Question

If a, b, c, and d are four different digits from 1 to 9, inclusive, then what's the largest possible value for the decimal a.b + c.d? will medaL

Answer 1

@Atsie

Answer 2

To be very honest, @Scott208 I've not heard of this and neither do I know how to do it. I've read it over and I don't really understand it. I don't want to be misguiding, so I'd trust someone else to be of help to you. I'm really sorry. :/

Answer 3

alright thanks anyways

Answer 4

You're welcome :)
I'll try to get someone else for ya ^.^

Answer 5

\(a.b+c.d\)
\(a+\large\frac{b}{10}\) + \(c +\large\frac{d}{10}\)

\(\large \frac{10a+b+10c+d}{10}\)
\(\large \frac{a+b+c+d +9a+9c}{10}\)

now if we want the maximum value of a.b+c.d then either of a/b/c/d must have values 9/8/7/6
so a+b+c+d is 9+8+7+6=30

so our equation becomes-

\(\large \frac{a+b+c+d +9a+9c}{10}\)
\(\large \frac{30+9(a+c)}{10}\)
\(\large\frac{30}{10}+\large\frac{9(a+c)}{10}\)
\(3+\large\frac{9(a+c)}{10}\)

now maximum value of a+c can be 9+8=17

so

\(3+\large\frac{9(17)}{10}\)

Answer 6

thanks

Answer 7

np

Answer 8

so 18.3

Answer 9

yeah

Answer 10

cool

Answer 11

you can also directly say that for a.b+c.d to be maximum a and c must be the biggest numbers that can be possible
so clearly a=9 or 8 and c=8 or 9
similarly d=7 or 6 and b=6 or 7
so the sum would be = 18.3