Problems & Puzzles: Puzzles

 Puzzle 983. Happy New Year 2020 In a few days we will close this year 2019 and welcome the next year 2020, in almost all the world around. It is a common practice to wait awake the born of the new year until the 12pm, and say cheers with our family and friends for the new born baby-year. Perhaps this is a good excuse for bring a puzzle about the twelve hour integers on the face of the clocks. This makes me use the following nice picture sent by my old friend Alberto Hernández, a few days ago: Accordingly our rules are: 1) We may use at the much four instances of just one of the four prime integers as "2", or "3" or "5" or "7" - instead the integer "9"- and 2) We may only use the following four arithmetic operators & five auxiliary symbols: [+, -, /, * square root, factorial, decimal point, concatenation and parentheses]  Q1. Send your solutions (one for each one-digit-prime) using the minimal of these allowed operators and auxiliary symbols Q2. Is it possible to get a solution using only four instances of one prime composed of two digits and all (or part, or more) of the allowed operators and auxiliary symbols, mentioned above?

During the week from Dec 28 to Jan 3, 2020, contributions came from Fred Schneider, Jim Howell,
Emmanuel Vantieghem and Carlos Rivera

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Fred wrote:

I could find answers for all but 2 cases.

 2 3 5 7 1 2/2 3/3 5/5 7/7 2 2 (3+3)/3 (5+5)/5 (7+7)/7 3 2 + 2/2 3 (5+5+5)/5 (7+7+7)/7 4 2 * 2 3 + 3/3 5/5/05 77/7 - 7 5 2 * 2 + 2/2 3 + 3 - 3/3 5 7 - (7+7)/7 6 2 + 2 + 2 3! 5 + 5/5 7/7/07 7 2 + 2 + 2 + 2/2 (***) 3 + 3 + 3/3 5 + (5+5)/5 7 8 2 * 2 * 2 3 * 3 - 3/3 5 + (5+5+5)/5 (***) 7 + 7/7 9 22/2 - 2 3 * 3 5+5 - 5/5 7 + (7+7)/7 10 2 * 2 * 2 + 2 3 * 3 + 3/3 5+5 (77-7)/7 11 22/2 33 / 3 55/5 77/7 12 (2+2)!/2 3 * (3 + 3/3) (55 + 5)/5 (77 + 7)/7 *** uses more than 4 integers

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Jim wrote:

Here are my solutions for puzzle 983.

1:   2/2         3/3          5/5             7/7

2:   2           (3+3)/3      (5+5)/5         (7+7)/7

3:   2+(2/2)     3            (5+5+5)/5       (7+7+7)/7

4:   2+2         3+(3/3)      5-(5/5)         (77/7)-7

5:   2+2+(2/2)   3+(3+3)/3    5               7-((7+7)/7)

6:   2+2+2       3+3          5+(5/5)         7-(7/7)

7:   2/(2*.2)+2  3+3+(3/3)    5+((5+5)/5)     7

8:   2*2*2       3*3-(3/3)    (5-(5/5))/.5    7+(7/7)

9:   (22/2)-2    3*3          5+5-(5/5)       7+((7+7)/7)

10:  2/.2        3/.3         5+5             7/.7

11:  22/2        33/3         55/5            77/7

12:  (22+2)/2    3*3+3        (55+5)/5        (77+7)/7

Note: Some of the parentheses are not needed, but I included them for clarity.

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Emmanuel wrote:

This are the only results I could find :

With four three's
1 = 3/3
2 = (3+3)/3
3 = 3
4= 3 +(3/3)
5 = 3 + 3 - (3/3)
6 = 3 + 3
7 = 3 + 3 + (3/3)
8 = 3*3 - (3/3)
9 = 3*3
10 = (33 - 3)/3
11 = 33/3
12 = 3*3 + 3

With four sevens
1 = 7/7
2 = (7 + 7)/7
3 = (7+7+7)/7
4 = (77/7) - 7
5 = 7 - (7 + 7)/7
6 = 7 - (7/7)
7 = 7
8 = 7 + (7/7)
9 = 7 +(7+7)/7
10 = (77 - 7)/7
11 = 77/7
12 = (77 + 7)/7

It was not asked, but I could not resist the temptation to work on a clock which used at mod two two's and two zeros (like in "2020") :
1 = 0!
2 = 2
3 = 2 + 0!
4 = 2 + 2
5 = 2 + 2 + 0!
6 = (2 + 0!)!
7 = (2 + 0!)! + 0!
8 = (2 + 0!)! + 2
9 = (2 + 0!)*(2 + 0!)
10 = 20/2
11 = 20/2 + 0!
12 = 2*((2 + 0!)!)

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Carlos Rivera wrote:

 Hour x=1 x=2 x=3 x=4 x=5 x=6 x=7 x=8 x=9 1 x/x x/x x/x x/x x/x x/x x/x x/x x/x 2 (x+x)/x (x+x)/x (x+x)/x (x+x)/x (x+x)/x (x+x)/x (x+x)/x (x+x)/x (x+x)/x 3 (x+x+x)/x (x+x+x)/x (x+x+x)/x (x+x+x)/x (x+x+x)/x (x+x+x)/x (x+x+x)/x (x+x+x)/x (x+x+x)/x 4 1+1+1+1 2+2 3+3/3 x x-x/x 6/.6-6 77/7-7 r(8+8) r(9)+9/9 5 1/(.1+.1) 2+2+2/2 3+3-3/3 x+x/x x x-x/x 7-(7+7)/7 r(8+8)+8/8 r(9)+r(9)-9/9 6 1/(.1+.1)+1 2+2+2 3+3 4+r(4) x+x/x x x-x/x 8/.8-r(8+8) 9-r(9) 7 (1+1+1)!+1 (2/2)/.2+2 3+3+3/3 4+r(4)+4/4 5+(5+5)/5 x+x/x x x-x/x 9+9/9-r(9) 8 1/.1-1-1 2*2*2 3*3-3/3 4+4 5+5*.5+.5 6+(6+6)/6 x+x/x x 9-9/9 9 x/.x-x/x x/.x-x/x x/.x-x/x x/.x-x/x x/.x-x/x x/.x-x/x x/.x-x/x x/.x-x/x x/.x-x/x 10 x/.x x/.x x/.x x/.x x/.x x/.x x/.x x/.x x/.x 11 xx/x xx/x xx/x xx/x xx/x xx/x xx/x xx/x xx/x 12 (xx+x)/x (xx+x)/x (xx+x)/x (xx+x)/x (xx+x)/x (xx+x)/x (xx+x)/x (xx+x)/x (xx+x)/x

Note: "r" goes for square root.

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