29th Eötvös Competition Problems 1925

1.  Given four integers, show that the product of the six differences is divisible by 12.

2.  How many zeros does the the decimal representation of 1000! end with?

3.  Show that the inradius of a right-angled triangle is less than 1/4 of the length of the hypoteneuse and less than 1/2 the length of the shortest side. 

Solutions

Problem 1

Given four integers, show that the product of the six differences is divisible by 12.

Solution

Two of the integers must have the same residue mod 3 and hence their difference must be a multiple of 3. Similarly, two of the integers, say A and B, must have the same parity and hence their difference is even. Now either one of the other two numbers, say C, has the same parity as A in which case A-C is even. Or both the other two numbers have opposite parity to A, in which case they have the same parity as each other and hence their difference is even. 

Problem 2

How many zeros does the the decimal representation of 1000! end with?

Solution

There are 200 multiples of 5 in X = {1, 2, ... , 1000}. 40 of them are multiples of 5², so each bring an additional power of 5. 8 are multiples of 5³, so each bring another, and 1 is a multiple of 5⁴, so bringing another. Thus the highest power of 5 dividing 1000! is 5²⁴⁹. Obviously 1000! is divisible by 2²⁴⁹, so the highest power of 10 divising 1000! is 10²⁴⁹. So it ends in 249 zeros. 

Problem 3

Show that the inradius of a right-angled triangle is less than 1/4 of the length of the hypoteneuse and less than 1/2 the length of the shortest side.

Solution

The circle lies between BC and the parallel line through A, so its diameter is less than AC. It also lies between AB and the parallel line through C. But the distance between these two lines is less than CM (where M is the midpoint of AB) which is half AB.