Math is radical 2
Consider the following equation 1 + 2 + 3 + 6 = a + b + c + d where a , b , c , d are integers. Find a + b + c + d .
The answer is 16593890.
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3 solutions
how did you do that?
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I can't neatly summarize here the method I used to find the solution, but it does start with the fact that a < 4 2 , if b , c , d > 0 , so I broke it down to a smaller problem, looking at each a < 4 2 . Then it was a tedious process of elimination. There is a way of converting
a + b + c
into the form
d + e + f
but not always, so it was a matter of checking all the possibilities out until something hit, i.e. something actually does convert to the 2nd form using integer d , e , f .
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Sorry but i couldn't get, please explain more
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@Ali Gh – Okay, square the left side, and you'll get
1 2 + 8 2 + 6 3 + 4 6
This is a starting point, we first assume a = 1 2 . Then we can convert pairs out of the 3 radical terms remaining, that is, there is a procedure (which doesn't always work) for converting
a + b
into the form
c + d
Then we can go to the next step as I've already described. It's very ad hoc, it's kind of difficult to systematically go through all the possible ways. As a matter of fact, I'm not even sure that this solution is unique, I only found one, and I posted the answer, and got, "Correct! Great effort." I don't have the proof that the answer is unique.
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@Michael Mendrin – Idk about unique. What if it were x,y,z, and t instead of a,b,c, and d? This would be a four-variable equation, and as we may know, you need at least 4 equations to get a unique solution for each.
Hey dude, i have contributed a lot of my time doing this problem and still i got 0 points even for giving the correct answer, just consider my solution and if you think it's totally correct then please help this poor kid gain 400 POINTS,,,
Here's what you should consider::
dd
pp
This implies that it has got many solutions, so please update the answer or do something and give my 400 points or at least tell me how i can reach the experts who can deal this situation!!
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Dude,
( − A + 8 2 + 6 3 + 4 6 ) 2
= A 2 + 8 2 ( 2 ) + 6 2 ( 3 ) + 4 2 ( 6 ) − 2 ( 8 ) A 2 − 2 ( 6 ) A 3 − 2 ( 4 ) A 6 + 2 ( 8 ) ( 6 ) 6 + 2 ( 8 ) ( 4 ) 1 2 + 2 ( 6 ) ( 4 ) 1 8
= A 2 + 3 3 2 − 1 6 A 2 − 1 2 A 3 − 8 A 6 + 9 6 6 + 1 2 8 3 + 1 4 4 2
= A 2 + 3 3 2 + ( 1 4 4 − 6 A ) 2 + ( 1 2 8 − 1 2 A ) 3 + ( 9 6 − 8 A ) 6
and not
A 2 + 3 3 2 + ( 1 4 4 − 6 A ) 2 + ( 6 4 − 1 2 A ) 3 + ( 4 8 − 4 A ) 6 .
Besides, I would like to say something about that part of your solution on the third to the last line,
2 1 + 4 1 3 + 5 8 0 8 − 2 ( 4 4 3 ) ( 1 2 6 ) + 8 6 4 .
This is equal to
2 1 + 4 1 3 + 6 6 7 2 − 3 1 6 8 2
However, for a real number a , you can only use the property a 2 = a if and only if a ≥ 0 . You cannot use it for − 3 1 6 8 2 because it is negative. In general, a 2 = ∣ a ∣ . Thus, ( − 3 1 6 8 2 ) 2 is not equal to − 3 1 6 8 2 but instead is equal to ∣ − 3 1 6 8 2 ∣ or 3 1 6 8 2 .
Thanks for the solution.I was getting there,but the key was to get rid of root 2 or root 3, since root 6 contains an element of both.
Can anyone simplify the sollution?
F = 1 + 2 + 3 + 6 = ( 1 + 2 ) ( 1 + 3 )
F 2 = ( 1 + 2 ) 2 ( 1 + 3 ) 2 = ( 3 + 2 2 ) ( 4 + 2 3 )
= 1 2 + 8 2 + 6 3 + 4 6 → a = 1 2
F 4 = a 2 + b + c + d + 2 a b + c + d
F 4 = ( 3 + 2 2 ) 2 ( 4 + 2 3 ) 2 = ( 1 7 + 1 2 2 ) ( 2 8 + 1 6 3 )
a 2 + b = 1 7 ⋅ 2 8 but a = 12
@Michael Mendrin Is it fixed that for a=21, b must be equal to 413? Cause if we fix a=21 then b= 17*28-441, ie 35. So is it possible??
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I don't know, I'll have to get back to my pile of notes about this problem. But my guess would be that this solution isn't unique.
The solution is not unique. I got a=12, b=332, c=145920, d=84297669153 which satisfies the equation and gives an alternative answer of a+b+c+d=84297815417.
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As far as I can see, your suggested solution is incorrect. I didn't check it algebraically, but numerically as it appears to suffice in this case:
1 + 2 + 3 + 6 ≈ 6 . 5 9 5 7 5 4 1 1 2 7 2 5 1 5 0
while
1 2 + 3 3 2 + 1 4 5 9 2 0 + 8 4 2 9 7 6 6 9 1 5 3 ≈ 6 . 5 9 5 7 5 4 1 1 2 7 2 4 9 0 6
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I stand corrected, thank you. My calculator's precision led me to believe the value I got for d was an integer; in fact it was d=84297669153.1542
Our approach will be straightforward: We keep writing expressions [ . . . ] as the square root of their square: 1 + 2 + 3 + 6 = ( 1 + 2 + 3 + 6 ) 2 = 1 2 + 8 2 + 6 3 + 4 6 = 1 2 + C + [ − C + 8 2 + 6 3 + 4 6 ] = 1 2 + C + 3 3 2 + C 2 + ( 1 4 4 − 1 6 C ) 2 + ( 1 2 8 − 1 2 C ) 3 + ( 9 6 − 8 C ) 6 To make the term 1 4 4 − 1 6 C disappear, we let C = 9 ; then we can easily finish the job: 2 1 + 4 1 3 + [ 2 0 3 + 2 4 6 ] = 2 1 + 4 1 3 + 4 6 5 6 + 9 6 0 1 8 = 2 1 + 4 1 3 + 4 6 5 6 + 1 6 5 8 8 8 0 0
çok süper comrade
Nice solution. I learned a new method. Thank you.
This is a BEAST.
1 + 2 + 3 + 6 = 2 1 + 4 1 3 + 4 6 5 6 + 1 6 5 8 8 8 0 0
I did not know that any such solution even existed until I tried solving this posted problem, wondering all the time, "is there really a solution here?"
Once this problem has been solved, it's easy to prove that the two sides are identical, by successive squaring of both sides with some re-arranging along the way.