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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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15y ago

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Related Questions

What is thesmallest 3digit number with unique digits?

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


Using only the digits 4 5 7 8 find the greatest product and the least product possible where one factor is a 3digit number?

4 x 578 = 2312 8 x 754 = 6032


What number has 24 as the product of its digits and the sum of its digits is a prime number?

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


How many digits can the product of 2-digit numbers have?

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.


Find 5 four digit numbers such that the product of the digits of each four digit prime number are also prime?

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


How do you multiply three digits in a logical way?

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


If the sum of a two digit number is 12 and the digits interchange their places the new number exceeds the original number by 25 what is the product?

-- "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)


What fraction of all 4 digits natural numbers have a product of their digits that is even?

To find the fraction of 4-digit natural numbers with a product of their digits that is even, we first need to determine the total number of 4-digit natural numbers. There are 9000 such numbers (from 1000 to 9999). Next, we consider the conditions for the product of digits to be even. For a number to have an even product of digits, at least one of the digits must be even. There are 5 even digits (0, 2, 4, 6, 8) and 5 odd digits (1, 3, 5, 7, 9). Therefore, the fraction of 4-digit natural numbers with an even product of digits is 5/10 * 9/10 * 9/10 * 9/10 = 3645/9000 = 809/2000.


Between 70 and 80 the product of its digits is 21?

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


How may digits are there in a 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.


What is the greatest number of digits a product could have if a 4-digit number is multiplied by a 2-digit number?

2


How to multiply numbers while considering significant digits?

When multiplying numbers with significant digits, count the total number of significant digits in each number. Multiply the numbers as usual, but round the final answer to match the least number of significant digits in the original numbers.