Solving Maths Quiz: Numbers 1-50 with 2,5,1,2

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The discussion centers around creating numbers from 1 to 50 using the digits 2, 5, 1, and 2, with all digits required to be used. Participants share various mathematical expressions and solutions, successfully generating many numbers but struggling with 41, 42, and 43. There are debates about the legality of certain operations, including factorials and the use of decimals. Some creative solutions are proposed, including complex expressions that involve factorials and decimals, while others express skepticism about the validity of certain notations. The thread concludes with a collaborative effort to finalize the remaining numbers, particularly focusing on finding alternatives for the challenging 41.
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simple I think not - my son has homework/quiz for fun - try to make numbers from 1 to 50 using the numbers 2,5,1,2 (xmas date)- I think all of them must be used.

for example

25+21=46

(5+1)2-2=34 five plus one squared minus two equals 34

15x2+2=32


weve got some we can't do even usng factorials(dbl/trpl) - decimals and fractions

the ones we can't do are 41/42/43

any help with these appreciated
 
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5-(2)(2)(1)=1
5-[2+(2-1)]=2
5-[(2/2)+1]=3
(5-1)+2-2=4
12-(5+2)=5
5+(2/2)(1)=6
5+(2/2)+1=7
5+2+(2-1)=8
5+(2+2)1=9
5+2+2+1=10
5(2)+(2-1)=11
5(2)+2(1)=12
 
are you allowed to use trig functions? not that it would necessarily help
in fact i doubt it would
 
im assuming they cannot be used more than once because you could just say:
5*2*2*2+1
 
hahaha [5!/(2+1)]+2=42
 
estoydemoda said:
im assuming they cannot be used more than once because you could just say:
5*2*2*2+1

lol if it could be used more than once u can just do 1^5-1^2+1... all the way to whatever and making all numbers you could posibbly want.
 
No lol they can't be used more than once - posted this on other sites - maths goodies did best we just need number 41 now and we have one for number 36 that we don't understand but we hope the tutor will .

(((1+5/2)!)^2 = 36
 
36 = (12 \times .5)^2

41 = \left(\frac{2}{.1} + .5\right) \times 2
 
51 = 52 + 1 - 2

52 = (25 + 1) \times 2

53 = 52 + 2 -1

54 = 52 + 2 \times 1

55 = 52 + 2 + 1

56 = \frac{\frac{2}{.5(recurring)} + 2}{.1}

57 = 52 + \frac{1}{.2}

58 = 5 \times 12 - 2

59 = 5 \times (12 - .2)

60 = 15 \times 2 \times 2

:-p
 
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  • #10
ceptimus said:
36 = (12 \times .5)^2

41 = \left(\frac{2}{.1} + .5\right) \times 2

Good one,ceptimus...It's cheating... :cry: It sould have been "0" before the dot. :-p There's my share of cheating:
36=[(5-2)!]^{2\cdot1}
41=[5-(2-2)!]1

Daniel.

PS.In the second line,there is no multiplication between 4 and 1. :-p
EDIT:41=\frac{5}{.(1)}-(2\cdot2)
 
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  • #11
I couldn't find answers for 69, 79, 83 and 87 but otherwise I have all the numbers up to 113.

Only functions I've used are +, -, *, /, ^, the decimal point, and (recurring)
 
  • #12
OK then. Without the 0. 'cheat'

36 = 12 \times (5 - 2) D'Oh!
 
  • #13
ceptimus said:
I couldn't find answers for 69, 79, 83 and 87 but otherwise I have all the numbers up to 113.

Only functions I've used are +, -, *, /, ^, the decimal point, and (recurring)

69=(5+2)1-2

Daniel.
 
  • #14
dextercioby said:
69=(5+2)1-2

Daniel.

You're not seriously expecting anyone to accept that 71 = (5+2)1 is an acceptable notation are you? :smile:
 
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  • #15
These are the answers we came up with mostly by ourselves and with help from this forum - while we have checked these they may not all be correct - tutor will mark this week. Number 44 was a double factorial and we hope this is legal (as well as the .5 in number 41 lol). The alternatives where afterthoughts while checking when typing.

1- 5-(2+2)x1
2- 5-(2+2)+1
3- 5-2x1^2
4- 5-(1+2)+2
5- (5x1)+2-2
6- 5+1+2-2
7- 5+2x1^2
8- 5+1^2+2
9- (5+2^2)x1
10- 5+2+2+1
11- 5+(2x(2+1))
12- (2x5)+2x1
13- (2x5)+2+1
14- (5+1)x2+2
15- (5+2)x2+1
16- (5+2+1)x2
17- (2+1)x5x2
18- ((2x2)!)-5-1
19- 5x(2+2)-1
20- 5x(2+2)x1
21- 5x(2+2)+1
22- ((5x2)+1)x2
23- 5^2-2x1
24- (2x2)x(5+1)
25- 5x(2+2+1)
26- 5^2-2+1
27- 25+2x1
28- 25+2+1
29- (((2+2)!)+5)x1
30- (5x(2+1))x2
31- ((2+5)!)+2+1 alternative ((5!)/(2+2))+1
32- 15x2+2 alternative((5-1)^2)x2
33- 2^5+2-1
34- (5+1)^2-2
35- 2^5+2+1
36- (((1+5)/2)!)^2
37- 21x2-5
38- (5+1)^2+2
39- 5!/2-21
40- 2^(2+1)x5
41- (21-.5)x2
42- (5+2)x((2+1)!)
43- 2x21.5
44- ((5+1)!)-2-2
45- (2+1)^2x5
46- ((5-1)!)*2-2
47- 21x2+5
48- (5+2)^2-1
49- 25x2-1
50 25x2x1

alternative to 41/44 welcome just in case
Many thanks to all who helped and a Happy Xmas and New Year to all from here in the UK.
 
  • #16
44 = 2^5 + 12

see other answers above.
 
  • #17
ceptimus said:
You're not seriously expecting anyone to accept that 71 = (5+2)1 is an acceptable notation are you? :smile:

I would call that an identity.At least in the decimal basis... :approve:
Read and weep:
41=[(2+1)!]^{2} +5 :-p

Daniel.
 
  • #18
beautifully done dexter
 

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