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

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The digits product of two digit has number four

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.

Take the smallest 2-digit number (10), and see how many digits you get when you multiply it by itself.Also, take the largest possible 2-digit number (99), and see how many digits you get when you multiply it by itself.

2

It can have up to 5 digits.

from 3 digits (10x10) to 4 digits (99X99)

The greatest number is 7.

-- "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)

73 is the largest two-digit number that is prime and has prime numbers for both of its digits.

Six.

81

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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