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.
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).
99 Or how about (9)9 ?
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
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.
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
1 to 9 have 1 digit and 10 to 99 have 2 digits so (9 x 1) + (90 x 2) = 189
99 Or how about (9)9 ?