Question 876961
Maybe the people at the artofproblemsolving website think this is an easy question.
For me, it is hard enough just to try to interpret the problem.
Here is my interpretation in table form:
{{{matrix(8,6,day,fossils,taken,fraction,fraction,fossils,
number,found,to_David,donated,kept,kept,
1,40,40,3/5,2/5,16,
2,8, 24,5/8,3/8, 9,
3,26,35,2/7,5/7, 25,
4,29,54,5/9,4/9, 24,
5,6, 30,5/6,1/6, 5,
6,10,15,2/3,1/3, 5)}}}
{{{TOTAL}}}{{{found=40+8+26+29+6+10=119}}}
{{{Donated}}}{{{to}}}{{{the}}}{{{museum=119-5=highlight(114)}}}
Ellen and Angela found {{{40}}} fossils the first day.
At the end of the day they checked in with David, the chief paleontologist.
He helped them separate the fossils and they ended having to give 3/5 of the total they had to the museum.
They could keep the remaining {{{1-3/5=2/5}}} of the 40 fossils,
and that was {{{(2/5)*40=16}}}.
 
On the second day, Ellen and Angela found {{{8}}} fossils.
Added to the {{{16}}} fossils thy had kept, they had {{{16+8=24}}} fossils.
when they took those {{{24}}} fossils to David,
he selected {{{5/8}}} of them to give to the museum.
That meant that Ellen and Angela could keep the other {{{1-5/8=3/8}}} of the {{{24}}} fossils.
So they kept {{{(3/8)*24=9}}} fossils.
 
On the third day, Ellen and Angela found {{{26}}} fossils.
Added to the {{{9}}} fossils they had kept the night before, they had a total of {{{9+26=35}}} fossils.
When they took those {{{35}}} fossils to David,
he selected {{{2/7}}} of them to give to the museum.
That meant that Ellen and Angela could keep the other {{{1-2/7=5/7}}} of the {{{35}}} fossils.
So they kept {{{(5/7)*35=25}}} fossils.
 
On the fourth day, Ellen and Angela found {{{29}}} fossils.
Added to the {{{25}}} fossils they had kept the night before, they had a total of {{{25+29=54}}} fossils.
When they took those {{{54}}} fossils to David,
he selected {{{5/9}}} of them to give to the museum.
That meant that Ellen and Angela could keep the other {{{1-5/9=4/9}}} of the {{{54}}} fossils.
So they kept {{{(4/9)*54=24}}} fossils.
 
On the fifth day, Ellen and Angela found {{{6}}} fossils.
Added to the {{{24}}} fossils they had kept the night before, they had a total of {{{246=30}}} fossils.
When they took those {{{30}}} fossils to David,
he selected {{{5/6}}} of them to give to the museum.
That meant that Ellen and Angela could keep the other {{{1-5/6=1/6}}} of the {{{30}}} fossils.
So they kept {{{(1/6)*30=5}}} fossils.
 
On the sixth day, Ellen and Angela found {{{10}}} fossils.
Added to the {{{5}}} fossils they had kept the night before, they had a total of {{{5+10=15}}} fossils.
When they took those {{{15}}} fossils to David,
he selected {{{2/3}}} of them to give to the museum.
That meant that Ellen and Angela could keep the other {{{1-2/3=1/3}}} of the {{{15}}} fossils.
So they kept {{{(1/3)*15=5}}} fossils.
 
The problem gives you some of the numbers for that table.
You start with
{{{matrix(8,6,day,fossils,taken,fraction,fraction,fossils,
number,found,to_David,donated,kept,kept,
1,"?","?",3/5,2/5,"?",
2,8, "?",5/8,3/8, "?",
3,26,"?",2/7,5/7, "?",
4,29,"?",5/9,4/9, "?",
5,6, "?",5/6,1/6, "?",
6,10,"?",2/3,1/3, 5)}}}
 
I see three options to build that table:
 
1) The table can be constructed from the bottom up.
Knowing that they ended up with {{{5}}} fossils,
and that those {{{5}}} fossils were {{{1-2/3=1/3}}} of what Ellen and Angela had took to David on the 6th evening,
you figure that on the evening of the sixth day they had taken {{{3*5=15}}} fossils to David.
Since they had collected {{{10}}} fossils on the sixth day,
that means that they had started the sixth day with {{{15-10=5}}} fossils that David had allowed them to keep on the night of the fifth day.
Those {{{5}}} fossils that David had allowed them to keep on the night of the fifth day were {{{1-5/6=1/6}}} of what they had shown David on the 5th evening.
That means that on the 5th evening, they had brought {{{6*5=30}}} fossils to David.
{{{matrix(8,6,day,fossils,taken,fraction,fraction,fossils,
number,found,to_David,donated,kept,kept,
1,"?","?",3/5,2/5,"?",
2,8, "?",5/8,3/8, "?",
3,26,"?",2/7,5/7, "?",
4,29,"?",5/9,4/9, "?",
5,6,6*5=30,5/6,1/6,15-10=5,
6,10,3*5=15,2/3,1/3, 5)}}}
You can keep back-calculating that way to fill the entire table.
 
2) Since I get easily mixed up, I chose to start filling the table from the top, by smart trial and error.
I used a spreadsheet program on the computer to make it easier for me.
I realized that to get integers on the first row, the first {{{"?"}}} had to be a multiple of 5.
Using {{{" 0 "}}} for the first {{{"?"}}} , I would get integers on the twotop rows,
but I do not get integers for all spots on the other rows.

Using {{{5}}} for the first {{{"?"}}} , I get integers on the top row,
but the second row ends up in {{{3.75=3&3/4}}}.
Using {{{10}}} for the first {{{"?"}}} , the second row ends up in {{{4.5=4&1/2}}} .
Using {{{15}}} for the first {{{"?"}}} , the second row ends up in {{{5.25=5&1/4}}} ,
and using {{{20}}} for the first {{{"?"}}} , the second row ends up in {{{6}}} .
Unfortunately, I do not get integers for all spots on the other rows.
However, I see a pattern:
{{{"0"}}} , {{{20}}}, {{{40}}} , {{{60}}} , etc
will results in integers all across the first and second rows.
I tried {{{40}}} for the first {{{"?"}}} , and I did get integers for all spots on all rows.
Better yet, the last number on the table was {{{5}}} so I knew I had a result.
 
3) The table can be constructed from the top down, using a little algebra and a lot of arithmetic:
{{{matrix(8,6,day,fossils,taken,fraction,fraction,fossils,
number,found,to_David,donated,kept,kept,
1,x,x,3/5,2/5,2x/5,
2,8, 2x/5+8,5/8,3/8,(3/8)(2x/5+8)="?",
3,26,26+"?",2/7,5/7, "!?",
4,29,29+"!?",5/9,4/9, "!!??",
5,6, "??",5/6,1/6, "???",
6,10,"????",2/3,1/3, 5)}}}
I could try to calculate
{{{"?"=(3/8)(2x/5+8)=6x/40+3=3x/20+3}}}
Then {{{26+"?"=26+3x/20+3=3x/20+29}}} , and {{{"!?"}}} is {{{5/7}}} of that, so
{{{"!?"=(5/7)(3x/20+29)=3x/28+145/7}}}
At best I could say that
{{{"!?"=3x/28+145/7=3x/28+20+5/7=20+3x/28+20/28=20+(3x+20)/28}}} needs to be an integer, and that the smallest natural number {{{x}}} that would work is {{{40}}} .
I know I would make mistakes going on with calculations like that.