2
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So here's how it works: Take the digits from 1991. You can use the digits in any order, only once (you can't make a 19 or 91), with any operation sign to get answers between 1-100. For example: $4=(1+1)^2-9+9$. You cannot use the zero power, but any other powers are good. Powers don't count as using one of the digits. Again, you may not form two digit numbers using your numbers. Logs may be used three times.

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9
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Nobody is going to check this anyway, but here it goes. I hope all the math is right.

$1 = 9 / 9 + 1 - 1$
$2 = 9 / 9 * 1 + 1$
$3 = 9 / 9 + 1 + 1$
$4 = (1+1)^2 - 9 + 9$
$5 = (1+1)^2 + 9 /9$
$6 = (9*9)^{1/4} * (1 + 1) $
$7 = 9^{1/2} + 9^{1/2} + 1 * 1$
$8 = (1+1*9/9)^3$
$9 = (9 / 9 + 1 + 1)^2$
$10 = (9*9)^{1/2} + 1*1$
$11 = (9*9)^{1/2} + 1+1$
$12 = (9*1*1)^{1/2} + 9$
$13 = (9*1)^{1/2} + 9 + 1$
$14 = (9)^{1/2} + 9+1+1$
$15 = 9*(1+1) - 9^{1/2}$
$16 = (1+1*9/9)^4$
$17 = 9*1 + 9 - 1$
$18 = 9 + 9 +1 -1$
$19 = 9 + 9 +1 *1$
$20 = 9 + 9 +1 +1$
$21 = (9 -1 -1) * 9^(1/2)$
$22 = 9^{3/2} - (9^{1/2} +1+1)$
$23 = 1 * 9^{3/2} - (9^{1/2} +1)$
$24 = 1 * 1 * 9^{3/2} - 9^{1/2})$
$25 = ((1+1)^2 + 9 /9)^2$
$26 = (9*9)^{3/4} -1*1$
$27 = (9 / 9 + 1 + 1)^3$
$28 = (9 * 9)^{3/4} + 1 * 1$
$29 = (9 * 9)^{3/4} + 1 + 1$
$30 = 9^{1/2} * 1 * (9+1)$
$31 = 9^{1/2} * (9+1) + 1$
$32 = (1+1*9/9)^5$
$33 = 9^{1/2} * (9 + 1 + 1)$
$34 = 9^{3/2} + 9 - 1 - 1$
$35 = 9^{3/2} * 1 + 9 - 1$
$36 = ((9*9)^{1/4} * (1 + 1))^2$
$37 = 9^{3/2} + 9 + 1 * 1$
$38 = 9^{3/2} + 9 + 1 + 1$
$39 = 9^{3/2} + 9^{1/2} * (1+1)^2$
$40 = (9^{1/2} + 1) * (9 + 1)$
$41 = (1+1)^5 + (9*9)^{1/2}$
