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

[tunesurfer]

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

 

[estoydemoda]

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

 

[estoydemoda]

are you allowed to use trig functions? not that it would necessarily help
in fact i doubt it would

 

[estoydemoda]

im assuming they cannot be used more than once because you could just say:
5*2*2*2+1

 

[estoydemoda]

hahaha [5!/(2+1)]+2=42

 

[futb0l]

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.

 

[tunesurfer]

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

 

[ceptimus]

$$36 = (12 \times .5)^2$$

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

 

[ceptimus]

$$51 = 52 + 1 - 2$$

$$52 = (25 + 1) \times 2$$

$$53 = 52 + 2 -1$$

$$54 = 52 + 2 \times 1$$

$$55 = 52 + 2 + 1$$

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

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

$$58 = 5 \times 12 - 2$$

$$59 = 5 \times (12 - .2)$$

$$60 = 15 \times 2 \times 2$$

 

[dextercioby]

$$36 = (12 \times .5)^2$$

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

Good one,ceptimus...It's cheating... It sould have been "0" before the dot. 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.
EDIT:$$41=\frac{5}{.(1)}-(2\cdot2)$$

 

[ceptimus]

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)

 

[ceptimus]

OK then. Without the 0. 'cheat'

$$36 = 12 \times (5 - 2)$$ D'Oh!

 

[dextercioby]

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.

 

[ceptimus]

$$69=(5+2)1-2$$

Daniel.

You're not seriously expecting anyone to accept that $$71 = (5+2)1$$ is an acceptable notation are you?

 

[tunesurfer]

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.

 

[ceptimus]

44 = 2^5 + 12

see other answers above.

 

[dextercioby]

You're not seriously expecting anyone to accept that $$71 = (5+2)1$$ is an acceptable notation are you?

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

Daniel.

 

[estoydemoda]

beautifully done dexter