1747 Getting a Super Bowl Ring Isn’t Easy

Today’s Puzzle:

On Wednesday I read an article in my local paper titled From undrafted free agent to the biggest stage: Britain Covey’s ‘unique ride’ to Super Bowl LVII. It got me quite excited for today’s game. Unfortunately, between Wednesday and Sunday, Covey suffered a hamstring injury.  Getting a Super Bowl Ring certainly isn’t easy. Being able to contribute to your team’s winning the game isn’t easy either. UPDATE: Good News! He was able to make at least one play in the second quarter! Further update: Even though his team lost the game in the final minutes, this rookie played well when he was on the field. I think both teams played exceptionally well and gave us all an exciting game to watch.

Before the game, it is a mystery which team will win the game. The difficulty level of this puzzle is a mystery also.

If you look at today’s puzzle just right, I think it looks a little like a super bowl ring, but I forwarn you, it will not be easy to get this puzzle either. You will need to place the numbers from 1 to 12 both in the first column and in the top row so that the given clues are the products of the numbers you place. Logic and practice will get you there. Good Luck!

You also might enjoy this next Super Bowl puzzle I saw on Twitter:

Factors of 1747:

  • 1747 is a prime number.
  • Prime factorization: 1747 is prime.
  • 1747 has no exponents greater than 1 in its prime factorization, so √1747 cannot be simplified.
  • The exponent in the prime factorization is 1. Adding one to that exponent we get (1 + 1) = 2. Therefore 1747 has exactly 2 factors.
  • The factors of 1747 are outlined with their factor pair partners in the graphic below.

How do we know that 1747 is a prime number? If 1747 were not a prime number, then it would be divisible by at least one prime number less than or equal to √1747. Since 1747 cannot be divided evenly by 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, or 41, we know that 1747 is a prime number.

More About the Number 1747:

1747 is a palindrome in base 15.
It’s 7B7₁₅ because 7(15²) + 11(15) + 7(1) = 1747.

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