As an occasional diversion from work, I like to try maths puzzles, and to this end, I recently purchased a couple of high school maths textbooks (for use with students aged from about 14 to 16). Not that I can't do the puzzles (I think my maths is okay), but because sometimes the questions and puzzles in them some interesting ideas about extension activities - for example, the recent question about the circle in the corner of a circle and a square is probably intended to be solved with trigonometry, and instead, I solved it in terms of square roots, using just Pythagoras' theorem. That led to me solving the situation for the circle in the corner of a hexagon (which wasn't in the textbook, but which had an interesting solution too).

Anyway, I found a great question in the extension section at the end of one of the textbooks, and it goes like this:

Creating Numbers: a task requiring imagination

Your task is to create every number from 1 to 50.

You can use only the

You can use the

Here are some examples:

1 = (4-3) / (2-1)

20 = 4

68 = 34 * 2 * 1

75 = (4+1)

So, here goes... 1 to 50, using only 1, 2, 3 and 4 and the basic maths operators. Some of the answers seem a little repetitive or derivative (look at 38 through 43), and in some cases I found alternative answers afterwards.

1 = (4-3) * (2-1)

2 = (4-3) + (2-1)

3 = (4+2) / (3-1)
Anyway, I found a great question in the extension section at the end of one of the textbooks, and it goes like this:

Creating Numbers: a task requiring imagination

Your task is to create every number from 1 to 50.

You can use only the

**digits**1, 2, 3 and 4 once in each and the operations + - * / .You can use the

**digits**as powers, and you must use*all*of the**digits**1, 2, 3, 4Here are some examples:

1 = (4-3) / (2-1)

20 = 4

^{2}+3 +168 = 34 * 2 * 1

75 = (4+1)

^{2}* 3So, here goes... 1 to 50, using only 1, 2, 3 and 4 and the basic maths operators. Some of the answers seem a little repetitive or derivative (look at 38 through 43), and in some cases I found alternative answers afterwards.

1 = (4-3) * (2-1)

2 = (4-3) + (2-1)

4 = (4 * 3) / (2+1)

5 = (2 * 4) – (1 * 3)

6 = (2 * 4) – (3-1)

7 = (3+4) * (2-1)

8 = (3+4) + (2-1)

9 = (3+4) + (2 * 1)

10 = 1 + 2 + 3 + 4

11 = (4 * 3) – (2 - 1)

12 = (4 * 3) * (2-1)

13 = (4 * 3) + (2-1)

14 = (4 * 3) + (2 * 1)

15 = (4 * 3) + 2 + 1

16= 4 ^ ((3+1)/2)

17 = 3(4+1) + 2

18 = 4

^{2}+ (3-1)

19 = 4

^{2}+ (3 * 1)

20 = 4

^{2}+ 3 + 1

21 = (4+3) * (2+1)

22 = (4+1)

^{2}– 3

23 = 3

^{2}+ 14

24 = 1 * 2 * 3 * 4

25 = 31 – (2+4)

26 = 13 * (4-2)

27 = 3

^{2}* (4-1)

28 = 32 – (4 * 1)

29 = 31 – (4 -2)

30 = (4+1) * 3 * 2

31 = 34 – (1+2)

32 = 4

^{(3-1)}* 2

33 = 34 – (2 -1)

34 = 34 * (2-1)

35 = 34 + (2-1)

36 = (4 * 3) * (2+1)

37 = 34 + 1 + 2

38 = 42 – (3+1)

39 = 42 – (3 * 1)

40 = 41 – (3-2)

41 = 43 – (1 * 2)

42 = 43 – (2-1)

43 = 41 + (3-1)

44 = (14 * 3) + 2

45 = 43 + (2 * 1)

46 = 42 + 1 + 3

47 = 41 + (3 * 2)

48 = 24 * (3-1)

49 = ((4 * 1) + 3)

^{2}

50 = 41 + 3

^{2}

As the logical extension, I attempted to carry on past 50. It becomes increasingly difficult, since 1, 2, 3 and 4 are all small numbers, and the combinations of those small digits become less useful in making specific larger values (especially the prime numbers).

However, if we expand the rules to allow ! (factorial) and decimal points, then this enables us to find solutions for 57 (for example). I'd like to thank the free math help forum community (especially Denis for his initial suggestion to extend the rules), for the additional solutions, comments, corrections and suggestions. They've been very friendly in quickly adopting my idea and sharing their comments and solutions. An additional rule that's been introduced is the use of decimals - by doing this, we can include dividing by .2 (for example) as a way of multiplying by 5.

