8
$\begingroup$

This puzzle relates to Prime to Prime: Get all first 25 Prime Numbers using up to 4 Primes and its sequel Prime to Prime Sequel

Using any three of the first 4 prime numbers (2,3,5 and 7) and the folllowing math operations get the first 23 numbers (from 1 to 23).

x / + - ^ ! !! Square root

Other Rules

Once you select the three primes all the three primes must appear (once only) in every equation. Use of only 1 or 2 primes not permitted. All three primes must appear once only. ( e.g. 4=7-3 not allowed. it could be 4=2+7-5)

Same three primes must appear in every equation.

Any prime that appears anywhere in the equation is counted. If you used 3^2 then you have used up 2 and 3

Multiple roots not allowed ( Sq root of sq root)

Concatenation is forbidden.

Parentheses are permitted.

Please no partial answers.

$\endgroup$
5
  • $\begingroup$ All three primes must appear only once? But you've given us four primes... $\endgroup$
    – Galen
    Commented May 4, 2020 at 14:41
  • 1
    $\begingroup$ Yes. You can pick any three out of 4 . Those three must appear once in every equation. Say you use 2 , 3 and 5 then only those 3 must be used once in every equation. $\endgroup$
    – DrD
    Commented May 4, 2020 at 15:17
  • $\begingroup$ For multiple roots, does that disqualify taking the square root of a square? For example, could I do \sqrt{5^2} ? $\endgroup$
    – Herb
    Commented May 4, 2020 at 16:40
  • $\begingroup$ Yes it does @Herb Wolfe $\endgroup$
    – DrD
    Commented May 4, 2020 at 17:42
  • $\begingroup$ This is a fun puzzle (I got here a bit late unfortunately). I doubt that allowing multiple roots would create any additional solutions in this case, since rounding isn't allowed. $\endgroup$
    – user63779
    Commented May 5, 2020 at 1:40

2 Answers 2

11
$\begingroup$

It might be more fun to fill the whole table. I've started with a few examples:

$$\require{begingroup}\begingroup \def\*{\times } \begin{array}{|c|c|c|c|c|} \hline n& 2,3,5 & 2,3,7 & 2,5,7 & 3,5,7 \\ \hline 1& 2\*3-5 & 7-2\*3 & \frac{2+5}{7} & 3+5-7 \\ 2& (3!-5)\*2 & 7-3-2 & \sqrt{(7-5)\*2} & \frac{3+7}5 \\ 3& (5-2)!-3 & \frac{7+2}{3} & 2\*5-7 & \sqrt{3^{7-5}} \\ 4& 2-3+5 & 2^{\sqrt{7-3}} & 2-5+7 & \frac{7+5}{3} \\ 5& (3-2)\*5 & \frac{7+3}{2} &\frac{7!!}{5!!}-2& 3-5+7 \\ 6& -2+3+5 & 2-3+7 & \frac{7+5}{2} & (7-5)\*3 \\ 7& 2\*5-3 & (3-2)\*7 & \sqrt{(2+5)\*7} & \sqrt{7^{5-3}} \\ 8& (\frac{3+5}{2})!! & -2+3+7 & 5+\sqrt{2+7} & 3\*5-7 \\ 9& (5-2)!+3 & (7-2)!!-3! & 2\*7-5 & -3+5+7 \\ 10& 2+3+5 & 7+\frac{3!}{2} & -2+5+7 & 5\*\sqrt{7-3} \\ 11& 2\*3+5 & 2\*7-3 & \frac{5!!+7}{2} & 5!!+3-7 \\ 12& (5-2)!+3! & 2+3+7 & \sqrt{(5+7)^2} & (7-5)\*3! \\ 13& 2\*5+3 & 2\*3+7 & (5-2)!+7 & {7!!\over5!!}+3!\\ 14& 3^2+5 & 2\*\frac{7!}{(3!)!} & 2+5+7 & (5-3)\*7 \\ 15& (3-2)\*5!! & (7-2)\*3 & 5\*\sqrt{2+7} & 3+5+7 \\ 16& (5+3)\*2 & 2^{(7-3)} & (5!!-7)\*2 & 3\*7-5 \\ 17& 3\*5+2 & 2\*7+3 & 2\*5+7 & 5!!+\sqrt{7-3} \\ 18& (5-2)!\*3 & (7-2)!!+3 & 5^2-7 & 3+5!-7!! \\ 19& 5!!+3!-2 & 3\*7-2 & 2\*7+5 & 5!!+7-3 \\ 20& 5!!+3+2 & 2\*(3+7) & 5!!+7-2 & (7-3)\*5 \\ 21& (5+2)\*3 & (7-2)!!+3! & 7\*(5-2) & 3*{7!!\over5!!} \\ 22& 5^2-3 & (7-3)!-2 & & 3\*5+7 \\ 23& 5!!+3!+2 & 3\*7+2 & 5!!\*2-7 & 5!!+(7-3)!! \\ \hline \end{array} \endgroup$$

