Question

Find all triples of natural numbers (a,b,c) such that a, b and c are in geometric progession, and a+b+c=111.

Answer 1

If a,b , c are in GP then :

b = ar
c = br

\(b^2 = ac\) --(1)

Answer 2

ok

Answer 3

We have to find, all the solns for a, b and c such that a, b and c are in GP , right?

Answer 4

yes..

Answer 5

Any ideas?

Answer 6

@UnkleRhaukus @.Sam. @Callisto @mathstudent55 - any of u can help?

Answer 7

No idea but can we solve it using pascal triangle ?

Answer 8

It is for determining Binomial coefficients... what is the use of binomial theorem related topics in this question?

Answer 9

ok

Answer 10

I got something : \(\bf{a + c = 111 - \sqrt{ac} \\ (a+c)^2 = (111-\sqrt{ac})^2 }\)

Answer 11

Ya ok

Answer 12

\(\cfrac{111}{2} \ge b^{\frac{3}{2}} \)

Answer 13

ok

Answer 14

@UnkleRhaukus

Answer 15

@hartnn

Answer 16

Correction : \(\cfrac{a+b+c}{3} \ge( {\sqrt{abc}})^{\frac{1}{3}}\)

Answer 17

ok

Answer 18

37+37+37 = 111
and they all r in GP too :P :)

Answer 19

a=b=c=37

Answer 20

Ok Thank you

Answer 21

1, 10, 100

Answer 22

^

Answer 23

Many numbers are there

Answer 24

\(a(1+r+r^2) = 3 *37\)

a = 3 , then 1 + r + r^2 = 37

similarly factorize 111 and do that

Answer 25

yea and this is a olympiad problem it wud be tough

Answer 26

Yep.

Answer 27

thats a brilliant logic :)

Answer 28

Thank you for the help :)

Answer 29

111, -111, 111

Answer 30

http://openstudy.com/study#/updates/51bdcdaee4b0a49e8c8e9a41

Answer 31

@ganeshie8 , -111 is not a natural number. So, 111,-111,111 is not possible.

We only have 2 sets of solns :

(37,37,37) and (1,10,100)

\(a+ar + ar^2 = 111\)

\(a(1 + r + r^2) = 111\)

Case 1 : \(a(1+r+r^2) = 1 \times 111\)

a = 1 or \(1 + r + r^2 = 1 \)
and
a = 111 or \(1+ r + r^2 = 111\)

\(1 + r + r^2 \) can not be equal to 1.

So, a = 1 and \(1 + r + r^2 = 111\)

\(1 + r + r^2 = 111\)
\(r^2 + r -110 = 0\)
This gives, r = 10 (r can not be negative)

If r = 10, then we have : a = 1 , b = ar = 10, c = ar^2 = 100

\(\bf{(1,10,100)}\)

Case 2 : \(a(1+r+r^2) = 3 \times 37\)

Either a =3 or \(1 + r+r^2 =3\) and
a = 37 or \(1 + r + r^2 = 37\)

\(1 + r +r^2 \ne 37\) as , r can not be zero.

Therefore, a = 37 and \(1 + r + r^2 =3\)

\(r^2 + r + 1 = 3\)
Which gives, r = 1 (r can not be negative)

Therefore , we get : a = 37, b = ar = 37 , c = ar^2 = 37

\(\bf{(37,37,37)}\)