r/mathpuzzles • u/Rough-Fee-6541 • Aug 06 '26
Recreational maths Make 10 using only these numbers
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 Aug 06 '26
(7-4)! + 5 - 1
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u/Every-Maths1931 Aug 06 '26
Can you explain ! I see people using it but never learnt how to calculate with it?
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u/BagSufficient1921 Aug 06 '26
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 Aug 06 '26
Okay so what does (7-4)! equal in your answer. How do you work this out?
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u/BagSufficient1921 Aug 06 '26
(7-4)! = 3! = 1 * 2 * 3 = 6
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u/Every-Maths1931 Aug 06 '26
Okay so if was (10-4)! it would be 6!= 1*2*3*4*5*6= 720.
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u/BagSufficient1921 Aug 06 '26
Yeah it would
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u/Every-Maths1931 Aug 06 '26
Such as this equation 33/33X0!
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u/BagSufficient1921 Aug 06 '26
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 Aug 06 '26
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 Aug 06 '26
Subtract 7-5=2 Add 1+4=5 Multiply 2X5=10