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7th Apr 2011, 04:42 PM #1OPMember
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
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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7th Apr 2011, 04:50 PM #2MemberWebsite's:
KWWHunction.com Google.coma=16
b=12
c=8
abc= 16x12x8 =1536 and a+b+c= 16+12+8=36.
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7th Apr 2011, 04:51 PM #3MemberWebsite's:
techtipsportal.comif you have 3 variables then you need 3 equations to solve the math ..
Otherwise trial and error method can be used ...SEO Tips 1- [TUT] Site Ping-Make your site indexed more faster
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7th Apr 2011, 04:55 PM #4Banned
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.
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7th Apr 2011, 04:55 PM #5
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7th Apr 2011, 05:04 PM #6Teh 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.
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7th Apr 2011, 05:06 PM #7Banned
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
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7th Apr 2011, 05:11 PM #8BannedWebsite's:
Khambaty.com iMotivate.net@Trusted Uploader - Please teach me 10th standard math , I will pay you monthly
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7th Apr 2011, 05:14 PM #9Banned
Sure at only @only 10000$ per month......
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7th Apr 2011, 05:16 PM #10Member
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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