here, then what are the prime numbers consecutive to each other??

2 3 and 5 or 3 5 and 7

Not more than that I think @hartnn

Answer 29

Any othere better way @mukushla can tell you if any..

But for now, Hit and Trial method is what you can do..

Answer 30

so one of these must be seven because xyz is divisible by 7 so say x=7
\[yz=7+y+z\]\[y=\frac{z+7}{z-1}=\frac{z-1+8}{z-1}=1+\frac{8}{z-1}\]so 8 is divisible by z-1 so\[z-1=1,2,4,8\]it gives\[z=2,3,5\]we are done

Answer 31

Great..