THE FOUR DIGITS PROBLEM

The Four Digits Problem by Andy Pepperdine 2007-01-12 (re-formatted 2010-01-02) based on notes made 50 years ago.

This work is released under the Creative Commons Attribution – Share-alike 2.5 license [1].

Make all integers from 0 upwards using the simplest mathematical operators, as defined in some way, and the digits 1, 2, 3 and 4. W. W. Rouse Ball gives this in his book Mathematical Recreations and Essays[2] as a more interesting exercise than the normal Four Fours where all the digits are 4. He states on page 15, that with the digits 1, 2, 3 and 4, and allowing the notation of the denary scale (including decimals), as also algebraic sums, products, and positive integral powers, all numbers from 1 up to 88 can be made. If the use of symbols for square roots and factorials, repeated a finite number of times, is also permitted, then we can get to 264; if negative indices are also permitted, to 276; and if fractional indices are permitted, to 312. NB: I interpret the last comment about “fractional indices” to mean the extraction of roots other than the square root by saying which one to extract. I cannot find any other solution to 277 and 307 otherwise. The 89 values I cannot find up to 1000 are: 313, 331, 391, 407, 419, 421, 437, 439, 443, 446, 454, 455, 457, 461, 467, 493, 557, 559, 586, 587, 614, 617, 633, 653, 659, 661, 662, 667, 772, 773, 779, 787, 788, 791, 807, 813, 815, 817, 823, 831, 833, 850, 851, 853, 857, 859, 861, 867, 871, 874, 877, 879, 881, 883, 885, 886, 887, 893, 897, 906, 907, 908, 911, 913, 914, 915, 917, 919, 921, 922, 923, 927, 929, 931, 932, 933, 935, 941, 942, 946, 947, 949, 951, 953, 956, 979, 983, 986, 989. Examples of each number from 0 upwards are given in the following table. I've tried to keep to the order in which Rouse Ball postulated the use of operators, but also, I've used factorial in preference to the square root sign, and allowed recurring decimals and integral powers in preference to all other powers, factorials and roots. Some interesting alternative methods are noted, too. Five digits Rouse Ball also gives the limits for 5 digits (1, 2, 3, 4 & 5) of 3832 without negative and fractional powers, and of 4282 with. My investigations indicate the same.

Digits-1234.odt 1 THE FOUR DIGITS PROBLEM

No. Method No. Method 0 14−3−2 33 341−2 1 32−4×1 34 34×2–1 2 123−4 35 342−1 3 2×31−4 36 324×1 4 4×2−3−1 37 2314 5 43−2×1 38 42–3−1 6 431−2 39 42–3×1 7 31−24 40 12×34 8 432−1 41 43–2×1 9 432×1 42 413−2 10 4321 43 43×2– 1 11 42−31 44 4×13 –2 12 4×3×2−1 45 423×1 13 34−21 46 3214 14 21–3−4 47 412×3 15 134−2 48 3×142 16 4×32−1 49 4123 17 3×142 50 13×4−2 18 32−14 51 12×43 19 23–4×1 3 52 2× −4 or 43−12 20 213−4 ·1  21 24–3×1 2 53 3 22 43−21 ·1×·4 23 31–2×4 54 4×132 24 4×123 55 4213 3 25 31–4−2 56 41 ·2 26 234−1 4 27 234×1 57 3× −1  ·2  28 32–4×1 58 31×2−4 29 42−13 59 31×3−4 30 13×24 60 4×312 31 43−12 13 61 −4 or 43 –2−1 32 3×12−4 ·2

Digits-1234.odt 2 THE FOUR DIGITS PROBLEM

No. Method No. Method 31×4 88 4×23 –1 62 2 13 4! 4! −31 4 89 or 63 23 ·2 ·2 1 · 3×12 90 64 2341 · 4 13×2 65 91 23×4−1 · 4 92 23×4×1 66 31×24 93 23×41 134 67 94 2×413! 2 3 68 34×2×1 2 95 ·4 or 4!×3!−1−1 69 3×24−1 ·1 70 2×314 96 4×213 71 24×3−1 97 14!–23 72 4×21 –3 98 3!–1!–4!2 73 3241 3×4 99 or 123–4! 21 ·1˙ 2˙ 74 4 ·3 3×4 100 or 4−3×·1−2 75 3×214 ·12

