Problem:
How many of the first 1000 positive integers can be expressed in the form
⌊2x⌋+⌊4x⌋+⌊6x⌋+⌊8x⌋,
where x is a real number, and ⌊z⌋ denotes the greatest integer less than or equal to z?
Solution:
Introduce the notation
f(x)=⌊2x⌋+⌊4x⌋+⌊6x⌋+⌊8x⌋(1)
and observe that if n is a positive integer, then from (1)
f(x+n)=f(x)+20n(2)
follows. In particular, this means that if an integer k can be expressed in the form f(x0​) for some real number x0​, then for n=1,2,3,… one can express k+20n similarly; i.e., k+20n=f(x0​)+20n=f(x0​+n). In view of this, one may restrict attention to determining which of the first 20 positive integers are generated by f(x) as x ranges through the half-open interval (0,1].
Next observe that as x increases, the value of f(x) changes only when either 2x,4x,6x or 8x attains an integral value, and that the change in f(x) is always to a new, higher value. In the interval (0,1] such changes occur precisely when x is of the form nm​, where 1≤m≤n and n=2,4,6 or 8. There are 12 such fractions; in increasing order they are:
81​,61​,41​,31​,83​,21​,85​,32​,43​,65​,87​ and 1.
Therefore, only 12 of the first 20 positive integers can be represented in the desired form. Since 1000=(50)(20), in view of (2), this imp1ies that in each of the 50 sequences,
1,2,3,…,20;21,22,23,…,40;…;981,982,983,…,1000, of 20 consecutive integers only 12 can be so expressed, leading to a total of (50)(12) or 600​ positive integers of the desired form.
The problems on this page are the property of the MAA's American Mathematics Competitions