Monday, December 29, 2014
100
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Find at least four ways of writing the number 100, each time using only one digit repeated five times.
For example: (999 / 9) - 9 = 102 (but you must get 100, not 102!!!)
Good luck!
For example: (999 / 9) - 9 = 102 (but you must get 100, not 102!!!)
Good luck!
100 Puzzle Solution
- 111 - 11 = 100
- (3 * 33) + (3 / 3) = 100
- (5 * 5 * 5) - (5 * 5) = 100
- (5 + 5 + 5 + 5) * 5 = 100
- (11 - 1) ^ ( 1 + 1) = 100 [Thanks to Steven Renich for that one!]
- ((2 * 2 * 2) + 2) ^ 2 = 100 [Thanks to David Cohen for this other one!]
- ((99 * 9) + 9) / 9 = 100 [Thanks to Taylor Lowry for this other one!]
- ((22 - 2) / 2) ^ 2 = 100 [Thanks to Karen D. Miller for this other one!]
- 6! / 6 - 6! / (6 * 6) = 100 [Thanks to Karen D. Miller for this other one!]
- 5! - 5 - 5 - 5 - 5 = 100 [Thanks to Karen D. Miller for this other one!]
- 5! - (5 + 5 + 5 + 5) = 100 [Thanks to Jim St. Clair for this other one!]
- 5 * 5 * (5 - 5 / 5) = 100 [Thanks to Rishi Mohan Sanwal for this one!]
- 4! + 4! + 4! + 4! + 4 = 100 [Thanks Karen D. Miller and Saurabh Gupta!]
- 99 + 9 ^ (9 - 9) = 100 [Thanks Gopalakrishnan Thirumurthy for this one!]
- 5! - 5 * (5 - 5 / 5) = 100 [Thanks Gopalakrishnan Thirumurthy for it!]
- (4! + 4 ^ (4 - 4)) * 4 = 100 [Thanks Gopalakrishnan Thirumurthy for it!]
- 4! * 4 + 4 - 4 + 4 = 100 [Thanks Gopalakrishnan Thirumurthy for it!]
- (4! * 4) + (4 * 4 / 4) = 100 [Thanks Gopalakrishnan Thirumurthy for it!]
- 5 * 5 * (5 - (5 - 5)!) = 100 [Thanks Gopalakrishnan Thirumurthy for it!]
- ((9 - 9) */ 9)! + 99 = 0! + 99 = 1 + 99 = 100 [Thanks to Bala Neerumalla!]
- (3 - 3)! + 33 * 3 = 100 [Thanks Bala Neerumalla for this other one!]
- (5 + 5) ^ ((5 + 5) / 5) = 100 [Thanks Bala Neerumalla for this one!]
- ((2 ^ 2) * 2 + 2) ^ 2 = 100 [Thanks Bala Neerumalla for this one!]
- (4! + ((4 - 4) */ 4)!) * 4 = 100 [Thanks Bala Neerumalla for this one!]
Glen Parnell points out that, in any solution in which the digit 2 is used, it could be replaced by 2!, whenever the 2 is not only a digit but also a number, as 2 = 2! (but of course, 22 is not equal to 22!).
Bala Neerumalla has suggested some alternative solutions using trigonometric functions Sin() and Cos():
- Sin(99 - 9) + 99 = Sin(90) + 99 = 1 + 99 = 100
- Cos((9 - 9) */ 9) + 99 = Cos(0) + 99 = 1 + 99 = 100
- Cos(3 - 3) + 33 * 3 = 100
- (4! + (Sin(4-4)*4)!)*4 = 100
- 88 + 8 + CR(8) + CR(8) = 100
- (8 + CR(8)) ^ (CR(8) * (8 / 8)) = 100
- ((CR(8) * CR(8) * CR(8)) + CR(8)) ^ CR(8) = 100
- ((8 * (8 / 8)) + CR(8)) ^ CR(8) = 100
Bala Neerumalla smartly noticed that he can produce number 100 by using a different number base than the normal base 10 (b10). The following example is in base 5 (b5):
- 444 - 44 = (4 * 5^2 + 4 * 5^1 + 4 * 5^0) - (4 * 5^1 + 4 * 5^0) = 100
- 111 - 11 = 100 [Wolfgang Solfrank pointed out that this works in any number base, thanks!]
- b16 4 ^ 4 +- 4 * (4 - 4) = 100 [Thanks to Glen Parnell for this one!]
- b16 (2 * 2 * 2 * 2) ^ 2 = 100 [Thanks to Glen Parnell for this one!]
- b8 4 * 4 * 4 +- (4 - 4) = 100 [Thanks to Glen Parnell for this one!]
- b8 (4 ^ 4) / 4 +- (4 - 4) = 100 [Thanks to Glen Parnell for this one!]
- b8 4 * 4 * 4 * (4 / 4) = 100 [Thanks to Glen Parnell for this one!]
- b8 (2 ^ (2 * 2)) * 2 * 2 = 100 [Thanks to Glen Parnell for this one!]
- b2 11 - 1 + 1 + 1 = 100 [Thanks to Glen Parnell for this one!]
- bX+1 ((X - X) */ X)! + XX = 100
- b16 ((F - F) */ F)! + FF = 100
- b12 ((B - B) */ B)! + BB = 100
- b10 ((9 - 9) */ 9)! + 99 = 100 [As already shown by Bala Neerumalla]
- b8 ((7 - 7) */ 7)! + 77 = 100
- b4 ((3 - 3) */ 3)! + 33 = 100
- b2 ((1 - 1) */ 1)! + 11 = 100
- ((L / L) + (L / L)) * L = ((50 / 50) + (50 / 50)) * 50 = 100
- (C / C) * (C / C) * C = (100 / 100) * (100 / 100) * 100 = 100
- (C / C) - (C / C) + C = (100 / 100) - (100 / 100) + 100 = 100
- CCC - CC = 300 - 200 = 100
- CC - CC + C = 200 - 200 + 100 = 100
- (L + L) * L ^ (L - L) = (50 + 50) * 50 ^ (50 - 50) = 100
- (L + L) / L ^ (L - L) = (50 + 50) / 50 ^ (50 - 50) = 100
- XX * X - X * X = 20 * 10 - 10 * 10 = 100
- (X + (X / X)) * X - X = (10 + (10 / 10)) * 10 - 10= 100
- X * X * (X ^ (X - X)) = 10 * 10 * (10 ^ (10 - 10) = 100
- X * X / (X ^ (X - X)) = 10 * 10 / (10 ^ (10 - 10) = 100
- C * C * C / C * C = 100 * 100 * 100 / 100 * 100 = 100
- (C - C) + (C - C) + C = (100 - 100) + (100 - 100) + 100 = 100
- L * (L - L) + L + L = 50 * (50 - 50) + 50 + 50 = 100
- C * (CC / C) - C = 100 * (200 / 100) - 100 = 100
- C * (C / C) + (C - C) = 100 * (100 / 100) + (100 - 100) = 100
Now, how about some solution using Roman numerals 'D' (500), or 'M' (1000)? If you find any alternative solutions, get in touch with us by email!
Easy!
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