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

Answer: 1050, 1200, 2550, 5100.

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About mvtrinh

Retired high school math teacher.
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