Michigan Papyrus

The Michigan papyrus 620 is a Greek papyrus written in the second century. It contains three arithmetic problems. One of these problems follows:
Four numbers: their sum is 9900; let the second exceed the first by one-seventh of the first; let the third exceed the sum of the first two by 300; and let the fourth exceed the sum of the first three by 300. Find the four numbers.
Source: mathcontest.olemiss.edu 4/30/2007

SOLUTION
Let $a,b,c,d$ be the four numbers. Let’s represent the first number $a$ as an apple pie divided into 7 equal pieces as follows:

We don’t know the numerical value of each piece. If we do, we will know the value of $a$ . All we know is that
$a=7\;pieces$

Because the second number $b$ exceeds the first number by one-seventh of the first,
$b=8\;pieces$

The third number $c$ exceeds the sum of the first two by 300,
$c=\left (7\;pieces+8\;pieces\right )+300$
$=15\;pieces+300$

The fourth number $d$ exceeds the sum of the first three by 300,
$d=\left (7\;pieces+8\;pieces+15\;pieces+300\right )+300$
$=30\;pieces+600$

The sum of the four numbers is 9900,
$9900=7\;pieces+8\;pieces+\left (15\;pieces+300\right )+\left (30\;pieces+600\right )$
$=60\;pieces+900$

Subtracting 900 from both sides,
$60\;pieces=9000$
$1\;piece=9000\div 60=150$

Now we can compute the value of the numbers:
$a=7\left (150\right )=1050$
$b=8\left (150\right )=1200$
$c=1050+1200+300=2550$
$d=1050+1200+2550+300=5100$

Check
$1050+1200+2550+5100=9900$