Cubes

Number Theory Level 2

$2008=a^{3}+b^{3}+c^{3}$

where $a$ , $b$ and $c$ are distinct positive integers, find the value of

$a+b+c.$

The answer is 22.

2 solutions

Niranjan Khanderia
Jul 8, 2014

13 > Cube root of 2008 >12............So make a table of N^3, N = 1,2, . . . 12.

Start with 12^3 = 1728. So value left is 2008 - 1728 = 280
See the table and find 6^3 = 216......280 - 216 = 64 = 4^3.
So 12^3 + 6^3 + 4^3 =2008.....>a + b + c = 12+6 + 4 = 22

It's funny when I did 10^3 + 10^3 + 2^3 = 2008 and the sum is also 22, even though the problem says distinct integers. Something weird there, if not coincidence.

Anh Vu - 6 years, 11 months ago

The number 2008 is 8 x 251. Since 8 is a cube, the problem might be reduced to find a, b, c where a^3 + b^3 + c3 =251.

Incidentally, Wiki says the smallest cube that is the sum of 3 cubes is 216: 3^3 + 4^3 + 5^3 = 6^3 (wow!!) (movie Pi says 216 is the face of God)

Now 251 is also a weird and wonderful number. According to wiki: 251 is a regular prime, an Eisenstein prime, a Chen prime, a Gaussian prime, a Sophie Germain prime, and a sexy prime.

and some of that means that 2(251)+1 is prime, and that (251) + 6 is prime.

251 is also sum of three consecutive primes (79 + 83 + 89) and the sum of seven consecutive primes (23 + 29 + 31 + 37 + 41 + 43 + 47).

and now here finally they acknowledge but do not explain your observation:

251 is also the smallest number that can be written as the sum of three cubes in two ways: 251 = 1^3 + 5^3 + 5^3 = 2^3 + 3^3 + 6^3.

Member Wilcox - 6 years, 11 months ago

Very informative! It's always good to know these side notes :)

Anh Vu - 6 years, 11 months ago

Problem says distinct integers, and reject when it is not distinct integers. So your answer is OK but that is not what is asked.

Niranjan Khanderia - 6 years, 11 months ago

Yah i know what the problem asks, but I am not concerned with that in my comment though. My question of curiosity is why 10,10,2 is also a solution and its sum is also 22. Is that a coincidence or is there some deeper insight into this?

Anh Vu - 6 years, 11 months ago

I assume (10;10;2) and (12;6;4) would share the same pattern. Says if I have the same given cubic equation and another equation which is the sum of those integer roots a;b;c equals 22, you will be able to find the general solution for a, b, and c, for example, in terms of an arbitrary integer t. Sounds like some Fermat's problem to me. Any comment? :)

Anh Vu - 6 years, 11 months ago

2008=2^3(p^3+q^3+r^3) where a=2p, b=2q,c=2r So p^3+q^3+r^3=251 which cube root is less than 7 , hence options avaialble are 1^3,2^3,3^3,4^3,5^3,6^3 i.e( 1,8,27,64,216) 8+27+216 gives 251 hence a=4,b=6,c=12

SANGRAM DAS - 6 years, 11 months ago

why is it 1728?

Adrian Delos Santos - 6 years, 11 months ago

It is the floor of cube root of 2008.

Niranjan Khanderia - 6 years, 11 months ago

DAAAAAAA i d´ont see ^3 i do ^2 sorry

Vasco Estrada Marques Marques - 6 years, 11 months ago

what is correct answer?

pradeep sharma - 6 years, 11 months ago

Nice solution.

Panya Chunnanonda - 6 years, 7 months ago

ok but 30^2+28^2+18^2=2008 and 30 + 28+18=76

Vasco Estrada Marques Marques - 6 years, 11 months ago

Its so simple 10^3+10 ^3+2^3= 2008.....10+10+2 =22

Sai Krishna Yetukuri - 6 years, 11 months ago

7895421*9999999

pruthvi malvi - 6 years, 11 months ago

What is this ? What has this to do with the problem on hand?

Niranjan Khanderia - 6 years, 11 months ago

78954202104579

saif khan - 6 years, 9 months ago

Dhrupal Shah
Jul 18, 2014

13 > Cube root of 2008 >12............So make a table of N^3, N = 1,2, . . . 12.

10^3 = 1000

2^3 =8

hence 10^3 + 10^3 + 2^3 = 2008

i.e. a+b+c=10+10+2=22

Cubes

The answer is 22.

2 solutions

0 pending reports