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MCIII Number Theory 1.12
MCIII Number Theory 1.12
Hello,
For this problem (1.12), I used the sum of cubes: 1728000001=(2^2*3*10^2)^3+1^3, which gave me 1201*1442401. How to factor 1442401? I tried 7, 11, 13, 17, 19, 23, 29. None of them work.
Thank you,
Tina Jin
Re: MCIII Number Theory 1.12
You're correct that the large numbers are very hard to factor!
In fact, the trick here is to use algebra and factoring a little more. Applying difference of cubes you should get $(1200+1)(1200^2 - 1200 + 1)$. Now, however, note we can write $$1200^2 - 1200 + 1 = 1200^2 + 2\cdot 1200 + 1 - 3\cdot 1200.$$Now we can apply difference of squares here.
Hope this helps!
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