You had me until "product." The product of 4 digits can't be prime.
2
It can have up to 5 digits.
It can have one . . . 1.0 x 1.0 = 1 It can have two . . . 1.0 x 0.1 = 0.1 It can have three . . . 10 x 10 = 100 It can have four . . . 99 x 99 = 9801 and .01 x .01 = .0001
73 is the largest two-digit number that is prime and has prime numbers for both of its digits.
81
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.
2
-- "The sum of a two-digit number" is unclear. I took it to mean"The sum of the digits of a two-digit number."-- "... the product ?" is unclear. Are you looking for the product of the two digits,or the product of the forward and backward numbers ?-- It's not possible to write a number whose two digits sum to 12 and whosereverse exceeds it by 25. The tens digit would have to be 4.611... and the unitsdigit would have to be 7.388... .Then(4.611...) + (7.388...) = 12(46.111...) + (7.388...) = 53.5(73.888...) + (4.611...) = 78.5Difference = 25So, the number can't be written, but ...The product of its two digits is 34.071 (rounded)The product of the forward and reverse numbers is 4,199.75 (rounded)
To find the last but one digit in the product of the first 75 even natural numbers, we need to consider the units digit of each number. Since we are multiplying even numbers, the product will end in 0. Therefore, the last but one digit (tens digit) will depend on the multiplication of the tens digits of the numbers. The tens digit will be determined by the pattern of the tens digits of the even numbers being multiplied.
It can have up to 5 digits.
The four numbers are: 113, 131, 151 and 191. For the product of the digits to be prime, the number must contain 2 ones - which greatly simplifies the exercise.
It can have one . . . 1.0 x 1.0 = 1 It can have two . . . 1.0 x 0.1 = 0.1 It can have three . . . 10 x 10 = 100 It can have four . . . 99 x 99 = 9801 and .01 x .01 = .0001
73 is the largest two-digit number that is prime and has prime numbers for both of its digits.
It can have 4 digits, because the highest possible two digit numbers 99*99=9801.
Six.
81
In the numbers 1-9 each number has 1 digit and there are 9 of them, so that's 9.In 10-99 each number has 2 digits, and there are 90 of them: 2x90 = 180There are 900 three digit numbers [100 through 999]: 2700 digits.There are 9000 four digit numbers: 36000 digits.90,000 numbers with five digits: 450,000 digits.900,000 numbers with six digits: 5,400,000 digits.Then 1 number with seven digits: 7 digits.Add them up and you have 5,888,896 digits.