76 2×41–3 4 2 101 3  or ·2−3– 4! 31−·2 ·1 77 ·4 102 34×21 78 13×24 103 3!!×·2−41 79 41×2−3 104 13×4×2 32×1 42 80 or 34−21 105 ·4 ·3·1 81 21×4−3 3×4 106 −2 41 ·1˙ 82 ·2·3 4! 107 −13 2×4 ·2 83 3 ·1 108 4×123 84 14×3×2 4!–1 109 –3! 85 41×23 ·2 2×43 86 43×2×1 110 87 21×43 ·1

Digits-1234.odt 3 THE FOUR DIGITS PROBLEM

No. Method No. Method 111 14!−32 31–4 135 ·2 3 112 4× −2  ·1  136 1342 3 1 ·4 3 113 !−4−3 137 ·1˙ or 4 ·2 ·2×·1˙ ·2 4 114 3× −2 4  ·1  2 4 138 ·3 or 3!–1! ·2˙ 115 23×41 ·1˙ 116 4×31 –2 139 142−3 117 3×41 –2 42×1 140 3×4 ·3 118 −2 ·1 141 143−2 119 123−4 142 3!×4!–2×1 24 2 120 143 3×4 −1 ·3−·1 144 12×4×3 121 124−3 145 1423 122 31×4−2 3 146 −4 123 3×42–1 ·1×·2 124 4×32 –1 147 21×34 125 41×32 3 −·4 126 42×3×1 148 ·1 3!× 4!1−2 127 1234 ·2 128 132−4 149 3!!×·241 3×2 129 3×412 150 ·1×·4 41–2 130 or 1243! 31 ·3 151 −4 ·2 3 131 −4 3 ·2×·1˙ ·4 152 ·1 132 134−2 ·2 3 −·4 4! 31−·4 133 ·1˙ or 13 153 ·2 ·2 ·2 3 4 154 4 ·2 ·1×·2 134 ·1 or 32!14 ·3

Digits-1234.odt 4 THE FOUR DIGITS PROBLEM

No. Method No. Method 31×2 4 155 177 −3 ·4 ·2×·1˙ 156  ! 4! 132 4 −2 178 ˙ 314 ·13 157 2 31 179 4! ·2˙ ·2 158 −4 or 3!!×·214 ·1˙ 3 3×42 180 31 ·1 159 4 ·2 3×2! 181 1 160 32×41 4 4 4! 161 2×3 −1 or 32!41 182 2 ·13˙ 162 1×2×34 4 4 183 3 163 2×3 1 ·2×·1˙ 4 164 2×3 1 184 23×4!−1 34–1 4−·3 165 185 ·2 ·1×·2 ·2˙ 166 4 186 31×24 1˙ 3 · 4!–3 187 −2 167 3!!×·24!−1 ·1˙ 168 42×31 4! 188 3!!×·1˙ 34 ·2˙ 169 −1 ·2 3×14 189 34×1 ·2˙ 170 ·2 41–3 190 34 ·2 171 1 ·2 191 4!×23−1 4 3 172 3!!×·2˙ 192 4 ×21 ·1˙ 193 142−3 173 1324 43·1˙ 3 3!! 194 174 4! or –21! .2˙ ·1×·2 4 4 3 4!–2 314 195 ·1˙ or −3 175 .1˙ ·2 ·2 ·2 – 4! ·2 176 3 196 −4 ·1 ·13

Digits-1234.odt 5 THE FOUR DIGITS PROBLEM

No. Method No. Method 4 24 197 −3 216 3 ·1×·2 ·1˙ 413 217 2134 198 ˙ ·2 218 3!4–12 2 199 14 3 24 219 3 4×32 ·1˙ 200 ·1 413 220 4! ·2 201 3 ·1˙ 2˙ 4! 221 32 41 ·1˙ 202 −3 ·2 24 222 3! 4 ·1˙ 203 3 ·1×·2 4!–2 223 3 ·2 ·1 204 3 4 ·1 224 42!×·31˙ 205 41×23 225 3×412 23 23 206 – 4! 226 −4 ·1 ·1 4 3 4! 2 227 231−4 207 ·2 or −3 ·1˙ 4 23 ·1˙ 228 3!× −2 or –4  ·1  ·1 41 4! 3 208 3 or −2 ·2 ·1˙ 229 231 –4 4 209 213−4 3 4!–1 230 ·2 or 14×3 ·3−·2 210 ·1 ·2 23 211 214−3 231 4! ·1˙ 4!–3 212 2 3! 4! 3 ·1 232 4× – 2 or −2  .1  ·1 24 213 −3 ·1˙ 233 234−1 43 234 234×1 214 −1 ·2 235 2341 43×1 236 2×3!–1!−4 215 ·2 24 237 −3 ·1

