Problem of the Month (July 2015)

Using the positive integers from 1 to n each used exactly once, along with any number of the 4 arithmetic symbols (+, – ·, ÷) and parentheses, what is the equation you can make with the largest number on both sides? What is the largest number a(n,k) that can be represented if there are k equal expressions, which together use the numbers 1 through n exactly once? How fast do these numbers grow? Is it true that a(n,k)k / n! → 1 as n → ∞?


ANSWERS

Solutions were received from Joe DeVincentis, Jon Palin, Bryce Herdt, and Brian Trial.

When k=1, Bryce Herdt pointed out the optimal expression is n·(n-1)·. . .·(2+1) = (3/2)n!

Here are the best known answers:

k=2k=3
na(n,k)a(n,k)k/n!expressionsAuthora(n,k)a(n,k)k/n!expressionsAuthor
3 31.5001 + 2 = 3  ----
4 51.041+1 + 4 = 2 + 3  ----
5 121.2002·(1+5) = 3·4  51.041+1 + 4 = 2 + 3 = 5 
6 301.2506·(1+4) = 2·3·5  7.476+1 + 6 = 2 + 5 = 3 + 4 
7 721.028+2·(7·5+1) = 3·4·6  12.342+1·(5+7) = 2·6 = 3·4 
8 168.7001·3·8·(2+5) = 4·6·7  24.342+1·2·(5+7) = 4·6 = 3·8 
9 504.7001·3·4·6·(2+5) = 7·8·9  721.028+2·(5·7+1) = 3·4·6 = 8·9 
10 19201.015+2·4·5·6·8 = 3·10·(7·9+1) JP 90.200+1·(2+4)·(7+8) = 3·5·6 = 9·10  
11 5040.636+3·5·6·7·8 = 4·9·10·(11+2+1) BH 180.146+(1+8)·(2+7+11) = 3·6·10 = 4·5·9  
12 15,840.523+4·5·6·12·(1+3+7) = 2·8·9·10·11   480.230+6·8·10 = 2·4·5·12 = 3·(11+9)·(7+1) JD
13 65,520.689+7·8·9·10·(11+2) = 3·4·5·(6+1)·12·13BH 1440.479+2·8·9·10 = 4·5·6·12 = (7+3)·(13·11+1) 
14 262,080.787+4·5·6·(11+2)·12·14 = (1+3)·7·8·9·10·13BH 2340.146+3·6·10·13 = 4·5·9·(11+2) = 12·(7+8)·(14–1)JD
15 982,800.738+5·6·(11+2)·12·14·15 = (1+4)·3·7·8·9·10·13BH 6480.208+3·15·(11·13+1) = 6·10·12·(7+2) = 5·8·9·(14+4) 
16 3,931,200.738+3·5·8·(11+2)·12·14·15 = (1+4)·6·7·9·10·13·16BH 11,520.073+5·16·(11·13+1) = 4·12·15·(7+9) = 8·10·(6+3)·(14+2)JD
17 7,983,360.145+3·7·10·11·12·16·(17+1) = 4·6·8·(9+2)·(13+5)·14·15BH 33,600.106+(17·13+3)·15·10 = (11·9+1)·8·7·6 = (16+4)·14·12·5·2JD
18 23,950,080.089+2·3·7·9·10·16·(17+5)·18 = 6·8·11·12·(13+4+1)·14·15BT 75,600.043+18·(16+4)·15·14 = (11·13–3)·9·(8+2)·6 = (17+1)·12·10·7·5BH
19 74,027,520.041+2·6·8·9·14·15·(16+1)·(19+5) = 3·4·7·10·12·(11+13)·17·18BH 77,760.003+1·2·3·4·5·6·9·12 = 8·10·(13+14)·(17+19) = 15·16·18·(11+7)BT
20 411,264,000.069+14·15·16·17·18·(19+1)·20 = 2·3·5·6·7·8·10·(9+11)·12·(13+4)BT

k=4k=5
na(n,k)a(n,k)k/n!expressionsAuthora(n,k)a(n,k)k/n!expressionsAuthor
7 7.476+1 + 6 = 2 + 5 = 3 + 4 = 7  ----
8 9.162+1 + 8 = 2 + 7 = 3 + 6 = 4 + 5  ----
9 12.057+1·(5+7) = 4 + 8 = 3 + 9 = 2·6  9.162+1 + 8 = 2 + 7 = 3 + 6 = 4 + 5 = 9 
10 18.028+3·6 = 2·9 = 10 + 8 = (1+4)·5 – 7JD 11.044+1 + 10 = 2 + 9 = 3 + 8 = 4 + 7 = 5 + 6 
11 40.064+2·(9+11) = 3·(6+7) + 1 = 4·10 = 5·8  14.013+3 + 11 = 4 + 10 = 5 + 9 = 6 + 8 = 1·2·7 
12 80.085+5·(12+4) = (1+9)·(2+6) = 7·11 + 3 = 8·10  20.006+4·5 = 2·10 = 11 + 9 = 12 + 8 = (6–1)·(7–3)JD
13 120.033+10·12 = 5·(11+13) = 7·(9+6) = 4·(7–1)·(3+2)JD 24.001+3·8 = 4·6 = 2·12 = 11 + 13 = 10·7 ÷ 5 + 9 + 1JD
14 240.038+3·8·10 = 4·5·12 = (14+6)·(11+1) = (13+2)·(7+9)JD 72.022+6·12 = 8·9 = 14·5 + 2 = 3·(7+11) = 3·(13+10+1)JD
15 720.205+4·12·15 = 8·9·10 = 5·(11·13+1) = 6·(7+3)·(14–2)JD 180.144+12·15 = 3·6·10 = 4·5·9 = 13·14 – 2 = (11+1)·(7+8)JD
16 1440.205+9·10·16 = 3·5·8·12 = 15·(14·7–2) = (6+4)·(11·13+1)JD 240.038+15·16 = 2·10·12 = 5·6·8 = (11+4)·(9+7) = (14+1)·(13+3)JD
17 3360.358+16·15·14 = 12·8·7·5 = (13+1)·10·6·4 = (17·11·9–3)·2JD
18 5040.100+14·9·8·5 = 12·10·7·6 = (17+4)·16·15 = (11·13–3)·18·2·1JD

k=6k=7
na(n,k)a(n,k)k/n!expressionsAuthora(n,k)a(n,k)k/n!expressionsAuthor
11 11.044+1 + 10 = 2 + 9 = 3 + 8 = 4 + 7 = 5 + 6 = 11  ----
12 13.010+1 + 12 = 2 + 11 = 3 + 10 = 4 + 9 = 5 + 8 = 6 + 7  ----
13 16.002+1·2·8 = 3 + 13 = 4 + 12 = 5 + 11 = 6 + 10 = 7 + 9  13.010+1 + 12 = 2 + 11 = 3 + 10 = 4 + 9 = 5 + 8 = 6 + 7 = 13 
14 24.002+2·12 = 3·8 = 4·6 = 10 + 14 = 11 + 13 = (9–5)·(7–1)  15.001+1 + 14 = 2 + 13 = 3 + 12 = 4 + 11 = 5 + 10 = 6 + 9 = 7 + 8 
15 24.000+3·8 = 4·6 = 15 + 9 = 14 + 10 = 13 + 11 = 12·2·(7–5–1)JD
16 60.002+4·15 = 5·12 = 6·10 = 7·9 – 3 = 2·(14+16) = (13–8)·(11+1)JD


If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 7/1/15.