Find the smallest positive integer which requires at least five 9's to be written in an expression which is equal to said number.
Allowed operations:
Disallowed operators. Everything else, including but not limited to:
An example of how to use these operators is using 4's to create 28 would be 2 8 = 4 × ( 4 + 4 + ( 4 ÷ 4 ) ) , which uses five 4's, but you can also do this with two 4's as follows: 2 8 = 4 ! + 4 , and this is the shortest method. Therefore, 28 can be written using at least two 4's.
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That's an impressive list!
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Yeah, it is quite large (and took me two sessions to do in latex), but I had the solutions written up. I have an iOS app called Tchisla which has similar exercises, and so far, no one has come up with a better solution for 94.
@Sharky Kesa By the way, you have a typo in your equation for 39. It uses a 3.
@Sharky Kesa - Man that was quite the challenge!! ,Here are some that i did differently,
2 2 = ( 9 ) ! ( 9 + 9 ( 9 ) ! ) 2 3 = 9 ( ( 9 ) ! ) 9 − 9 3 2 = 9 9 9 − 9 3 8 = 9 1 ( ( 9 ) ! ( ( 9 ) ! ) ! − ( 9 ) ! ) 5 7 = 9 1 ( ( ( 9 ) ! ) ! 9 ! + 9 ) 6 3 = 9 9 − ( ( 9 ) ! ( 9 ) ! ) 6 6 = 9 9 9 ( 9 ) ! 7 6 = 9 ( ( 9 ) ! ) ! − ( 9 ) ! ( 9 ) !
Also i think you have to correct 8 9 :)
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I assume in 38 and 57, you don't actually have the 1 there, but should be the expression in the following brackets? Also, great job! You'll find that my solutions give the lowest number of 9's required to create the number.
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Once again, I have no better method of solving this apart from going through all the numbers:
1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 2 0 2 1 2 2 2 3 2 4 2 5 2 6 2 7 2 8 2 9 3 0 3 1 3 2 3 3 3 4 3 5 3 6 3 7 3 8 3 9 4 0 4 1 4 2 4 3 4 4 4 5 4 6 4 7 4 8 4 9 5 0 5 1 5 2 5 3 5 4 5 5 5 6 5 7 5 8 5 9 6 0 6 1 6 2 6 3 6 4 6 5 6 6 6 7 6 8 6 9 7 0 7 1 7 2 7 3 7 4 7 5 7 6 7 7 7 8 7 9 8 0 8 1 8 2 8 3 8 4 8 5 8 6 8 7 8 8 8 9 9 0 9 1 9 2 9 3 9 4 = 9 9 = 9 ( 