Using 9,9,9,9 to make 1-100

Using   9, 9, 9, 9   and the basic operations make the numbers   1   to   100.

You may also use decimal points.
 
Here are the numbers   1   to   20:
 
1   =   (9 + 9) / (9 + 9)
2   =   (9 + √9) / (9 – √9)
3   =   (9 + 9 + 9) / 9
4   =   (√9 * √9) / 9   +   √9
5   =   (9 – √9)   –   (9 / 9)
6   =   (9 * √9) / 9   +   √9
7   =   (9 / 9)   +   √9   +   √9
8   =   (99) / 9   –   √9
9   =   (√9 * √9)   –   (9 – 9)

10   =   (√9 * √9)   +   (9 / 9)
11   =   99 / (√9 * √9)
12   =   (9 * 9) / 9   +   √9
13   =   9   +   √9   +   (9 / 9)
14   =   (99) / 9   +   √9
15   =   (√9 + √9) * √9   –   √9
16   =   (9 / .9)   +   √9   +   √9
17   =   (9 + 9)   –   (9 / 9)
18   =   ((√9 + √9) * 9) / √9
19   =   (9 + 9)   +   (9 / 9)
20   =   (99 / 9)   +   9

 
Please extend the list
 
 

Triangle (5, 7, 8)

 
A triangle with sides of length   5,   7,   and   8

Perimeter   =   20
Area   =   10 √3   ~   17.3205

one of its angles measures 60 degrees:

Applying the Law of cosines

7^2   =   8^2   +   5^2   –   (2*8*5) cos(60)

The triangle has a 60 degrees angle between the sides of length   5   and   8.

 

How many non-similar triangles with sides of integer length and angle 60 degrees can you find?
 
 
 

Primes p; n(p + n) is a square

Consider the first few prime numbers:           2,   3,  5,  7,  11,  13,  17,  19,  23,  29,  31,  37

And   f(p) = n(p + n),     where  p  is an odd prime

and find  n  so that  f(p)  is a square

1(3 + 1)  =  4  =  2^2                                                        4(5 + 4)  =  36  =  6^2
9(7 + 9)  =  144  =  12^2

25(11 + 25)  =  900  =  30^2                                 36(13 + 36)  =  1764  =  42^2
64(17 + 64)  =  5184  =  72^2                                81(19 + 81)  =  8100  =  90^2
121(23 + 121)  =  17424  =  132^2                  196(29 + 196)  =  44100  =  210^2
225(31 + 225)  =  57600  =  240^2              324(37 + 324)  =  116964  =  342^2

 

These results indicate that there exists a positive integer   n   such that   n(p + n)   is a perfect square

Could you show that for any odd prime p there’s a unique positive integer   n   such that  
n(p + n)   is a perfect square?

 
 
 
 
 

Using 1,2,3,4,5 only once to make 1 – 100

Game:

Using  1, 2, 3, 4, 5  only once and the basic operations (+, -, * , / , !) to make all the numbers from  1  to  100
 

0 = 5 + 3 – (4*2/1)                                                5 = 4*3/(5 – 1) + 2
1 = (4 + 5)/3 – 2*1                                                 6 = 4*3/(5 – (1 + 2))
2 = (3 + 5 – 4)/2*1                                                7 = (3*2/1) + 5 – 4
3 = 4 + 5 – (3*2/1)                                                 8 = 4*2/(5 – (3 + 1))
4 = (3 + 5)/( 4- 2*1)                                              9 = 4*3/1 – 5 + 2

10 = 5*(3 – 4/2 + 1)                                             15 = 5*(4 – 3 + 2)/1
11 = 5*2 + 4/2 – 1                                                 16 = (3 + 5) * (4 – 2)/1
12 = 5*2 + 4/(3 – 1)                                             17 = 5*3 + 4 – 2/1
13 = 5*3/1 – 4 + 2                                                18 = 3 * (5 + 4/2 – 1)
14 = 4*(5 + 3 – 1)/2                                             19 = 5*4 – 3/(2 + 1)

20 = (3 + 4/2 – 1) * 5                                          25 = (4 + 3 – 2/1)*5
21 = (4 + 5 – 2) * 3/1                                           26 = 4! + 2/(3! – 5*1)
22 = (3! + 5) * (4 – 2/1)                                      27 = (4 + 2)*5 – 3/1
23 = (5*4 + 3)/(2 – 1)                                          28 = 4!/(5 – 2)!*(3! + 1)
24 = (3 + 5 – 2/1)*4                                             29 = 5*3! – 4/2 + 1

30 = (5 + 3)*4 – 2/1                                             35 = 5*(4!/3 – 2 + 1)
31 = 5*3! + 4/2 – 1                                                36 = 3!*(5 + 4/2 – 1)
32 = 5*3! + 4/1 – 2                                               37 = 5*(4!/3 – 1) + 2
33 = 5*(4 + 3) – 2/1                                              38 =  5!/3 – 4 + 2*1
34 = 5*3! + 4/(2 – 1)                                            39 = 2*(5! – 4 + 1)/3!

