Sum of really Big CUBES !!!!!!!
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where 1 < a < 1 0 and n is a positive number. What is the value of n a 2 ?
Hint: ( A + B + C ) 3 = A 3 + B 3 + C 3 + 3 ( A + B ) ( B + C ) ( C + A )
The answer is 468.
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1 solution
Thank you for the solution, is this the only way to do it?
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The only easiest way I know of.
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I still don't get it.
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@Peter Bishop – You can of course brute force your way through, that's not the nicest way to do it.
When you want to evaluate C D − A B , consider setting C = 4 0 0 0 0 − 1 3 1 7 , D = 4 0 0 0 0 − 5 5 8 , A = 5 0 0 0 0 + 5 0 6 7 , B = 5 0 0 0 0 − 2 6 9 4 2 , this makes the calculation via algebra a little bit easier. And when you want to multiply by 1 0 5 − E , consider its factorization 5 7 , it's easier to multiply by 1 0 then divide 2 for each multiple of 5 . A few tricks from Vedic mathematics helps too.
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@Pi Han Goh – That seems like a lot of calculation for me. Thanks, anyway. You're really helpful.
By the way, from line 2 to line 3, I don't understand it.
I got 1 0 1 5 − 1 0 1 0 ( A + B + E ) + 1 0 5 ( A B + A E + B E ) − A B E
1 0 1 5 − 1 0 1 5 + 1 0 5 ( A B + A E + B E ) − A B E
= 1 0 5 ( A B + A E + B E ) − A B E
Is there any formula for this , or did I miss something?
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@Peter Bishop – I make mistakes everywhere all the time. I fixed it, again. Thanks, Peter Bishop
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@Pi Han Goh – You don't have to thank me. It's me who have to thank you. Thank you very much. Without you I wouldn't have a clue on how to do this problem. Thank you. Good night!!!
EDITED: Fixed some lines.
Note that 5 5 0 6 7 + 2 3 0 5 8 = 3 8 6 8 3 + 3 9 4 4 2 = 7 8 1 2 5 = 1 0 5 − 2 1 8 7 5
Let A , B , C , D , E denote the values 5 5 0 6 7 , 2 3 0 5 8 , 3 8 6 8 3 , 3 9 4 4 2 , 2 1 8 7 5 respectively. Then,
( A + B + E ) 3 = = = A 3 + B 3 + E 3 + 3 ( A + B ) ( B + E ) ( A + E ) A 3 + B 3 + E 3 + 3 ( 1 0 5 − E ) ( 1 0 5 − A ) ( 1 0 5 − B ) A 3 + B 3 + E 3 + 3 ( 1 0 5 ( A B + A E + B E ) − A B E )
Similarly, ( C + D + E ) 3 = C 3 + D 3 + E 3 + + 3 ( 1 0 5 ( C D + C E + D E ) − C D E )
Because, A + B = C + D ⇒ ( A + B + E ) 3 = ( C + D + E ) 3
A 3 + B 3 + E 3 + 3 ( 1 0 5 ( A B + A E + B E ) − A B E ) A 3 + B 3 − C 3 − D 3 = = = = C 3 + D 3 + E 3 + 3 ( 1 0 5 ( C D + C E + D E ) − C D E ) 3 ⋅ 1 0 5 ( C D + C E + D E − A B − A E − B E ) − E ( C D − A B ) 3 ( 1 0 5 − E ) ( C D − A B ) + 3 ⋅ 1 0 5 E ( C + D − A − B ) 3 ( 1 0 5 − E ) ( C D − A B ) = 6 ⋅ 1 0 1 3
So a = 6 , n = 1 3 ⇒ n a 2 = 4 6 8