$\endgroup$
11
  • $\begingroup$ Wow. OK. I did that but for me it did not work with all combinations. Only one. May be you will have better approach $\endgroup$
    – DrD
    Commented May 4, 2020 at 13:55
  • 1
    $\begingroup$ 2, 3, 5 is done $\endgroup$
    – Herb
    Commented May 4, 2020 at 17:19
  • 2
    $\begingroup$ Of course, now that @David G. has generated a complete chart, the challenge for new entries here is to differ from that chart $\endgroup$
    – humn
    Commented May 4, 2020 at 18:13
  • 1
    $\begingroup$ 2, 3, 7 is complete now as well. $\endgroup$
    – Herb
    Commented May 4, 2020 at 20:42
  • 1
    $\begingroup$ @oAlt; I've fixed 257:2 $\endgroup$
    – JMP
    Commented May 5, 2020 at 3:59
10
$\begingroup$

I tried writing a generator. I can get everything from 1 to 24 inclusive, for all 4 combinations. I can get 0 to 33 for one combination. The first ungeneratable counting number is 68.

$$\begin{array}{|c|c|c|c|c|} \hline n& 2,3,5 & 2,3,7 & 2,5,7 & 3,5,7 \\ \hline 0 & (5-(2+3)) & (3-\sqrt{(2+7)}) & (7-(2+5)) & - \\ 1 & (\frac{5}{(2+3)}) & (7-(2\times3)) & (\frac{7}{(2+5)}) & ((3+5)-7) \\ 2 & \sqrt{(5+(2-3))} & (7-(2+3)) & \sqrt{(7+(2-5))} & (\frac{(3+7)}{5}) \\ 3 & ({2}^{3}-5) & (\frac{(2+7)}{3}) & ((2\times5)-7) & \sqrt{(7-(3-5))} \\ 4 & (5+(2-3)) & \sqrt{(7+{3}^{2})} & (7+(2-5)) & (\frac{(5+7)}{3}) \\ 5 & (5\times(3-2)) & (\frac{(3+7)}{2}) & \sqrt{({2}^{5}-7)} & (7+(3-5)) \\ 6 & (5-(2-3)) & (7+(2-3)) & (\frac{(5+7)}{2}) & (3\times(7-5)) \\ 7 & ((2\times5)-3) & (7\times(3-2)) & \sqrt{(7\times(2+5))} & (\frac{{7}!!}{(3\times5)}) \\ 8 & (\frac{{5}!}{{(2+3)}!!}) & (7-(2-3)) & {(7+(2-5))}!! & ((3\times5)-7) \\ 9 & (3\times(5-2)) & (3\times\sqrt{(2+7)}) & ((2\times7)-5) & (7-(3-5)) \\ 10 & (5+(2+3)) & \sqrt{({7}!!-(2+3))} & (7-(2-5)) & (5\times\sqrt{(7-3)}) \\ 11 & (5+(2\times3)) & ((2\times7)-3) & (\frac{({5}!!+7)}{2}) & ({5}!!+(3-7)) \\ 12 & \sqrt{(\frac{{(2\times3)}!}{5})} & (7+(2+3)) & ({5}!!-\sqrt{(2+7)}) & (\frac{{5}!}{(3+7)}) \\ 13 & (5+{2}^{3}) & (7+(2\times3)) & (7+{(5-2)}!) & (5+{(7-3)}!!) \\ 14 & (5+{3}^{2}) & (7\times\sqrt{({3}!-2)}) & (7+(2+5)) & (7\times(5-3)) \\ 15 & \sqrt{({5}!!\times{(2+3)}!!)} & (7+{2}^{3}) & (\frac{{(2+5)}!!}{7}) & (7+(3+5)) \\ 16 & (2\times(3+5)) & (7+{3}^{2}) & (2\times({5}!!