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.
67/72 of them.
from 3 digits (10x10) to 4 digits (99X99)
144
-21
two
231
22
420
231
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.
67/72 of them.
If they are multiples of 9, the digits add up to a multiple of 9.
from 3 digits (10x10) to 4 digits (99X99)
15 or 35
You had me until "product." The product of 4 digits can't be prime.
As the numbers 1 to 99 are multiplied together, one factor of the product will be 10 which means the last digit must be 0 (a zero). Without working out the product, it can be seen that every multiple of 5 in the original numbers can be paired up with an even number (that is not a multiple of five) and multiplied together (which produces a multiple of 10) which are all factors of the product together; thus the product will ends with that number of zeros: there are 19 multiples of 5 in the numbers 1-99, so the last 19 digits of the product are all 0 (zero).