The largest digits number with only 2 digits alike is 99
99
45. There are only 10 digits: 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. So I am not sure where the "99 digits", in the question come from.
99 Or how about (9)9 ?
This question can be answered regardless of what digits are used. If there are four options for the first digit, then there are only 3 options for the second digit, and 2 options for the third. There are no options left for the fourth and final digit. So, the answer is 4! (or four factorial).
50
100*99*...*2*1= 93326215443944152681699238856266700490715968264381621468592963895217 59999322991560894146397615651828625369792082722375825118521091686400 0000000000000000000000
The divisible rule for 2 is if the last digit of the number is 0,2,4,6, or 8. The divisible rule for 3 is if the last 2 digits of the number is divisible by 3,03,06,09,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,90,93,96, or 99.
The largest digits number with only 2 digits alike is 99
99
99 + 99 = 198
If you mean, "What is the largest number of digits possible in the product of two 2-digit numbers" then 99 * 99 = 9801, or 4 digits. Anything down to 59 * 17 = 1003 will have 4 digits.
First number is 20 and last is 99 so 20 = 1 then 21 = 2 then 29 = 10 So from 29 take 19 to make 10 then apply to 99 as in 99 - 19 = answer 80
Oh, what a delightful question! If we take a look at the numbers from 1 to 99, we'll find that they have a total of 189 digits. Each number from 1 to 9 has one digit, numbers from 10 to 99 have two digits each. Just imagine all those lovely digits coming together to create a beautiful numerical landscape!
1680 any 4 from 8 is factorial 8 divided by factorial 4 8*7*6*5*4*3*2*1/4*3*2*1 = 8*7*6*5 = 1680
45. There are only 10 digits: 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. So I am not sure where the "99 digits", in the question come from.
99 Or how about (9)9 ?