Call an integer a two-three integer if it is greater than 1 and has the form 2
a * 3
b, where a ≥ 0 and b ≥ 0.
Express 19992000 as a sum of two-three integers so that none of the addends divides another.
Source: Eighth Korean Mathematical Olympiad (Crux Math. Dec 1999)
The wording for this problem was very familiar, I eventually tracked down
Interesting representation. This problem is just an implementation of the older one. So I will apply the method I described there to the number 19992000.
19992000
= 2^6 * 3 * 104125
= 2^6 * 3 * (3^10 + 45076)
= 2^6 * 3 * (3^10 + 2^2 * 11269)
= 2^6 * 3 * (3^10 + 2^2 * (3^8 + 4708))
= 2^6 * 3 * (3^10 + 2^2 * (3^8 + 2^2 * 1177))
= 2^6 * 3 * (3^10 + 2^2 * (3^8 + 2^2 * (3^6 + 448)))
= 2^6 * 3 * (3^10 + 2^2 * (3^8 + 2^2 * (3^6 + 2^6 * 7)))
= 2^6 * 3 * (3^10 + 2^2 * (3^8 + 2^2 * (3^6 + 2^6 * (3 + 2^2))))
= 2^6 * 3 * (3^10 + 2^2 * (3^8 + 2^2 * (3^6 + 2^6 * 3 + 2^8)))
= 2^6 * 3 * (3^10 + 2^2 * (3^8 + 2^2 * 3^6 + 2^8 * 3 + 2^10))
= 2^6 * 3 * (3^10 + 2^2 * 3^8 + 2^4 * 3^6 + 2^10 * 3 + 2^12)
= 2^6 * 3^11 + 2^8 * 3^9 + 2^10 * 3^7 + 2^16 * 3^2 + 2^18 * 3
Corrected a typo Charlie found.
Edited on October 15, 2017, 9:55 pm
Analytic Solution