51 = (12 * 4) + 3 OR (4

^{2}+1) x 3

52 = 4

^{3}- 12

53 = (1 + 4!) * 2 + 3

54 = (13 * 4) +2

55 = 34+ 21

56 = (1 + 3 + 4!)2

57 =(1+4)! / 2 - 3 OR (4 + 2) / .1 - 3

58 = (31 * 2) - 4

59 = (21 * 3) - 4

60 = 3

^{4}- 21

61 = 4

^{3}- (1 + 2)

62 = 4

^{3}- (1 * 2)

63 = 4

^{3}- (2 - 1)

64 = (2 - 1) * 4

^{3}

65 = (2 - 1) + 4

^{3}

66 = (2 * 1) +4

^{3}

67 = (34 * 2) -1

68 = 34 * 2 * 1

69 = (34 * 2) + 1

70 = 4

^{3}+ (1 + 2)!

71 = ((4! / 2) * 3!) - 1

72 = 24 * 3 * 1

73 = (3 * 4!) + (2 - 1)

74 = (3 * 4!) + (2 * 1)

75 = (4+1)

^{2}* 3

76 = (41 * 2) - 3!

77 = ((4! + 1) * 3) + 2

78 = (4! + 2) * 3 * 1

79 = 3

^{4}- (2 * 1)

80 = 3

^{4}- (2 -1)

81 = (4! + 1 + 2) * 3

82 = 3

^{4}+ (2 - 1)

83 = 3

^{4}+ (2 * 1)

84 = 3

^{4}+ 2 + 1

85 = (43 * 2) - 1

86 = 43 * 2 * 1

87 = (21 * 4) + 3

88 = (43 + 1) * 2

89 = (3!)! / (2 * 4) - 1

90 = (1! + 2!) * (3! + 4!)

91 = (23 * 4) - 1

92 = 23 * 4 * 1

93 = (23 * 4) + 1

94 = (1 + 3)4! - 2

95 = 3!*(2

^{4}) - 1

96 = (12 * 4!) /3

97 = 4(3! - 2)! + 1

98 = (1 + 3)4! + 2

99 = 123 - 4!

100 = (3 / .12) * 4

The original textbook question asked only for the numbers from 1 to 50 with only powers and basic operators, but by expanding the rules, many more numbers have become achievable. Dare we go above 100?

Further posts: Numbers 101-150 and Numbers 150-200.

101 = (3^2)*11+2

ReplyDeleteThanks, Jasmina - unfortunately the rules of the game are to use only each number once, and to use all the digits 1, 2, 3 and 4.

ReplyDelete101 = (1/.2)³ - 4!, but 103 seems to need use of a recurring decimal.

ReplyDeleteCorrection: 103 has a solution that uses a decimal point, but I think 109 needs use of a decimal point and a recurring decimal or a square root symbol.

DeleteI see you have 109 without recurring decimal, so now I am not sure when it is first needed - maybe for 137. It would be nice to avoid use of square root, and I am not sure when that is first needed either.

Deletei believe 11 = (4*3)-(2-1) and (4*3)-(2*1) would actually equal 10 :)

ReplyDeleteIt does, so I'll make the correction - thank you for spotting it! :-)

Deleteget 12.75 using 1,2,3 and 4

ReplyDelete12 + 3/4 perhaps?

DeleteThank you for response:) If it was only that easy!

ReplyDeleteCan't combine digits, use them twice or use decimals:(

Can't combine digits? According to whose rules?

ReplyDeleteIf you can't combine digits, and have only 1, 2, 3 and 4 then I think you're going to struggle.

The closest I can get, which I don't think you'll approve of either, is (4 + 1/2) * 3 = 13.5.

I got less than 0.25 away from 12.75 , but supposedly there is a way to get 12.75

ReplyDeletefactorials and square roots are allowed, just wonder if there are other math symbols/operations that can be used to increase/decrease result

Thanks for responding:)it will be a long weekend with math lol

Can you help me with a spreadsheet I am making?

ReplyDeleteI am doing it without combining numbers. Can't find one for 34 :/

You can use factorial, exponents, and the four basic ones. No decimals or combining numbers.

The spreadsheet is here, if you want to see it:

https://docs.google.com/a/coronadousd.org/spreadsheets/d/167DUN-2R1petAI2oYYV6BFDpNaBJzGwmktF1n1bs4rw/edit?usp=sharing

https://docs.google.com/spreadsheets/d/167DUN-2R1petAI2oYYV6BFDpNaBJzGwmktF1n1bs4rw/edit?usp=sharing

Delete43 doesn't look quite right. How about 4 / .1 + 3! / 2

ReplyDeleteHopefully I didn't double post

The 16th problem is a little convoluted. Thanks for making this

ReplyDeleteYou're right: in the interests of simplicity, how about 4* (3+(2-1)) as an improvement?

DeleteThis is a great improvement! Thanks for responding

DeleteThe 49th problem doesn't look quite right. It's somewhat off from the actual answer.

ReplyDelete(4*1) = 4

Delete4 + 3 = 7

7^2 = 49

Hence ((4*1)+3)^2 = 49

how to get 91 using only values of 1,2,3,4 and pemdas

ReplyDeleteHow to make 34 using 1,2,5,6 using all digits only once? Thanks in advance.

ReplyDelete(6*(5+1))-2 = 36-2= 34

ReplyDelete