Try this beautiful problem from the Pre-RMO, 2019 based on Greatest Integer.
Find the greatest integer not exceeding the sum \(\sum_{n=1}^{1599}\frac{1}{\sqrt{n}}\)
Largest integer
Divisibility
Integer
But try the problem first...
Answer: is 78.
PRMO, 2019, Question 22
Elementary Number Theory by David Burton
First hint
\(\int\limits_1^{1600}\frac{1}{x}{d}x < \sum_{x=1}^{1599}\frac{1}{\sqrt{n}}\)
\(< 1+\sum_{n=1}^{1599}\frac{1}{\sqrt{x}}{d}x\)
Second Hint
or, \([2\sqrt{x}]_{1}^{1600}< \sum_{n=1}^{1599}\frac{1}{\sqrt{n}}\)
\(< 1+|2{\sqrt{x}}|_1^{1599}\)
Final Step
or, 78<\(\sum_{n=1}^{1599}\frac{1}{\sqrt{n}} <2\sqrt{1599}-1\)
or, 78 < \(\sum_{n=1}^{1599}\frac{1}{\sqrt{n}}\)<78.97
or,\([\sum_{n=1}^{1599}\frac{1}{\sqrt{n}}]\)=78.
Try this beautiful problem from the Pre-RMO, 2019 based on Greatest Integer.
Find the greatest integer not exceeding the sum \(\sum_{n=1}^{1599}\frac{1}{\sqrt{n}}\)
Largest integer
Divisibility
Integer
But try the problem first...
Answer: is 78.
PRMO, 2019, Question 22
Elementary Number Theory by David Burton
First hint
\(\int\limits_1^{1600}\frac{1}{x}{d}x < \sum_{x=1}^{1599}\frac{1}{\sqrt{n}}\)
\(< 1+\sum_{n=1}^{1599}\frac{1}{\sqrt{x}}{d}x\)
Second Hint
or, \([2\sqrt{x}]_{1}^{1600}< \sum_{n=1}^{1599}\frac{1}{\sqrt{n}}\)
\(< 1+|2{\sqrt{x}}|_1^{1599}\)
Final Step
or, 78<\(\sum_{n=1}^{1599}\frac{1}{\sqrt{n}} <2\sqrt{1599}-1\)
or, 78 < \(\sum_{n=1}^{1599}\frac{1}{\sqrt{n}}\)<78.97
or,\([\sum_{n=1}^{1599}\frac{1}{\sqrt{n}}]\)=78.