-7)) & ((3\times7)-5) \\ 17 & (2+(3\times5)) & (3+(2\times7)) & (7+(2\times5)) & ({5}!!+\sqrt{(7-3)}) \\ 18 & ({5}!!+\sqrt{{3}^{2}}) & ({3}!\times\sqrt{(2+7)}) & ({5}^{2}-7) & ({3}!+(5+7)) \\ 19 & ({5}^{2}-{3}!) & ((3\times7)-2) & (5+(2\times7)) & ({5}!!-(3-7)) \\ 20 & (5+{(2+3)}!!) & (2\times(3+7)) & ({5}!!-(2-7)) & (5\times(7-3)) \\ 21 & (3\times(2+5)) & (\frac{{7}!!}{(2+3)}) & (7\times(5-2)) & (\frac{(3\times{7}!!)}{{5}!!}) \\ 22 & ({5}^{2}-3) & (7+{(2+3)}!!) & ({5}!!+\sqrt{{7}^{2}}) & (7+(3\times5)) \\ 23 & ({5}!!+{2}^{3}) & (2+(3\times7)) & (2+(\frac{{7}!!}{5})) & ({5}!!+{(7-3)}!!) \\ 24 & (\frac{{(2+3)}!}{5}) & {(7-\sqrt{{3}^{2}})}! & (2\times(5+7)) & {(\frac{(5+7)}{3})}! \\ 25 & (5\times(2+3)) & - & ({2}^{5}-7) & ({5}!!+(3+7)) \\ 26 & ({2}^{5}-{3}!) & (2+{(7-3)}!) & - & (5+(3\times7)) \\ 27 & {3}^{(5-2)} & (3\times(2+7)) & - & ({3}!+(\frac{{7}!!}{5})) \\ 28 & (3+{5}^{2}) & (7\times({3}!-2)) & - & ({3}!+({5}!!+7)) \\ 29 & ({2}^{5}-3) & ({{3}!}^{2}-7) & ({5}!!+(2\times7)) & ((5\times7)-{3}!) \\ 30 & (5\times(2\times3)) & ({3}!\times(7-2)) & ({5}!!+{(7-2)}!!) & (\frac{{5}!}{(7-3)}) \\ 31 & ({3}!+{5}^{2}) & (7+{({3}!-2)}!) & - & - \\ 32 & {2}^{(\frac{{5}!!}{3})} & \sqrt{{2}^{(3+7)}} & (7+{5}^{2}) & ((5\times7)-3) \\ 33 & ({(2\times3)}!!-{5}!!) & ((\frac{{7}!!}{3})-2) & ((5\times7)-2) & ((\frac{{5}!}{3})-7) \\ 34 & - & ({({3}!)}!!-(2\times7)) & ({7}^{2}-{5}!!) & - \\ 35 & (3+{2}^{5}) & (7\times(2+3)) & (\frac{{7}!!}{(5-2)}) & (\frac{{5}!!}{(\frac{3}{7})}) \\ 36 & ({3}!\times{(5-2)}!) & - & - & (3\times(5+7)) \\ 37 & - & (2+(\frac{{7}!!}{3})) & (2+(5\times7)) & (7+({3}!\times5)) \\ 38 & ({3}!+{2}^{5}) & - & - & (3+(5\times7)) \\ 39 & (3\times({5}!!-2)) & ({({3}!)}!!-(2+7)) & (7+{2}^{5}) & ({5}!!+{(7-3)}!) \\ 40 & (5\times{2}^{3}) & (\frac{{(7-2)}!}{3}) & (\frac{{5}!}{\sqrt{(2+7)}}) & (5\times{(7-3)}!!) \\ 41 & ({({3}!)}!!-(2+5)) & ({(2\times3)}!!-7) & - & ({3}!+(5\times7)) \\ 42 & ({3}!\times(2+5)) & (7\times(2\times3)) & ({7}!!\times(\frac{2}{5})) & (\frac{{7}!}{{(\frac{{5}!!}{3})}!}) \\ 43 & ({(2\times3)}!!-5) & ({7}^{2}-{3}!) & - & - \\ 44 & - & (2+({3}!\times7)) & ({7}^{2}-5) & - \\ 45 & (5\times{3}^{2}) & (3\times{(7-2)}!!) & (5\times(2+7)) & (3+(\frac{{7}!}{{5}!})) \\ 46 & - & ({7}^{2}-3) & - & ({({3}!)}!!+(5-7)) \\ 47 & (2+(3\times{5}!!)) & - & - & (7+(\frac{{5}!