Digits-1234.odt 6 THE FOUR DIGITS PROBLEM

No. Method No. Method 238 241−3 264 4!×13−2

1 −2  ·2 ·1 3! 239 3 −4 265 ·4 4×3×2 240 3! ·1 266 −4 ·2×·1˙ 241 41 3 −2 1 4! 3! 267  ·2 or  242 243−1 3 4! ·1 ·2˙ 243 243×1 268 2×134 244 2431 23! 269 −1 41 ˙ 245 3 2 ·4 234 246 41×3×2 270 1 ·1 247  ·2 3 4 23! 271 1 248 31×4×2 ·4˙

4! 2 4! 249 3 272 32 ·1 ·1 3 4! 250 2×41 273 3 ·2−·1˙ 251 3!×42−1 3! 252 21×4×3 274 4 ·2×·1˙ 253 3!×421 4!31 31 275 254 4 −2 ·2 255 2314! 23×4 276 41!×2·3 or 256 432−1 ·1˙ −·2 4!2 277 34 ·1˙ 257 −3  ·1 − 3!!×·4−·1 2 or 258 4312 278 −4! 4 ·1 –3!!−2 259  3! −1 ×·2 3·4˙ 13×4 279 260 ·1˙ 2 ·2 21×4 4!32 280 261 ·3 ·1˙ ·2−3−·1˙ 262 2 4!×·1˙ 3! or 3!!×·1˙4!−2 281 ·4˙ 4! 263 23 282 4!×12–3! ·1

Digits-1234.odt 7 THE FOUR DIGITS PROBLEM

No. Method No. Method 1 1 283 3!!×·4 – ! ·2−3 –4! ·2 303  ·2  or 3 ·1˙ 32 ·4 284 −4 ·1˙ 34 3! 304 −2 or 4 285 4!×12−3 ·1˙ ·2×·1 286 143×2 3!–1!2 305 3 4 4 · 287 −1 ·2˙ 34 324 306 or ˙ ˙ ˙ 34–2 ·2−·1 ·1 288 ·1˙ 307 ·3×·1 4−·2 289 1342 308 312−4 4!32 309 3!!×·4˙ 21 290 ·1 310 4!×13−2

291 4!×123 −2 311 3!!×·4˙ –·1˙ 32 292 4 312 314−2 ·1˙ 1 293 3!!×·4 ·2 294 42×3!1 295 3!!×·41−·2 3 296 −4 ·12 132 297 or 321–4! ·4˙ 3! −·4 4!3! 298 ·1 or −2 ·1 ·2 32! 299 −1 ·4 3 300 or 3!!×·412 ·14−2 32! 301 1 ·4 3! ·4 4!3! 302 ·1 or 2 ·1 ·2

Digits-1234.odt 8 THE FOUR DIGITS PROBLEM

Notation

The midpoint (·) is used for the decimal point. Recurring decimals are written with dots over the first and last digit of the recurring sequence. So ·1˙ is the same as ·1111 (i.e. one ninth), ·1˙ 2˙ is the same as ·121212 (i.e. twelve ninety-ninths, or four thirty-thirds), and ·13˙ is the same as ·133333 (i.e. twelve ninetieths, or two fifteenths). Some care is required to see these dots and some printers may not be able to print them well. Addition, subtraction and multiplication are as normal, but division is always given as a fraction. The factorial function is represented as usual by a ! giving 3! = 6 and 4! = 24 . Square roots are also as usual, with 4 = 2 and ·1˙ = 1/3 . To get 277 and 307, and some higher numbers, I have assumed that Rouse Ball was using the notation of an n-th root where n is not integral. So ·1 2 = 210 = 1024 , and negative roots are the equivalent of reciprocal powers, so −·2 ·3˙ = 1/3−5 = 35 = 243 . References

[1] Creative Commons License, http://creativecommons.org/licenses/by/2.5/ [2] W.W. Rouse Ball, revised by H.S.M. Coxeter, Mathematical Recreations and Essays, 11th edition, 1939, reprinted in 1967 by Macmillan, p15ff.

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