2 and 6
88
19
The number is 45. The sum of its digits i.e. 4+5=9 Five times the sum of its digits i.e. 5 times 9 which is 45
1 + 1 = 2 The sum of the digits is therefore 2.
Add the digits together. The sum of the digits of 23 is 5.
88
There are two numbers that satisfy the criteria. 38 and 83
12 and 2412 and 2412 and 2412 and 24
33
12 or 24
19
12
62
-- "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)
The 2-digit number must be 20, because it is the only 2-digit number whose sum of its two even digits, 2 + 0 = 2, is greater than the product of its two even digits, 2 x 0 = 0. Moreover, 20 is a product of the two consecutive integers 4 and 5.
28
the sum of my digits is 6? answer=60