Math: abc = 1536 and...

[jaya28inside]

Hello guys...!

It's very funny because my little brother asked me this question;

abc = 1536
and
a+b+c = 36

What are the possibilities of a, b, and c ? )

Could any body help me,
1st) What topic in math is this that I should look upon?
2nd) Step by step how to solve it (the way)...?
3rd)... your kindness....

 

[Glenn]

a=16
b=12
c=8

abc= 16x12x8 =1536 and a+b+c= 16+12+8=36.

Check it out with your Bro

 

[pompom]

if you have 3 variables then you need 3 equations to solve the math ..

Otherwise trial and error method can be used ...

 

[Maze]

8,12 and 16.

If you restrict a,b,c to positive integers the only solution (up to permutations) is 8, 12, 16.
One approach is to look at factorizations of 1536, rejecting possibilities that involve numbers greater than 36.

 

[devilboy23]

hmm.. let me try to solve :P


a=16
b=12
c=8

abc= 16x12x8 =1536 and a+b+c= 16+12+8=36.

Check it out with your Bro

its correct

 

[Nitish]

Just a slight correction in glenn's answer, maze is correct. Answer is 16,12 and 8 but we cannot assign any value to any variable. All 3 variables can have all 3 values.

 

[Trusteduploader]

a=bc=156- abc=156-a=1536-b-can so that we can conjucate abc=156- abc=156-a=1536-b-abc= 16x12x8 =1536 and a+b+c= 16+12+8=36.abc= 16x12x8 =1536 and a+b+c= 16+12+8=36.a=1536-b-c so abc=156
a=1536-b-c so abc=156
and bc=156- abc=156-a=1536-b-
bc=156- abc=156-a=1536-b-bc=156- abc=156-a=1536-b-
bc=156- abc=156-a=1536-b-

0=0 hence proved
I am bit weak in maths LMAO

 

[Netguy]

@Trusted Uploader - Please teach me 10th standard math , I will pay you monthly

 

[Trusteduploader]

Sure at only @only 10000$ per month......

 

[raflamo]

i am upto this much 8-)

(a+b+c)2=a2+b2+c2+2ab+2bc+2ca

(a+b+C)2=(a+b)2+c2+2bc+2ca
(a+b+C)2=(36-c)2+c2+2bc+2ca
(a+b+C)2=(36-c)2+c2+2c(b+a)
(a+b+C)2=(36-c)2+c2+2c(36-c)

(36)2=(36-c)2+c2+2c(36-c)

solving this second degree equation you will get c

substitute c in those two equation given in the question