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, Brian Trial, and Alex Rower.

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=2 k=3
n a(n,k) a(n,k)^k/n! expressions Author a(n,k) a(n,k)^k/n! expressions Author
3 3 1.500 1 + 2 = 3 - - - -
4 5 1.041+ 1 + 4 = 2 + 3 - - - -
5 12 1.200 2·(1+5) = 3·4 5 1.041+ 1 + 4 = 2 + 3 = 5
6 30 1.250 6·(1+4) = 2·3·5 7 .476+ 1 + 6 = 2 + 5 = 3 + 4
7 72 1.028+ 2·(7·5+1) = 3·4·6 12 .342+ 1·(5+7) = 2·6 = 3·4
8 168 .700 1·3·8·(2+5) = 4·6·7 24 .342+ 1·2·(5+7) = 4·6 = 3·8
9 504 .700 1·3·4·6·(2+5) = 7·8·9 72 1.028+ 2·(5·7+1) = 3·4·6 = 8·9
10 1920 1.015+ 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)·12·13 BH 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)·7·8·9·10·13 BH 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)·3·7·8·9·10·13 BH 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)·6·7·9·10·13·16 BH 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)·14·15 BH 33,600 .106+ (17·13+3)·15·10 = (11·9+1)·8·7·6 = (16+4)·14·12·5·2 JD
18 23,950,080 .089+ 2·3·7·9·10·16·(17+5)·18 = 6·8·11·12·(13+4+1)·14·15 BT 75,600 .043+ 18·(16+4)·15·14 = (11·13–3)·9·(8+2)·6 = (17+1)·12·10·7·5 BH
19 74,027,520 .041+ 2·6·8·9·14·15·(16+1)·(19+5) = 3·4·7·10·12·(11+13)·17·18 BH 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=2	k=3
n	a(n,k)	a(n,k)^k/n!	expressions	Author	a(n,k)	a(n,k)^k/n!	expressions	Author
3	3	1.500	1 + 2 = 3		-	-	-	-
4	5	1.041+	1 + 4 = 2 + 3		-	-	-	-
5	12	1.200	2·(1+5) = 3·4		5	1.041+	1 + 4 = 2 + 3 = 5
6	30	1.250	6·(1+4) = 2·3·5		7	.476+	1 + 6 = 2 + 5 = 3 + 4
7	72	1.028+	2·(7·5+1) = 3·4·6		12	.342+	1·(5+7) = 2·6 = 3·4
8	168	.700	1·3·8·(2+5) = 4·6·7		24	.342+	1·2·(5+7) = 4·6 = 3·8
9	504	.700	1·3·4·6·(2+5) = 7·8·9		72	1.028+	2·(5·7+1) = 3·4·6 = 8·9
10	1920	1.015+	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)·12·13	BH	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)·7·8·9·10·13	BH	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)·3·7·8·9·10·13	BH	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)·6·7·9·10·13·16	BH	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)·14·15	BH	33,600	.106+	(17·13+3)·15·10 = (11·9+1)·8·7·6 = (16+4)·14·12·5·2	JD
18	23,950,080	.089+	2·3·7·9·10·16·(17+5)·18 = 6·8·11·12·(13+4+1)·14·15	BT	75,600	.043+	18·(16+4)·15·14 = (11·13–3)·9·(8+2)·6 = (17+1)·12·10·7·5	BH
19	74,027,520	.041+	2·6·8·9·14·15·(16+1)·(19+5) = 3·4·7·10·12·(11+13)·17·18	BH	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=4 k=5
n a(n,k) a(n,k)^k/n! expressions Author a(n,k) a(n,k)^k/n! expressions Author
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 – 7 JD 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 + 1 JD
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)·2 JD
18 5040 .100+ 14·9·8·5 = 12·10·7·6 = (17+4)·16·15 = (11·13–3)·18·2·1 JD

	k=4	k=5
n	a(n,k)	a(n,k)^k/n!	expressions	Author	a(n,k)	a(n,k)^k/n!	expressions	Author
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 – 7	JD	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 + 1	JD
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)·2	JD
18	5040	.100+	14·9·8·5 = 12·10·7·6 = (17+4)·16·15 = (11·13–3)·18·2·1	JD

