Help needed with integer/algebra question

[bash2444]

I've been working with this question for about an hour now and I can't wrap my head around it.

'The integers from 1 to 9 are listed on a whiteboard. If an additional m 8s and n 9s are added to the list, then the mean of all the numbers is 7.3. Find m+n.'

I figured out that the mean before the m 8s and n 9s was added was 5, and now when the m 8s and n 9s are added I don't know what number to divide by. Will it be 11?

Please and thank you.

 

[Dr.Peterson]

I've been working with this question for about an hour now and I can't wrap my head around it.

'The integers from 1 to 9 are listed on a whiteboard. If an additional m 8s and n 9s are added to the list, then the mean of all the numbers is 7.3. Find m+n.'

I figured out that the mean before the m 8s and n 9s was added was 5, and now when the m 8s and n 9s are added I don't know what number to divide by. Will it be 11?

Please and thank you.

If you add in m 8's and n 9's, besides the 9 numbers you started with, then the total number of items is 9+m+n. Right?

You'll end up with a Diophantine equation, as I think you recognize. This is not "beginning algebra"!

 

[Steven G]

The sum of the numbers will be 45 + 8m + 9n = 45 + 8m + (8+1)n = 45 + 8(m+n) + n

The average 7.3 = (45 + 8(m+n) + n)/(9 + m + n)

This is just my opinion: This type of problem can certainly be considered beginning algebra. Just consider that Jacobson calls his 'advanced algebra textbook' basic algebra

 

[Steven G]

65.7 + 7.3(m+n) = 45 + n +(m+n)
n - 20.7 = 6.3(m+n)
m+n = (n - 20.7)/6.3
So you want the right hand side to be an integer. When does this happen?

 

[lex]

The numbers written on the board are:
1....9 [imath]\hspace4ex[/imath] 9 numbers totalling 45
[imath]\underbrace{8.......8}_\text{m}[/imath] [imath]\hspace2ex[/imath]m numbers totalling 8m
[imath]\underbrace{9.......9}_\text{n}[/imath] [imath]\hspace2ex[/imath]n numbers totalling 9n

Overall total = 45 + 8m + 9n
and there are 9 + m + n numbers

[imath]\therefore[/imath] Average = [imath]\frac{45+8m+9n}{9+m+n}[/imath]

So [imath]\frac{45+8m+9n}{9+m+n} = \frac{73}{10}[/imath]

i.e. for some positive integer [imath]k[/imath]

(1) [imath]45 + 8m + 9n =73k[/imath]
(2) [imath]9+m+n=10k[/imath]

Taking [imath]k[/imath] to be an unknown constant, m and n to be variables, solve the simultaneous equations, to find m and n in terms of [imath]k[/imath]:

[imath]9 \times (2) - (1) \rightarrow[/imath]
(A) [imath]\boxed{m=17k-36}[/imath]

[imath]8 \times (2) - (1) \rightarrow[/imath]
(B) [imath]\boxed{n=27-7k}[/imath]

Now [imath]k[/imath] is a positive integer, m and n are non-negative integers, therefore:

(A) [imath]\rightarrow[/imath] [imath]k[/imath] is at least 3
(B) [imath]\rightarrow[/imath] [imath]k[/imath] is at most 3
So [imath]k=3[/imath]

(A) m=15, (B) n=6
(Using these values we do get an average of 7.3)