The digits of a positive two-digit integer N are interchanged to form integer K. Find all possibilities for N under the conditions described: The sum of K and twice N is less than 60. How do I write an equation for this?
\[26 \text{ is a two digit number } \\ \text{ another way to express this is } 2 \cdot 10 +6 \\ \text{ So say } N=xy \text{ where } x \text{ is in the tens digit } \\ \text{ and } y \text{ is in the one's digit } \\ \text{ So } N=10x+y\] (by the way this xy I have written for N doesn't mean x and y multiplied; it is just the digits of N)
So can you write what K would look like
k = 10y + x
great
so you have N=10x+y and K=10y+x and you are given K+2N<60
just plug in your K and your N and combine like terms
and if you want to find N check a few pairs for (x,y)
10y+x + 2(10x+y) <60 10y+x+20x+2y < 60 12y+ 21x < 60 21x<60-12y x= 60/21 - 12/21y
Is this correct?
\[x<\frac{60}{21}-\frac{12}{21} y \text{ though I probably would leave \it like } \\ 12y+21x<60\]
\[21x+12y<60\]
so N is two digits and N in digit form looks like N=xy we don't want x to be 0 because basically that would mean N is a one digit number
so (1,0) would be the first pair I look at
then (1,1) and I would keep doing this to til I bust (you know go over 60)
21(1)+12(0)=21<60 so N=10 is good 21(1)+12(1)=21+12=33<60 so N=11 is good
see if 12 works
21(1)+12(2)= 45 < 60 21(1)+12(3)= 57<60 21(1)+12(4)= 69>60
ok but I don't think that is all
we can look at N=20,21,22
20 and 21 work
and 3something wouldn;t work because 21(3)=63
so now we are done
I see how to do this now. Thank you sooo much! I was stuck on this yesterday because I divided by 21 and kept getting decimals.
np i'm just glad my way wasn't too long :p
It's a good way. Thank you again!
np
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