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  1.     
    #1
    Member

    Red face Math: abc = 1536 and...

    Hello guys...!

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

    Code: 
    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....
    jaya28inside Reviewed by jaya28inside on . Math: abc = 1536 and... Hello guys...! It's very funny because my little brother asked me this question; abc = 1536 and a+b+c = 36 Rating: 5

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  3.     
    #2
    Member
    Website's:
    KWWHunction.com Google.com
    a=16
    b=12
    c=8

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

    Check it out with your Bro
    Intel Xeon E3 -1240 16 GB DDR3 Ram 4 x 1 TB HDD 1 Gbit 100 TB BW -
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  4.     
    #3
    Member
    Website's:
    techtipsportal.com
    if you have 3 variables then you need 3 equations to solve the math ..

    Otherwise trial and error method can be used ...

  5.     
    #4
    Banned
    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.

  6.     
    #5
    Member
    Quote Originally Posted by .:KnowinGlenn:. View Post
    hmm.. let me try to solve


    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

  7.     
    #6
    Teh GFX Whore! ^.^
    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.


    I joined KWWHunction before it was cool!
    My "AWESOME" Graphic designing Service


  8.     
    #7
    Banned
    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

  9.     
    #8
    Banned
    Website's:
    Khambaty.com iMotivate.net
    @Trusted Uploader - Please teach me 10th standard math , I will pay you monthly

  10.     
    #9
    Banned
    Sure at only @only 10000$ per month......

  11.     
    #10
    Member
    i am upto this much

    (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

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