k=6 k=7
n a(n,k) a(n,k)^k/n! expressions Author a(n,k) a(n,k)^k/n! expressions Author
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 18 .000+ 1·2·9 = 3+15 = 4+14 = 5+13 = 6+12 = 7+11 = 8+10 AR
16 60 .002+ 4·15 = 5·12 = 6·10 = 7·9 – 3 = 2·(14+16) = (13–8)·(11+1) JD 20 .000+ 16+4 = 15+5 = 14+6 = 13+7 = 12+8 = 11+9 = (1+3)/2·10 AR

	k=6	k=7
n	a(n,k)	a(n,k)^k/n!	expressions	Author	a(n,k)	a(n,k)^k/n!	expressions	Author
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	18	.000+	1·2·9 = 3+15 = 4+14 = 5+13 = 6+12 = 7+11 = 8+10	AR
16	60	.002+	4·15 = 5·12 = 6·10 = 7·9 – 3 = 2·(14+16) = (13–8)·(11+1)	JD	20	.000+	16+4 = 15+5 = 14+6 = 13+7 = 12+8 = 11+9 = (1+3)/2·10	AR

k=8 k=9
n a(n,k) a(n,k)^k/n! expressions Author a(n,k) a(n,k)^k/n! expressions Author
15 15 .000+ 1+14 = 2+13 = 3+12 = 4+11 = 5+10 = 6+9 = 7+8 = 15 AR - - - -
16 17 .000+ 1+16 = 2+15 = 3+14 = 4+13 = 5+12 = 6+11 = 7+10 = 8+9 AR - - - -
17 20 .000+ 1·2·10 = 3+17 = 4+16 = 5+15 = 6+14 = 7+13 = 8+12 = 9+11 AR 17 .000+ 1+16 = 2+15 = 3+14 = 4+13 = 5+12 = 6+11 = 7+10 = 8+9 = 17 AR
18 22 .000+ 18+4 = 17+5 = 16+6 = 15+7 = 14+8 = 13+9 = 12+10 = (1+3)/2·11 AR 19 .000+ 1+18 = 2+17 = 3+16 = 4+15 = 5+14 = 6+13 = 7+12 = 8+11 = 9+10 AR
19 - - - - 22 .000+ 1·2·11 = 3+19 = 4+18 = 5+17 = 6+16 = 7+15 = 8+14 = 9+13 = 10+12 AR
20 - - - - 24 .000+ 20+4 = 19+5 = 18+6 = 17+7 = 16+8 = 15+9 = 14+10 = 13+11 = (1+3)/2·12 AR

	k=8	k=9
n	a(n,k)	a(n,k)^k/n!	expressions	Author	a(n,k)	a(n,k)^k/n!	expressions	Author
15	15	.000+	1+14 = 2+13 = 3+12 = 4+11 = 5+10 = 6+9 = 7+8 = 15	AR	-	-	-	-
16	17	.000+	1+16 = 2+15 = 3+14 = 4+13 = 5+12 = 6+11 = 7+10 = 8+9	AR	-	-	-	-
17	20	.000+	1·2·10 = 3+17 = 4+16 = 5+15 = 6+14 = 7+13 = 8+12 = 9+11	AR	17	.000+	1+16 = 2+15 = 3+14 = 4+13 = 5+12 = 6+11 = 7+10 = 8+9 = 17	AR
18	22	.000+	18+4 = 17+5 = 16+6 = 15+7 = 14+8 = 13+9 = 12+10 = (1+3)/2·11	AR	19	.000+	1+18 = 2+17 = 3+16 = 4+15 = 5+14 = 6+13 = 7+12 = 8+11 = 9+10	AR
19	-	-	-	-	22	.000+	1·2·11 = 3+19 = 4+18 = 5+17 = 6+16 = 7+15 = 8+14 = 9+13 = 10+12	AR
20	-	-	-	-	24	.000+	20+4 = 19+5 = 18+6 = 17+7 = 16+8 = 15+9 = 14+10 = 13+11 = (1+3)/2·12	AR

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.