Is it enough information?
If a , b , c are three positive integers such that a b c + a b + b c + c a + a + b + c = 1 0 0 0 , then enter the value of a + b + c .
The answer is 28.
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3 solutions
Please can you elaborate how you factorized the expression?
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Umm... Its just basic prime factorizing.
7 1 1 1 3 1 0 0 1 1 4 3 1 3 1
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I don't understand this sorry. Plz elaborate!
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@Rishabh Tiwari – Um.. 1 0 0 1 ÷ 7 = 1 4 3 , 1 4 3 ÷ 1 1 = 1 3 . 1 3 ÷ 1 3 = 1 . So, 1 0 0 1 = 7 × 1 1 × 1 3 .
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@Ashish Menon – That's not what I asked , I was asking the factorization of abc + ab + bc + ca + a + b + c ?
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@Rishabh Tiwari – a ( b c + b + c + 1 ) + ( b c + b + c ) = a ( b + 1 ) ( c + 1 ) + ( b + 1 ) ( c + 1 ) − 1 = ( a + 1 ) ( b + 1 ) ( c + 1 ) − 1
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@Mateo Matijasevick – Thank you sir, its clear now.! Perfect sol. Btw.!
@Rishabh Tiwari – Ashish can you tell me if we can comment using Latex ? How so?
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@Rishabh Tiwari – Comment using LaTeX? We can always type latex codes between \ ( and \ ) or \[ and \text{ }\) are thecase maybe.
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@Ashish Menon – Can you give an example?
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@Rishabh Tiwari
–
Hi
. Now go to your icon in the top right part of the screen press toggle LaTeX and see this comment. You will see something like LaTeX: \ ( \text{Hi} \ ). Now remove the word
LaTeX
and the colon
:
and post the remaining of it in a new comment and watch the magic happen!
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@Ashish Menon – Hi .
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@Rishabh Tiwari – Bye .
@Ashish Menon – Got it thanks !!
Very nicely done +1!!!
Based on Mateo's solution, we can also construct more problems like this:
If abcd + abc + acd + abd + bcd + ab + bc + cd + ad + a + b + c + d = 21020, then find a + b + c +d... Ans: a + b + c + d = 31 or 48
or
If abcd + abc + acd + abd + bcd + ab + bc + cd + ad + a + b + c + d = 31030, then find a + b + c +d... Ans: a + b + c + d = 58
and so on..
The problem can also be done by using algebra but a bit longer method , @Hung Woei Neoh & @Ashish Siva please comment , thank you.!
Nice question (+1) let me think of that.
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Ok buddy thanx for commenting , as of now I am going to my coaching classes ,will be hearing from you in the evening. and yep I learnt basic latex with your help, thank you so much.!
Will you mind sharing algebraic way of tackling this problem ? :)
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Its a very long method & a waste of time to use , @Mateo Matijasevick has presented the best solution! However I can give you a hint , just write the eq.n as:-
a b + b c + a + c = 1 0 0 0 − a b c − c a − b ,
Now LHS can be factorised in (a+c)(b+1) , from here you will get the value of b or a+c , you can find either of the two & after writing cyclic expressions in a,b,c solve the system of equations, & finally you get the value of a+b+c as 28. Factorization is the key to solving these type of problems. Hope this helps , thank you!
From the given expression, ( a + 1 ) ( b + 1 ) ( c + 1 ) = 1 0 0 1 = 1 3 ⋅ 1 1 ⋅ 7 . As a , b , c are positive integers, then ( a + 1 ) , ( b + 1 ) , ( c + 1 ) > 1 , therefore, a + 1 = 7 , b + 1 = 1 1 and c + 1 = 1 3 , thus, a + b + c = 2 8 . The order of the factors does not matter.