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$pOnGeBoB : )
Lv 4
$pOnGeBoB : ) asked in Science & MathematicsMathematics · 1 decade ago

Math Problem (riddle) -hard?

only using the numbers 1953

make a equation to get the answers of any of the following

32

26

28

29

30

36

37

38

40

choose one you like to do or try all

you can use -, +, / x and parenthesis, square root, absolute value, exponents, ect

you can also put the number together ex: 53-19=

Update:

thanks everybody

5 Answers

Relevance
  • Omega_lx
    1 decade ago
    Favourite answer

    32 = (√9 + 1) * (5 + 3)

    26 = √(9 - 5) * 13

    28 = (9 + 5) * (3 - 1)

    29 = (1 + 3) * 5 + 9

    30 = (9 - 3) * 5 / 1

    36 = (1 + 5) * (9 - 3)

    37 = (9 - 1) * 5 - 3

    38 = 91 - 53

    For this one only way I found is to use cube root (hardest one imo)

    40 = (5 + 3)^(1/3) * (9 + 1)

  • The Gnostic
    Lv 7
    1 decade ago

    26 = 13 × √(9 - 5)

    30 = 5 × 3! × 1^9

    37 = (5! ÷ 3)^1 - √9

    40 = 5! ÷ 3 × 1^9

  • ?
    1 decade ago

    91 - 53 = 38

    [(3^5)/9]+1=28

    [(3^5)/9]-1=26

    {3^(square root 0f 9)]+5}/1=32

  • Anonymous
    1 decade ago

    (95+1)/3 = 32

    your right , these are hard ;P

  • viet fire
    1 decade ago

    well i only got some

    39-1......38

    39+1.....40

    35-9......26

    35+1.....36

    3^2 + 19.....28

    19+13.......32

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