Math Puzzles

# How many pennies?

Hi! It’s great to know that this puzzle got your attention. So let’s begin by revisiting the problem statement: Solution:

First, let’s assume that the poorest one has x number of pennies. Now, assume that the next child has the number of pennies equal to xy (where y is an integer, as per the given condition). Next, the ratio of the number of pennies that the third child possesses to the number of pennies possessed by the first two must be integer. A simple way of achieving this is to assume that the third child has xyz number of pennies (again, z is an integer). This way, the ratios will be:

xyz:xy = z:1 (integer ratio because z is assumed to be an integer)

xyz:x = yz:1 (integer, because multiplication of two integers will yield another integer).

Applying the same logic throughout, we can assume the fortunes of other children to be:

4th Child: xyzm (m is an integer)

5th Child: xyzmn (n is an integer)

6th Child: xyzmno (o is an integer)

7th Child: xyzmnop (p is an integer)

The sum of all fortunes is 2,879.

Mathematically,

x + xy + xyz + xyzm + xyzmn + xyzmno + xyzmnop = 2879

Taking out x as common:

x (1 + y + yz + yzm + yzmn + yzmno + yzmnop) = 2879

Note that 2879 is a prime number – which means that it can be divided only by itself and 1. In other words, 2879 has only two factors – 1 and 2879.

Therefore, x = 1 and

1 + y + yz + yzm + yzmn + yzmno + yzmnop = 2879

Thus,

y + yz + yzm + yzmn + yzmno + yzmnop = 2879-1

y + yz + yzm + yzmn + yzmno + yzmnop = 2878

Taking y out as common:

(1 + z + zm + zmn + zmno + zmnop) = 2878 = (2)(1439)

Once again, 1439 is a prime number. Therefore, y = 2

Thus,

1 + z + zm + zmn + zmno + zmnop = 1439

z + zm + zmn + zmno + zmnop = 1438

z (1 + m + mn + mno + mnop) = 1438 = (2)(719)

719 is a prime number. Therefore, z = 2

Solving like this, we get:

m = 2

n = 2

o = 2

Finally,

o (1 + p) = 178

(2)(1 + p) = 178

1 + p = 89

p = 88

Thus, the different fortunes will be:

1st Child: 1 penny

2nd Child: 1×2 = 2 pennies

3rd Child: 1x2x2 = 4 pennies

4th Child: 1x2x2x2 = 8 pennies

5th Child: 1x2x2x2x2 = 16 pennies

6th Child: 1x2x2x2x2x2 = 32 pennies

7th Child: 1x2x2x2x2x2x88 = 2816 pennies

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