1682 This Puzzle Is Not as Difficult as It Looks

Today’s Puzzle:

Three months ago I was inspired by a puzzle I saw on Twitter:

I enjoyed solving this complicated-looking system of equations, but let me tell you, Looks Can Be Deceiving! The puzzle is not as difficult as it looks.

I decided to make a similar puzzle, and I’ve waited for my 1682nd post to share it with you. If you can solve the Twitter puzzle, then you can solve my puzzle, too!

Why did I wait until my 1682nd post to share this puzzle? Because if you add the three equations together you get:
(x + y + y + z + x + z)(x + y + z) = 1682,
(2x + 2y + 2z)(x + y + z) = 1682,
2(x + y + z)(x + y + z) = 1682,
2(x + y + z)² = 1682.
The factors of 1682 will be quite helpful at this point. What is the greatest common factor of the numbers after the equal signs?

The numbers in one of 1682’s Pythagorean triples, 580-609-1682, are featured prominently in this puzzle.

I hope you enjoy solving my puzzle, and maybe you will make and solve some puzzles of your own!

Factors of 1682:

  • 1682 is a composite number.
  • Prime factorization: 1682 = 2 × 29 × 29, which can be written 1682 = 2 × 29².
  • 1682 has at least one exponent greater than 1 in its prime factorization so √1682 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1682 = (√841)(√2) = 29√2.
  • The exponents in the prime factorization are 1 and 2. Adding one to each exponent and multiplying we get (1 + 1)(2 + 1) = 2 × 3 = 6. Therefore 1682 has exactly 6 factors.
  • The factors of 1682 are outlined with their factor pair partners in the graphic below.

More About the Number 1682:

1682 is the sum of two squares in two different ways:
29² + 29² = 1682, and
41² + 1² = 1682.

1682 is the hypotenuse of two Pythagorean triples:
82-1680-1682, calculated from 2(41)(1), 41² – 1², 41² + 1², and
1160-1218-1682, which is (20-21-29) times 58.

1680, 1681, 1682, 1683, and 1684 are the second smallest set of FIVE consecutive numbers whose square roots can be simplified.

1680 square roots

1682/2 = 841, which is the second number in the smallest set of FIVE consecutive numbers whose square roots can be simplified.

269 and Five More Consecutive Square Roots

  • 269 is a prime number.
  • Prime factorization: 269 is prime.
  • The exponent of prime number 269 is 1. Adding 1 to that exponent we get (1 + 1) = 2. Therefore 269 has exactly 2 factors.
  • Factors of 269: 1, 269
  • Factor pairs: 269 = 1 x 269
  • 269 has no square factors that allow its square root to be simplified. √269 ≈ 16.401

How do we know that 269 is a prime number? If 269 were not a prime number, then it would be divisible by at least one prime number less than or equal to √269 ≈ 16.401. Since 269 cannot be divided evenly by 2, 3, 5, 7, 11, or 13, we know that 269 is a prime number.

As I have previously written, 844, 845, 846, 847, and 848 are the smallest FIVE consecutive numbers whose square roots can be simplified. Here are the second smallest FIVE with the same property.

1680 square roots

The first number in the second set, 1680, equals 2 x 840 which is very close to the first number in the first set. Will strings of five consecutive numbers with reducible square roots occur about once every 850 numbers?

We can find the number of factors for these numbers by examining their prime factorizations.

1680 prime factorization

The number of factors for each of the integers in this second set ranges from 3 to 40. Only two of the integers have the same number of factors. Finding another string of four or more numbers that have reducible square roots as well as the same number of factors may be difficult.