Skip to main content
  1. Problems/

2026 - Problem Set 01

Solutions Page
#

Problem 1
#

Evaluate the following:

$$ 2-3+10-15+26-35+50-63+82-99 \cdots +2026 $$

Hints
#

Hint 1The pattern is: $$ t_n = (-1)^{n-1}n^2+1 $$
Hint 2Can you do something about the \(1\) term? What does the sum look like visually without the 1 term?

Problem 2
#

If the function \(f(x)\) is quadratic, \(f(0)=2025\), \(f(1)=2026\), and \(f(2)=1984\), then what is \(f(10)\)

Hints
#

Hint 1What is the equation for f(0)?
Hint 2What happens if you substitute \(x\) and \(f(x)\) with their values at each point?

Problem 3
#

There are \(2026\) numbered lights with corresponding light switches in a hallway and \(2026\) people in a line. Each person goes down the hallway, switching every light switch that is a multiple of their place in line.

For example, the fifth person in line would press light switches \(\{5, 10, 15, 20 \cdots\} \).

After every person goes through the hallway, how many lights are on?

Hints
#

Hint 1If light \(n\) is on at after everyone goes, what is known about how many factors it has?
Hint 2For light \(n\), which people press its light switch?

Problem 4
#

For some positive integer $n$, the number \(110n^3\) has \(110\) positive integer divisors, including \(1\) and the number \(110n^3\). How many positive integer divisors does the number \(81n^4\) have?

Hints
#

Hint 1The amount of factors a number \(n\) has is the product of all the exponents of the number plus \(1\).

E.g., \(110\) has the prime factors \(2\cdot 5\cdot 11\) therefore \(110\) has \((1+1)\cdot (1+1)\cdot (1+1) = 8\) prime factors.

Hint 2\(2 \cdot 5 \cdot 11 \cdot n^3\)'s factors must be divisible by \(2, 5,\) and \(11\). If you make \(n\) something such that when substituded one exponent is \(10\) and another is \(4\), then the amount of factors will be a multiple of \(110\).

Problem 5
#

Let \(\tau(n)\) denote the number of positive integer divisors of \(n\). Find the sum of the six least positive integers \(n\) that are solutions to \(\tau (n) + \tau (n+1) = 7\).

Hints
#

Hint 1In order to obtain a sum of \(7\), we must have:
  • either a number with \(5\) divisors (a fourth power of a prime) and a number with \(2\) divisors (a prime), or
  • a number with \(4\) divisors (a semiprime or a cube of a prime) and a number with \(3\) divisors (a square of a prime). (No integer greater than \(1\) can have fewer than \(2\) divisors.)
Hint 2You can compute all the possibilities by hand.