Week 4

Example 1. How many pentagons and hexagons are there in any standard football?

Example 2. I am a number. I am equal to three times the sum of my digits. What number am I?

Of course, you can find the solution just below, but it is highly recommended that, you first try to solve it on your own.

Just remember the words of Paul Halmos, who says “the only way to learn mathematics is to do mathematics.

Solution 1. Let P be the number of pentagons and H be the number of hexagons present in any standard football. We can use Euler’s formula V E + F = 2 to find out the number of pentagons and hexagons.

Each pentagon has 5 vertices and 6 for each hexagon. Total number of vertices becomes (5P + 6H). But we have counted each vertex thrice, once for each adjacent polygon.

Similarly, each pentagon has 5 edges and 6 for each hexagon. Total number of edges becomes (5P + 6H). But we have counted each edge twice, once for each adjacent polygon.

There are P pentagons and H hexagons, each forming a face. Hence, total number of faces F = (P + H). Therefore, V E + F = 2 gives

Notice that each pentagon is surrounded by 5 hexagons. So, there should be 5P hexagons. But we have counted each hexagon thrice for each of its 3 adjacent pentagons.

Therefore, there are 12 pentagons and 20 hexagons in a standard football.

Solution 2. First note that the number can be of single digit, two digits, three digits, four digits, . . . Definitely, there is no single digit number apart from 0, with the given property. Whether there will be any number(s) of two digits, three digits etc., we need to consider the following cases:

Case 1. Let the number be of two digits. We can take the number as ab = 10a + b, where a, b are the digits of the given number. According to the symmetry of the given question

10a + b = 3(a + b)      or, 7a = 2b      or, (a, b) = (2, 7).

Hence the only two-digit number satisfying the given condition is 27.

Case 2. Let the number be of three digits. We can take the number as abc = 100a + 10b + c, where a, b, c are the digits of the given number. According to the symmetry of the given question

100a + 10b + c = 3(a + b + c)        or, 97a + 7b = 2c.             

Since a, b, c are the digits (maximum value of the RHS is 18 and the minimum value of LHS is more than 100), above relation is not possible. No such three-digit number exist.

Case 3. Let the number be of four digits. We can take the number as abcd = 1000a + 100b + 10c + d, where a, b, c, d are the digits of the given number. According to the symmetry of the given question

1000a + 100b + 10c + d = 3(a + b + c + d)        or, 997a + 97b + 7c = 2d.             

Since a, b, c, d are the digits, above relation is not possible. No such four-digit number exist.

Following the similar lines of argument, we can say that the possible non-trivial number is of two-digits only.

Only possible numbers satisfying the given condition are 0 and 27.

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