}{3})) \\ 48 & {(5-(2-3))}!! & {(7+(2-3))}!! & (\frac{{7}!}{{(2+5)}!!}) & {((3\times7)-{5}!!)}!! \\ 49 & ({2}^{{3}!}-{5}!!) & \sqrt{{7}^{({3}!-2)}} & (7\times(2+5)) & {7}^{(5-3)} \\ 50 & - & - & (\frac{({7}!!-5)}{2}) & (5\times(3+7)) \\ 51 & (3\times(2+{5}!!)) & (\frac{({7}!!-3)}{2}) & - & - \\ 52 & - & (3+{7}^{2}) & - & (7+(3\times{5}!!)) \\ 53 & (5+{(2\times3)}!!) & ({({3}!)}!!-(2-7)) & ((\frac{{5}!}{2})-7) & - \\ 54 & ((\frac{{5}!}{2})-{3}!) & ({3}!\times(2+7)) & (5+{7}^{2}) & - \\ 55 & ({({3}!)}!!+(2+5)) & (7+{(2\times3)}!!) & (\frac{(5+{7}!!)}{2}) & ({({3}!)}!!+(\frac{{7}!!}{{5}!!})) \\ 56 & ({5}!-{2}^{{3}!}) & (7\times{2}^{3}) & - & (7\times(3+5)) \\ 57 & ((\frac{{5}!}{2})-3) & ({7}!!-{(2\times3)}!!) & - & ({5}!!+({3}!\times7)) \\ 58 & ({({3}!)}!!+(2\times5)) & - & - & ({({3}!)}!!+\sqrt{({7}!!-5)}) \\ 59 & ({2}^{{3}!}-5) & ((2+{7}!!)-{({3}!)}!!) & - & - \\ 60 & \sqrt{(5\times{(2\times3)}!)} & \sqrt{({({3}!)}!\times(7-2))} & (\frac{({5}!!+{7}!!)}{2}) & (\frac{{5}!}{\sqrt{(7-3)}}) \\ 61 & ({({3}!)}!!-(2-{5}!!)) & - & - & - \\ 62 & - & ({({3}!)}!!+(2\times7)) & - & ((5+{7}!!)-{({3}!)}!!) \\ 63 & ({5}!!+{(2\times3)}!!) & (7\times{3}^{2}) & (\frac{{(2+7)}!!}{{5}!!}) & ({7}!!\times(\frac{3}{5})) \\ 64 & {(3+5)}^{2} & {{(7-3)}!!}^{2} & \sqrt{{2}^{(5+7)}} & {(5-7)}^{{3}!} \\ 65 & ({({3}!)}!!+(2+{5}!!)) & - & - & ({7}!!-(\frac{{5}!}{3})) \\ 66 & ({3}!+(\frac{{5}!}{2})) & - & - & (3\times({5}!!+7)) \\ 67 & - & - & (7+(\frac{{5}!}{2})) & - \\ 68 & - & - & - & - \\ 69 & (5+{2}^{{3}!}) & ({7}!!-{{3}!}^{2}) & - & ({({3}!)}!!+(\frac{{7}!!}{5})) \\ \hline \end{array}$$

Edit: Line Break issue (that someone else also noticed). Corrected ${{3}!}!!$ to ${{(3}!)}!!$ and ${{3}!}!$ to ${{(3}!)}!$. (The mathjax source is OK, but it doesn't combine right.)

$\endgroup$
3
  • $\begingroup$ Ah I was just about to try this when I saw your answer. +1 $\endgroup$
    – user63779
    Commented May 5, 2020 at 1:37
  • $\begingroup$ Impressive... even if some items are disqualified as per author's comment (e.g. 22 with 2,5,7) $\endgroup$
    – frarugi87
    Commented May 5, 2020 at 7:47
  • $\begingroup$ @David G. Are there any answers if we do not use !! double factorial? $\endgroup$
    – DrD
    Commented May 5, 2020 at 12:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.