9 ) ! = 9 = 9 + 9 9 = 9 + 9 ( 9 ) ! = ( 9 ) ! = 9 − 9 ( 9 ) ! = 9 − 9 9 = 9 = 9 + 9 9 = 9 9 9 = 9 + 9 = 9 + 9 + 9 9 = 9 + 9 + 9 ( 9 ) ! = 9 + ( 9 ) ! = 9 + 9 − 9 ( 9 ) ! = 9 + 9 − 9 9 = 9 + 9 = 9 + 9 + 9 9 = ( 9 ) ! ( 9 ) ! ( ( 9 ) ! ) ! = 9 + 9 + 9 = 9 ( ( ( 9 ) ! ) ! + ( 9 ) ! ) × ( 9 ) ! = ( 9 ) ! ( 9 ) ! ( ( 9 ) ! ) ! + 9 = 9 × 9 − 9 = 9 × 9 − 9 ( 9 ) ! = 9 × 9 − 9 9 = 9 × 9 = 9 × 9 + 9 9 = 9 × 9 + 9 ( 9 ) ! = 9 × 9 + 9 = 9 9 9 − ( 9 ) ! = 9 ( 9 ) ! 9 ( ( 9 ) ! ) ! = 9 9 9 = ( 9 ) ! × ( 9 ) ! − 9 ( 9 ) ! = ( 9 ) ! × ( 9 ) ! − 9 9 = ( 9 ) ! × ( 9 ) ! = ( 9 ) ! × ( 9 ) ! + 9 9 = ( 9 ) ! × 9 ) ! + 9 ( 9 ) ! = ( 9 ) ! × ( 9 ) ! + 9 = ( 9 ) ! 9 ( ( 9 ) ! ) ! = 9 ( 9 ) ! ( ( 9 ) ! ) ! + 9 = ( 9 ) ! × ( 9 ) ! + ( 9 ) ! = ( 9 ) ! 9 ( ( 9 ) ! ) ! + 9 = 9 ( ( 9 ) ! ) ! − ( 9 ) ! × ( 9 ) ! = ( 9 ) ! × 9 − 9 = ( 9 ) ! 9 ( ( 9 ) ! ) ! + ( 9 ) ! = ( ( 9 ) ! ) ! × 9 9 ! − 9 = 9 × ( 9 ) ! − ( 9 ) ! = ( 9 ) ! 9 ( ( 9 ) ! ) ! + 9 = ( ( 9 ) ! ) ! × 9 ) 9 ! − ( 9 ) ! = 9 × ( 9 ) ! − 9 = 9 × ( 9 ) ! − 9 ( 9 ) ! = 9 × ( 9 ) ! − 9 9 = 9 × ( 9 ) ! = 9 × ( 9 ) ! + 9 9 = ( ( 9 ) ! ) ! × 9 9 ! = 9 × ( 9 ) ! + 9 = ( 9 ( 9 ) ! ) ( 9 ) ! − ( 9 ) ! = ( ( 9 ) ! ) ! × 9 9 ! + 9 = 9 × ( 9 ) ! + ( 9 ) ! = ( 9 ( 9 ) ! ) ( 9 ) ! − 9 = 9 ( ( 9 ) ! ) ! − 9 − 9 = 9 × ( 9 ) ! + 9 = ( 9 ( 9 ) ! ) ( 9 ) ! = 9 ( ( 9 ) ! ) ! − 9 − ( 9 ) ! = ( ( ( 9 ) ! ) ! + ( 9 ) ! ) × ( 9 ) ! = ( 9 ( 9 ) ! ) ( 9 ) ! + 9 = 9 ( ( 9 ) ! ) ! − 9 − 9 = 9 × 9 − 9 − 9 = ( 9 ( 9 ) ! ) ( 9 ) ! + ( 9 ) ! = 9 ( ( 9 ) ! ) ! − 9 = 9 × 9 − 9 = 9 ( ( 9 ) ! ) ! − 9 − ( 9 ) ! = 9 ( ( 9 ) ! ) ! − ( 9 ) ! = 9 × 9 − ( 9 ) ! = 9 ( ( 9 ) ! ) ! − 9 − 9 = 9 ( ( 9 ) ! ) ! − 9 = 9 × 9 − 9 = 9 ( ( 9 ) ! ) ! − 9 = 9 ( ( 9 ) ! ) ! = 9 × 9 = 9 × 9 + 9 9 = 9 ( ( 9 ) ! ) ! + 9 = 9 × 9 + 9 = 9 ( ( 9 ) ! ) ! − 9 + ( 9 ) ! = 9 ( ( 9 ) ! ) ! + ( 9 ) ! = 9 × 9 + ( 9 ) ! = 9 ( ( 9 ) ! ) ! − 9 + 9 = 9 ( ( 9 ) ! ) ! + 9 = 9 9 − 9 = 9 ( ( 9 ) ! ) ! + 9 9 = 9 ( ( 9 ) ! ) ! + 9 + 9 = 9 9 − ( 9 ) ! = 9 9 − 9 − 9 ( 9 ) !
For 94, I have not found a better solution (and I doubt there is), and I found that for all numbers under 100, all numbers but 94 require at most four 9's.