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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.

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Q: What four 3-digit prime numbers less than 200 each have a prime number for the sum of their digits AND a prime number for the product of their digits?

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0.12 Or 102 if you do not want to include non integers.

4 x 578 = 2312 8 x 754 = 6032

There are two numbers that satisfy the criteria. 38 and 83

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.

You had me until "product." The product of 4 digits can't be prime.

Just multiply one pair of your numbers to give you a product, and then multiply their product by your third number.

Digits is how many numbers you have in a number. If you have the number 4 it has one digit if you have the number 20 it has two digits and if you have the number 558 it has three digits. So basically in the number 1085 it has 4 digits because there is 4 numbers in it, the numbers are 1,0,8 and 5. Hoped you understand.

-- "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)

Try the different numbers - there are only 11 after all - and multiply their digits. Or analyze the factors of the number 21.

2

52 is a number whose digits have a product of 10.

88

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