SOLUTION: The tens digits of a 2-digit number is 3 more than the ones digit. if the product of the number and the sum of its digits is 841, find the number.
This is what I did :)
Tens= x
Question 1033581: The tens digits of a 2-digit number is 3 more than the ones digit. if the product of the number and the sum of its digits is 841, find the number.
This is what I did :)
Tens= x+3
DISABLED_event_Ones= x
Number is 10(x+3)+x
10x+30+x
11x+30
Sum of the digits 2x+3
So (11x+20)(2x+3) = 22x^2+93x-751=0
I'm stuck there. I don't know what I did wrong..
841 = 29*29 = .
This is the ONLY possible factoring of the number 841 in non-trivial smaller factors.
Do they satisfy the condition? - Obviously, not.
What is the conclusion? - The conclusion is: the condition is wrong.
