108
There are exactly 20 Nines. 9,19,29,39,49,59,69,79,89,90,91,92,93,94,95,96,97,98 which totals 18 and the last number 99 has 2 nines and therefore 20 nines in total.
It could be: 5*90 = 5
(9*9)+9+9+(9/9) = 100
73 = |- (sqrt(9!))/9 -| + 9 - sqrt 9 [ceiling]= |_ sqrt sqrt sqrt sqrt sqrt sqrt sqrt 139! _|{the rounded-down 128th-root of 139!; since we can make 139 with fournines, this gives us another way to make 73 with four nines}Another option that you may find is simpler:Noting that 9 to the power zero equals 1, then:73 = (9 x 9) - 9 + 9 0= 81 - 9 + 1 = 73
About 100,000 of each.
Not counting the 3 in the ones place, there are 100229 threes and 100106 nines.
Zeros: 488,895 Ones: 600,001 Twos, Threes, Fours, Fives, Sixes, Sevens, Eights, Nines: . . . . . . 600,000 each Total: 5,888,896 digits. Start cutting, Kim !
9x9+9+9+9/9
Very interesting !If you write the counting numbers from 1 to 1,000,000,you'll write the digit '9' exactly 600,000times.
The number of strings of four decimal digits that have exactly three digits which are nine is the number of ways to choose three places to choose the nines multiplied by the number of possibilities for the other digits (because you can put all of the possibilities in an array so that each column has a different way of arranging the nines and that each row has a different set of other digits (because they do not affect each other)). The number of ways to choose the three places to choose the nines is the same as the number of ways of choosing 4-3=1 spot for there not to be a nine. The number of ways of choosing 1 spot where there are 4 spots is 4 (the first, the second, the third, or the fourth). Since there are 9 digits that are not nine, the number of choices for the other digits is 9 (since there is only one digit). Thus, there are 4*9=36 strings of four decimal digits that have exactly three digits which are nine.
30 and 6 18 and 18 four nines six sixes nine fours twelve threes eighteen twos thirty-six ones
The nines received their memory wiped as a part of their final ceremony.
If that is a terminating decimalthen put the digits after the decimal point over 1 followed by the same number of zeros as digits after the decimal point and simplify:0.363636363636 has 12 digits, so put over 1 followed by 12 zeros:0.363636363636 = 363636363636/1000000000000= 90909090909/250000000000If that is a repeating decimal (as in 0.363636...)then put the repeating digits over the same number of nines and simplify: 0.363636... has 2 repeating digits (36), so put them over 2 nines (99):0.363636... = 36/99= 4/11
They receive their first Bicycle.
There are infinitely many numbers between 14 and 15. Just type "14." then continue typing any arbitrary digits (except all zeroes, or all nines), and you have a number between 14 and 15.There are infinitely many numbers between 14 and 15. Just type "14." then continue typing any arbitrary digits (except all zeroes, or all nines), and you have a number between 14 and 15.There are infinitely many numbers between 14 and 15. Just type "14." then continue typing any arbitrary digits (except all zeroes, or all nines), and you have a number between 14 and 15.There are infinitely many numbers between 14 and 15. Just type "14." then continue typing any arbitrary digits (except all zeroes, or all nines), and you have a number between 14 and 15.
The answer is infinite. There is no collection of nines so large that you can't add another nine to it.