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**