40 = 5!/(4 – 3 + 2) * 1                                         45 = 5*(4!/3 + 2 – 1)
41 = 2*(5! + 4 – 1)/3!                                           46 =  5*(4!/2 – 3) + 1
42 = 5!/3 + 4 – 2*1                                               47 =  5!/3 + 4*2 – 1
43 = (4! – 2*5)*3 + 1                                            48 =  (1 + (5 – 3)/2)*4!
44 = 4 * (3! + 5/(2 – 1))                                       49 = (3! + 1)*(4!/2 – 5)

50 = (5!/4 – 3! + 1) * 2                                        55 = 5!*2/4 – 3! + 1
51 = 1 * (5! + 3! – 4!)/2                                       56 = (5 + 1)!/(2*3!) – 4
52 = 4!*(2 + 1) – 5!/3!                                         57 = (5!/(2 + 4) – 1)*3
53 = 5!*2/4 – (3! + 1)                                           58 = (5 + 1)!/(4*3) – 2
54 = 1*(5! – 4!)/2 + 3!                                         59 = 5!/2 + 3 – 4*1

60 = 5!*(3 – 2 + 1)/4                                            65 = 2*(5! – 4!)/3 + 1
61 = 5!*(4 – 3)/2 + 1                                             66 = 4!*(5 + 1)/2 – 3!
62 = 5!/3 + 4! – 2*1                                              67 = 3*4! + 5/(1 – 2)
63 = 5!/(4 – 2) + 3*1                                            68 = 5!/(3 – 1) + 4*2
64 = 1*(5! – 4)/2 + 3!                                           69 = 5!/2 + (4 – 1)*3

70 = 3*4! + (1 – 5)/2                                            75 = 2*5!/3 – (4 + 1)
71 = 5!/2 + 4*3 – 1                                                76 = 4!*3!/2 + 5 – 1
72 = (2 + 1) * (5!/4 – 3!)                                     77 = 3*4!/(2 – 1) + 5
73 = 3*(4! + 2) – 5/1                                            78 = 3*(4! + (5 – 1)/2)
74 = 4*5!/3! – (2 + 1)!                                         79 = 4*5!/3! – 2 + 1

80 = 5!*(3! – 4)/(1 + 2)                                       85 = 3*(4! + 5) – 2/1
81 = 4*5!/3! + 2 – 1                                               86 = 3!*(4! + 5)/2 – 1
82 = 4*(5!/3! + 1) – 2                                           87 = 4*(5!/3! + 2) – 1
83 = 2*5!/3 + 4 – 1                                               88 = (5 + 3) * (4!/2 – 1)
84 = 3*5!/4 – (2 + 1)!                                          89 = (5!*3)/4 – 2 + 1

90 = (4 + 1) * (5!/3! – 2)                                     95 = (5!*4)/(2 + 3) – 1
91 = (5!*3)/4 + 2 – 1                                             96 = (5! – 4!) * 3/(2 + 1)
92 = 5*(4! – 3!) + 2/1                                         97 = (3!)!/5 + 1 – 2*4!
93 = 5! – (4! + 1*3!/2)                                        98 = (5!*4)/(3! – 1) + 2
94 = 5! – (4*3! + 2/1)                                          99 =  5! + 3!/2 – 4!*1

100 = 5*(4! – 3! + 2)/1

 
 
 
 
 
 
 
 
 

Num3ers with repeated digits divisible by a prime they contain

Start with a 2-digit prime number

11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

and find numbers that repeat the digits of the chosen prime and divisible by this prime
 

For example,

19,   1919

11191   is divisible by 19,        11191   =   19 * 19 * 31

17,   1717

22333   =   23 * 971
44333   =   43 * 1031
555533333   =   53 * 10481761

13,   1131,   1313,   3133

Divisible by   11:
11,   1111,   111111,   11111111,   1111111111, …

N.B.   numbers of the form   abababab…   are not the examples I’m looking for.

E.g.   1313,   131313…   1717,   171717…,   191919…

What I consider to be good examples are:

11191,   1131,   3133,   22333,   44333,   555533333

Find larger ones (if possible) and find other numbers using the remaining 2-digit numbers.

 
 
 
 
 
 

Concatenation | (x – y + z)^3 = x || y || z

Older post:   Concatenation: x||y||z = x^3 + y^3 + z^3

 

Find integers   x,   y   and   z   such that

                                                 (x – y + z)^3   =   x || y || z

                                                 (-x + y + z)^3   =   x || y || z

                                                 (x + y + z)^3   =   x || y || z

                                                 (x + y – z)^3   =   x || y || z

 

For example,

(9 + 11 + 25)^3    =    91125    =    9 || 11 || 25    =    45^3

(786 – 330 + 467)^3   =   786330467   =   786 || 330 || 467

(188 – 132 + 517)^3    =    188132517    =    188 || 132 || 517

(258 – 474 + 853)^3    =    258474853    =    258 || 474 || 853

(-360 + 944 + 128)^3    =    360944128    =    360 || 944 || 128

 
 
And,

(13 – 3 + 1)^3   =   1331
(103 – 03 + 01)^3   =   1030301
(1003 – 003 + 001)^3   =   1003003001
(10003 – 0003 + 0001)^3   =   1000300030001
(100003 – 00003 + 00001)^3   =   1000030000300001
(1000003 – 000003 + 000001)^3   =   1000003000003000001

(106 – 12 + 08)^3   =   1061208
(1006 – 012 + 008)^3   =   1006012008
(10006 – 0012 + 0008)^3   =   1000600120008
(100006 – 00012 + 00008)^3   =   1000060001200008
(1000006 – 000012 + 000008)^3   =   1000006000012000008

 
 
 
 
 

Using Primes 2,3,5,7 only once to make 1-100

Game:

You can use the primes   2, 3, 5, 7   and the basic operations   +, -, *, /   and   !   (factorial) only once to make numbers from   1   to   100.
 

N.B.    you may use decimal points
 

For example,

0   =   (7 + 3)/5 – 2
1   =   (7 – 5) – (3 – 2)     or     1   =   (3 * 5) – (2 * 7)
2   =   7 – 3! + (3 – 2)
3   =   (7 – 2) – (5 – 3)
4   =   2*3 – (7 – 5)

 
How many other numbers can you express?