Problem of the Month (December 1999)

An old puzzle is: Write every number from 1 to 100 using four 4's, and the symbols +, -, *, /, and ^. This month we explore writing numbers with copies of one digit. Since we are about to enter year 2000, how many of each digit are needed to make 2000?

Let D(d,n) be the minimum number of copies of digit d that are needed to make the number n. Can you find D(d,n) for small n? What's the largest n for which D(d,n)=k? What's the smallest n for which D(d,n)=k? Can you find any bounds on the function D(d,n)? For which digit d is D(d,n) usually the lowest?


ANSWERS

Joe DeVincentis wrote a computer program that calculated all the numbers that could be made with 8 or less of a digit. Therefore most of the results below are due to him.

Philippe Fondanaiche found many minimal solutions to the 2000 problem, and gave many of the values of D(d,n) for small n.

The shortest ways to make 2000 with copies of a single digit are:

(1+1)×(11–1)1+1+1

2222–222

((33+3)/3)3×(3–(3/3))
(3×3+3/3)3×(3–(3/3))
((33–3)/3)3×(3–(3/3))
((33+3)/3)3×(3+3)/3
(3×3+3/3)3×(3+3)/3
((33–3)/3)3×(3+3)/3
(3+3)×333+(3+3)/3
(3+3)×333+3–3/3
(3/3/3+333)×(3+3)
((333)/3+3)/(3+3)+3

(44–4)×(4+4)–4×4
44×(4+4)–44–4
(4×4+4)(4–4/4)/4
44×44+44/4
44×44+4×4×4

5×5×(55+5×5)
(5+5)5/(5×(5+5))
(5+5)(5–5/5)/5
(5+5)5/(55–5)

((6+6+6)×666+6+6)/6 (EF)

(7×7+7/7)×(7×7–7–(7+7)/7)
7×(7×(7×7–7)–7)–7–(7+7)/7
((77/7+7)×777+7+7)/7
((7+7)/7)77/7–7×7+7/7

(88–8)×(8+8+8+8/8)
((((8+8)/8)8)–8)×8+8+8
8×((8+8)×(8+8)–8)+8+8

(999+9/9)×(9+9)/9
(99/9+9)×(99+9/9)
((9+9)×999+9+9)/9

Here are the shortest known ways to make n with copies of the digit d:

Fewest Number of Digits d to Make n

n \ d123456789
11 2/23/34/45/56/67/78/89/9
21+1 2(3+3)/3(4+4)/4(5+5)/5 (6+6)/6(7+7)/7(8+8)/8(9+9)/9
31+1+1 2+(2/2)34-(4/4) (5+5+5)/5(6+6+6)/6(7+7+7)/7(8+8+8)/8(9+9+9)/9
41+1+1+1 2+23+(3/3)4 5-(5/5)6-(6+6)/677/7-78*8/(8+8) (9+9+9+9)/9
51+1+1+1+1 2+2+(2/2)3+3-(3/3)4+(4/4) 56-(6)/67-(7+7)/7 8-(8+8+8)/8(99-9)/(9+9)
6(1+1)(1+1+1) 2+2+23+34+(4+4)/4 5+(5/5)67-(7/7) 8-(8+8)/8(99+9)/(9+9)
711-1-1-1-1 22/2-2-23+3+(3/3)4+4-(4/4) 5+(5+5)/56+(6/6)7 8-(8/8)9-(9+9)/9
811-1-1-1 2(2+2)3*3-(3/3)4+4 5+(5+5+5)/56+(6+6)/67+(7/7) 89-(9/9)
911-1-1 22/2-23*34+4+(4/4) 5+5-(5/5)6+(6+6+6)/67+(7+7)/7 8+(8/8)9
1011-1 (22-2)/2(33-3)/3(44-4)/4 5+5 (66-6)/6(77-7)/7 (88-8)/89+(9/9)
1111 22/233/344/455/566/677/788/899/9
1211+1 (22+2)/23*3+34+4+4(55+5)/5 6+6(77+7)/7(88+8)/8(99+9)/9
1311+1+1 22/2+23*3+3+(3/3)4+4+4+(4/4) (55+5+5)/56+6+(6/6)7+7-(7/7) (88+8+8)/8(99+9+9)/9
1411+1+1+1 (2+2)2-233/3+3(44-4)/4+4 5+5+5-(5/5)6+6+(6+6)/6 7+7 8+8-(8+8)/8(99+9+9+9)/9
1511+1+1+1+1 22+2-(2/2)3*3+3+344/4+4 5+5+56+6+(6+6+6)/6 7+7+(7/7)8+8-(8/8)9+9-(9+9+9)/9
16(1+1+1+1)1+1 22+233-(33/3) 4*455/5+5(66-6)/6+6 7+7+(7+7)/78+8 9+9-(9+9)/9
17(1+1)(11-1-1)-1 22+2+(2/2)3*(3+3)-(3/3) 4*4+(4/4)(55+5)/5+566/6+6 (77-7)/7+78+8+(8/8)9+9-(9/9)
18(1+1)(11-1-1) 22-2-23*(3+3)4*4+(4+4)/4 (55+5+5)/5+56+6+6 77/7+78+8+(8+8)/89+9
19(1+1)(11-1)-1 22-2-(2/2)3*(3+3)+(3/3) 4*4+4-(4/4)5*5-5-(5/5) 6+6+6+(6/6)(77+7)/7+7 88/8+89+9+(9/9)
20(1+1)(11-1) 22-233/3+3*3 4*4+45*5-5 6+6+6+(6+6)/67+7+7-(7/7) (88+8)/8+899/9+9
21 11+11-1 22-(2/2) 33-3-3 4*4+4+(4/4) 5*5-5+(5/5) (66+66-6)/6 7+7+7 (88+88-8)/8 (99+9)/9+9
22 11+11 22 (33+33)/3 (44+44)/4 (55+55)/5 (66+66)/6 (77+77)/7 (88+88)/8 (99+99)/9

The largest number we can make with n copies of d seems to be one of the following forms:

For d=1, 11111111 for n even and 111111111 for n odd. For d=2 or d=3, 22222. For d=4 through d=9, 4444.

Kit Vongmahadlek asked what the smallest number that could not be made with seven 7's was. This inspired Joe DeVincentis to crank these out for other values as well. The smallest numbers requiring n copies of d are:

Smallest Numbers Requiring n copies of d

n \ d123456789
1 123456789
2 211111111
3 332222222
4 455633333
5 571313891654
6 72740231815232114
7 172946783850302924
8 4114913098838713114943
9 76????160???


Philippe Fondanaiche asked whether 2000 could be made from 8 different digits in increasing order without using concatenation. Here are some solutions:

(2+3)4/5×(6–7+8+9)
(–1+34)×(–5+6+7+8+9)
1×2×45–6×(7–8+9)
1×(2(3+5)–6)×(7–8+9)
1×23+4×(–6+7×8×9)
–1+2+3+4×(–5+7×8×9)
–1+((2–3+4)5+6)×8×9
1×(2+3)(4+5–6)×(7+9)
(1–2+3)×4×5×(6×7+8)

Luc Kumps points out that in the old four 4's puzzle, you need other symbols like "√" or "!" to make the numbers from 1-100.


If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 12/15/02.