Monday, December 29, 2014

100

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!

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!]
Forum member Bealzbob reminds us that, whenever we have a subtraction like A - B, we can rewrite it as an addition with a negative number, like A + (-B). In the case of the first solution, 111 - 11 = 111 + (-11).

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
Gopalakrishnan Thirumurthy has intelligently played around with cubic root functions. If we were to represent this function with a name like, say, CR(), the following alternative solutions could be achieved:
  • 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
However! If the cubic root function is represented as it should be, ie ³Ö, then there is the extra "3" which invalidates the solution. These alternative solutions using CR(), have therefore been given as possible examples that will not be accepted in the future. Unless, of course, the "3" in ³Ö is actually one of the five digits used in a solution. Same goes with any other root or power. This also applies to any usage of the symbol "%", as it really involves the number 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
However! The above equation mixes up two different number bases: the left hand side is b5, while the right hand side is b10 (100 b5 is equal to 25 b10). This is the reason why we will not be posting anymore solutions that mix different number bases. We will, however, accept solutions that use number bases different from 10, as long as they are consistent on both sides of the equation. Therefore, we will accept:
  • 111 - 11 = 100 [Wolfgang Solfrank pointed out that this works in any number base, thanks!]
Here are some more examples of equations using different number bases:
  • 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!]
Furthermore, Glen Parnell noticed that there exists a general rule that applies to equations using any number base:
  • bX+1 ((X - X) */ X)! + XX = 100
Here are some examples of it:
  • 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
Jeff Shall cleverly discovered that he can produce 100 using the Roman numeral 'L', which corresponds to number 50:

  • ((L / L) + (L / L)) * L = ((50 / 50) + (50 / 50)) * 50 = 100
Unashamedly based on the above solution by Jeff Shall, we found that we can also use the Roman numeral 'C', which corresponds to number 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
Gopalakrishnan Thirumurthy has discovered a lot of new combinations with Roman numerals, including numeral 'X', which is number 10:

  • (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
Gopalakrishnan Thirumurthy has also noticed that all the solutions involving five times the number 5, could be rewritten by substituting it with Roman numeral 'V'. This is also true if, instead of number 5 and numeral 'V', we were talking about number 1 and Roman numeral 'I'.
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!