$42 = 9^{3/2} - 1 + (9^{1/2} + 1)^2$
$43 = (1+1)^4 + (9/(9^{1/2}))^3$
$44 = (9^{1/2} + 9^{1/2})^2 + (1+1)^3$
$45 = (9^{1/2} + 1 + 1) * 9$
$46 = 9^2 - 9^{3/2} - (1+1)^3$
$47 = (9^{1/2} + 1)^2 * 9^{1/2} - 1$
$48 = (9^{1/2} + 9^{1/2} + 1)^2 - 1$
$49 = (9^{1/2} + 9^{1/2} + 1 * 1)^2$
$50 = 9^2 - 9^{3/2} - (1+1)^2$
$51 = 9^{3/2} + 9^{1/2} * (1+1)^3$
$52 = 9^2 - 9^{3/2} - 1 - 1$
$53 = 9^2 - 9^{3/2} - 1 * 1$
$54 = 9^{1/2} * (1 + 1) * 9$
$55 = (1+1)^6 - (9*9)^{1/2}$
$56 = (9-1)^2 - 9 + 1$
$57 = (9^{3/2} - (1 + 1)^3) * 9^{1/2}$
$58 = (1+1)^6 - 9^{1/2} - 9^{1/2}$
$59 = (9^{1/2} - 1)^5 + 9^{3/2} * 1$
$60 = (9^{1/2} + 1)^3 - 9^{1/2} - 1$
$61 = (9^{1/2} + 1)^3 - 9^{1/2} * 1$
$62 = (9^{1/2} + 1)^3 - 9^{1/2} + 1$
$63 = 9* (9-1-1)$
$64 = (1+1*9/9)^6$
$65 = (1+9/9)^6 + 1$
$66 = (9^{1/2} + 1)^3 + 9^{1/2} - 1$
$67 = (9^{1/2} + 1)^3 + 9^{1/2} * 1$
$68 = (9^{1/2} + 1)^3 + 9^{1/2} + 1$
$69 = (9^{3/2} - (1 + 1)^2) * 9^{1/2}$
$70 = 9^2 - 9 - 1 - 1$
$71 = 9* (9-1) - 1$
$72 = 9* 1* (9-1)$
$73 = 9* (9-1) + 1$
$74 = 9^2 - 9 + 1 + 1$
$75 = (9^{3/2} - 1 - 1) * 9^{1/2}$
$76 = 9^2 - 9^{1/2} - 1 - 1$
$77 = (9^{3/2} - 1) * 9^{1/2} - 1$
$78 = (9^{3/2} - 1) * 9^{1/2} * 1$
$79 = 9 * 9 - 1 - 1$
$80 = 9 * 9 * 1 - 1$
$81 = 9 * 9 * 1 * 1$
$82 = 9 * 9 * 1 + 1$
$83 = 9 * 9 + 1 + 1$
$84 = (9^{3/2} + 1) * 9^{1/2} * 1$
$85 = (9^{3/2} + 1) * 9^{1/2} + 1$
$86 = 9^2 + 9^{1/2} + 1 + 1$
$87 = ((1+1)^5 - 9^{1/2}) * 9^{1/2}$
$88 = 9^2 + 9 - 1 - 1$
$89 = 9* (9+1) - 1$
$90 = 9* 1* (9+1)$
$91 = 9* (9+1) + 1$
$92 = 9^2 + 9 + 1 + 1$
$93 = (1+1)^5 * 9^{1/2} - 9^{1/2}$
$94 = (1+1)^6 + 9^{3/2} + 9^{1/2}$
$95 = 9^2 + (\log_39)^4 - 1 -1$
$96 = (1+1)^5 * (9*9)^{1/4} $
$97 = 9*9 + (1+1)^4$
$98 = 9^2 + (1+9^{1/2})^2 + 1$
$99 = 9* (9+1+1)$
$100 = (9+1)*(9+1)$

