## 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!!!)

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

**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!]

**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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