 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Two-three integers (Posted on 2017-10-15) Call an integer a two-three integer if it is greater than 1 and has the form 2a * 3b, 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)

 No Solution Yet Submitted by Ady TZIDON No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Analytic Solution | Comment 4 of 6 | 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
 Posted by Brian Smith on 2017-10-15 17:58:37 Please log in:

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