A couple of days ago on Twitter, I saw an interesting puzzle posed by Sunil Singh @Mathgarden.
This is probably the greatest algebra problem that can be given to ALL students. The bridges represent SUMS of adjacent fields. There are 8 UNKNOWNS! Without using Algebra–which is accessible to MS students–this problem is hard. Algebra makes it easier…but how?;) pic.twitter.com/C7L7TJgBrt
— Sunil Singh (@Mathgarden) May 11, 2021
After I found one of several of its solutions, I wondered if I could add a bridge that would use all nine numbers from 1 to 9 in the solution, so I tweaked it. I decided to move the puzzle to the ocean when I added that extra bridge.
I was able to solve this problem using logic and addition facts, rather than algebra. Try solving it yourself. If you want to see any of the steps I used to solve the puzzle, scroll down to the end of the post.
Factors of 1649:
- 1649 is a composite number.
- Prime factorization: 1649 = 17 × 97.
- 1649 has no exponents greater than 1 in its prime factorization, so √1649 cannot be simplified.
- The exponents in the prime factorization are 1 and 1. Adding one to each exponent and multiplying we get (1 + 1)(1 + 1) = 2 × 2 = 4. Therefore 1649 has exactly 4 factors.
- The factors of 1649 are outlined with their factor pair partners in the graphic below.
More About the Number 1649:
1649 is the sum of two squares in TWO different ways:
32² + 25² = 1649, and
40² + 7² = 1649.
1649 is the hypotenuse of FOUR Pythagorean triples:
399-1600-1649, calculated from 32² – 25², 2(32)(25), 32² + 25²,
560-1551-1649, calculated from 2(40)(7), 40² – 7², 40² + 7²,
776 1455 1649, which is (8-15-17) times 97, and
1105 1224 1649, which is 17 times (65-72-97).
Some Logical Steps to Solve Today’s Puzzle:
I found four different ways to write the solution but they are all rotations or reflections of each other. Here are the steps to one of those four ways:
1st Step: The biggest number that can be used is 9. Since every island must be included in at least two sums, neither 8 nor 9 can be an addend; they both must be sums. 7 must be an addend in their sums because adding any number to 7 will yield 8, 9, or some larger forbidden number. Thus 8 and 9 are bridges that connect to island 7.
I chose to write 7, 8, and 9 in the top right section, but 8 and 9 could change places with each other. I could have also chosen to write those numbers in the bottom left area.
2nd Step: 1 + 7 = 8, and 7 + 2 = 9.
3rd Step: 1 + 2 = 3.
4th Step: The last island must be the smallest remaining number (4) because the smallest remaining number can’t be the sum of a bigger number and the number on either adjacent island.
Final Step: 1 + 4 = 5, and 4 + 2 = 6.
Did you enjoy this puzzle? How did my steps compare to the steps that you took?
Please, check the comments for another solution.
2 thoughts on “1649 Tweaking a Puzzle Posed by Sunil Singh @Mathgarden”
Fun puzzle! I started with a different assumption, and found a different solution. I made 7, 8, and 9 all bridges coming off the 6 island, and everything else fell into place. The other islands of course had to be 1, 2, 3 — with 1 and 2 NOT sharing a bridge.
Oh yes! I followed what you wrote and was able to solve it using your assumption, too! (Not on my first try, mind you, because 1 and 2 shared a bridge on my first attempt, but as soon as I put 9 on the center bridge, and 7 and 8 on the other bridges coming off the 6 island, everything worked beautifully.)