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  • $\begingroup$ Nice. I am currently working under the assumption that multiplication and division are not allowed, as the question specifies "operation sign" - although I did think about asking. $\endgroup$ – Jonathan Allan Mar 9 '16 at 11:15
  • $\begingroup$ I fixed some MathJax typos and checked your mathematics. Only 95 is incorrect (currently it makes 105) - I'll let you change that. $\endgroup$ – Jonathan Allan Mar 9 '16 at 15:03
  • $\begingroup$ @JonathanAllan. Damn it:). Thanks for this. actually 95 was the last one I did. I think the adrenaline rush of finishing this made me screw up. $\endgroup$ – Marius Mar 9 '16 at 15:09
4
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Using only addition and subtraction as the operators, only integer powers, and only employing $\log_39=2$ and $\log_99=1$ (avoiding the temptation to use $\log1=0$):

\begin{align} (1+1)^3+\log_39-9&=1\\ 9+1+1-9&=2\\ 9-(1+1)^2-\log_39&=3\\ 9+(1+1)^2-9&=4\\ 9-\log_39-1-1&=5\\ \log_39+\log_39+1+1&=6\\ 9+1-\log_39-1&=7\\ 9+(1+1)^3-9&=8\\ 9+1+1-\log_39&=9\\ 9+9-(1+1)^3&=10\\ 9+\log_3{9}+1-1&=11\\ (1+1)^3+\log_3{9}+\log_3{9}&=12\\ 9+\log_3{9}+1+1&=13 \\ 9+9-(1+1)^2&=14\\ 9+\log_39+(1+1)^2&=15\\ 9+9-1-1&=16\\ (\log_39+1)^2+9-1&=17\\ 9+9+1-1&=18\\ 9+(1+1)^3+\log_39&=19\\ 9+9+1+1&=20\\ (1+1)^5-9-\log_39&=21\\ 9+9+(1+1)^2&=22\\ (1+1)^4+9-\log_39&=23\\ (\log_39+1)^3-\log_39-1&=24\\ (1+1)^5+\log_39-9&=25\\ 9+9+(1+1)^3&=26\\ (1+1)^4+9+\log_39&=27\\ (\log_39+\log_39-1)^3+1&=28\\ (1+1)^5-\log_39-\log_99&=29\\ (\log_39+1)^3+\log_39+1&=30\\ (1+1)^5-\log_39+\log_99&=31\\ (1+1)^5+9-9&=32\\ (\log_39)^5+\log_99+1-1&=33\\ (\log_39)^5+\log_39+1-1&=34\\ (\log_39+1)^3+9-1&=35\\ (9+1)^2-(9-1)^2&=36\\ (\log_39+1)^3+9+1&=37\\ (\log_39)^6+1-(\log_39+1)^3&=38\\ (1+1)^5+9-\log_39&=39\\ (9+1+1)^2-9^2&=40\\ (\log_39)^5+9+1-1&=41\\ (1+1)^5+9+\log_99&=42\\ (\log_39)^5+9+1+1&=43\\ (\log_39)^5+(1+1)^3+(\log_39)^2&=44\\ (\log_39)^5+9+(1+1)^2&=45\\ (1+1)^6-9-9&=46\\ (9-1-1)^2-\log_39&=47\\ (\log_39)^5+(\log_39)^5-(1+1)^4&=48\\ (9-\log_39)^2+1-1&=49\\ (1+1)^5+9+9&=50\\ (9-1-1)^2+\log_39&=51\\ (1+1)^5+(\log_39)^4+(\log_39)^2&=52\\ (\log_39)^6-9-1-1&=53\\ (9-1)^2-9-1&=54\\ (\log_39)^6+1-9-1&=55\\ (9-1)^2+1-9&=56\\ (\log_39)^6+1+1-9&=57\\ (9-1-1)^2+9&=58\\ (9-1)^2-(\log_39)^2-1&=59\\ (\log_39)^5+(\log_39+1)^3+1&=60\\ (9-1)^2-\log_39-1&=61\\ (9-1)^2-log_99-1&=62\\ (9-1)^2+1-\log_39&=63\\ 9^2-9-(1+1)^3&=64\\ 9^2-(\log_39)^3-(1+1)^3&=65\\ (9-1)^2+\log_99+1&=66\\ (9-1)^2+\log_39+1&=67\\ 9^2-9-(1+1)^2&=68\\ 9^2-(1+1)^3-(\log_39)^2&=69\\ 9^2-9-1-1&=70\\ (\log_39)^6+9-1-1&=71\\ 9^2+1-9-1&=72\\ (\log_39)^6+9+1-1&=73\\ 9^2+1+1-9&=74\\ (\log_39)^6+9+1+1&=75\\ 9^2+(1+1)^2-9&=76\\ 9^2-\log_39-1-1&=77\\ 9^2-\log_99-1-1&=78\\ 9^2+1-\log_39-1&=79\\ 9^2+1-\log_99-1&=80\\ 9^2+1+1-\log_39&=81\\ 9^2+9-(1+1)^3&=82\\ 9^2+\log_39+1-1&=83\\ 9^2+\log_99+1+1&=84\\ 9^2+\log_39+1+1&=85\\ 9^2+9-(1+1)^2&=86\\ 9^2+(1+1)^2+\log_39&=87\\ 9^2+9-1-1&=88\\ 9^2+(\log_39)^2+(1+1)^2&=89\\ (9+1)^2-9-1&=90\\ 9^2+(1+1)^3+\log_39&=91\\ 9^2+9+1+1&=92\\ 9^2+(\log_39)^3+(1+1)^2&=93\\ 9^2+9+(1+1)^2&=94\\ (9+1)^2-(\log_39)^2-1&=95\\ 9^2+(1+1)^4-\log_99&=96\\ (9+1)^2-\log_39-1&=97\\ 9^2+9^2-(1+1)^6&=98\\ (9+1)^2+1-\log_39&=99\\ (9+1)^2+\log_99-1&=100\\ \end{align}

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  • $\begingroup$ Nice answer, but one thing puzzles me. The question said " Logs may be used three times". You have more than 3 log in there. Was that 3 in total or 3 per number? $\endgroup$ – Marius Mar 9 '16 at 15:18
  • $\begingroup$ Ah, I interpreted that as three times per equation, not three times for all one hundred equations. $\endgroup$ – Jonathan Allan Mar 9 '16 at 15:21

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