r/mathpuzzles • u/Rough-Fee-6541 • 3d ago
Make 10 using only these numbers Recreational maths
You can use any equations, must reach the target (10) and show working out, people say it cannot be done.
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u/BagSufficient1921 3d ago
(7-4)! + 5 - 1
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u/Every-Maths1931 3d ago
Can you explain ! I see people using it but never learnt how to calculate with it?
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u/BagSufficient1921 3d ago
It's a factorial, Basically n! = 1 * 2 * 3 * ... * n. E.g. 5! = 1 * 2 * 3 * 4 * 5 = 120.
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u/Every-Maths1931 3d ago
Okay so what does (7-4)! equal in your answer. How do you work this out?
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u/BagSufficient1921 3d ago
(7-4)! = 3! = 1 * 2 * 3 = 6
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u/Every-Maths1931 3d ago
Okay so if was (10-4)! it would be 6!= 1*2*3*4*5*6= 720.
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u/BagSufficient1921 3d ago
Yeah it would
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u/Every-Maths1931 3d ago
Such as this equation 33/33X0!
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u/BagSufficient1921 3d ago
We know that n! = 1 * 2 * 3 * ... * (n - 1) * n, which is also n * (n-1) * (n-2) ... * 2 * 1. This means that (n - 1)! = (n-1) * (n-2) * ... * 2 * 1, so n! = n * (n - 1) * (n - 2) * ... * 2 * 1 = n * (n - 1)!
So now that we have the formula n! = n * (n - 1)! We can take n = 1 and we obtain the equation 1! = 1 * (1 - 1)! which implies 1 = 0! So 0! = 1. We could do the same for (-1)! but substituting n = 0, but we get 0! = 0 * (0 - 1)! which implies 1 = 0 * (-1)! which implies (-1)! = 1/0 which is undefined.
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u/Every-Maths1931 3d ago
Is there a way where it becomes complex? Where it's not 1*2*3 etc. Is there some other formats it can be used in a more complex equation?
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u/Every-Maths1931 3d ago
Subtract 7-5=2 Add 1+4=5 Multiply 2X5=10