Here are a few facts about the puzzle number, 1339:
1339 is a composite number.
Prime factorization: 1339 = 13 × 103
1339 has no exponents greater than 1 in its prime factorization, so √1339 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 1339 has exactly 4 factors.
The factors of 1339 are outlined with their factor pairs in the graphic below.
1339 is the hypotenuse of a Pythagorean triple:
515-1236-1339 which is (5-12-13) times 103
Now I’ll write a little bit about the number 1329:
1329 is a composite number.
Prime factorization: 1329 = 3 × 443
The exponents in the prime factorization are 1 and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1) = 2 × 2 = 4. Therefore 1329 has exactly 4 factors.
Factors of 1329: 1, 3, 443, 1329
Factor pairs: 1329 = 1 × 1329 or 3 × 443
1329 has no square factors that allow its square root to be simplified. √1329 ≈ 36.45545
1329 is divisible by 3 because it is made with three consecutive numbers (1, 2, 3) and 9, a number divisible by 3.
1329 looks interesting in some other bases: It’s 929 in BASE 12, 369 in BASE 20 234 in BASE 25
Regardless of its size, an evergreen tree is a mighty symbol at Christmastime. Today’s factoring puzzle features a couple of relatively small Christmas trees, but don’t think for even one minute that these little trees make for an easy puzzle. It’s a level 6 puzzle so there are several places that the clues might trick you. Use logic throughout the entire process, and you should be able to solve it!
That factoring puzzle has nothing to do with the factors of 1320. In case you are looking for factor trees for the number 1320, here are a few of the MANY possible ones:
Factors of 1320:
1320 is a composite number.
Prime factorization: 1320 = 2 × 2 × 2 × 3 × 5 × 11, which can be written 1320 = 2³ × 3 × 5 × 11.
The exponents in the prime factorization are 3, 1, 1, and 1. Adding one to each and multiplying we get (3 + 1)(1 + 1)(1 + 1) )(1 + 1) = 4 × 2 × 2 × 2 = 32. Therefore 1320 has exactly 32 factors.
Taking the factor pair with the largest square number factor, we get √1320 = (√4)(√330) = 2√330 ≈ 36.331804
Sum-Difference Puzzles:
330 has eight factor pairs. One of those pairs adds up to 61, and another one subtracts to 61. Put the factors in the appropriate boxes in the first puzzle.
1320 has sixteen factor pairs. One of the factor pairs adds up to 122, and a different one subtracts to 122. If you can identify those factor pairs, then you can solve the second puzzle!
The second puzzle is really just the first puzzle in disguise. Why would I say that?
1320 is the third number that is at the top of more than one Sum-Difference Puzzle. This next one is a primitive.
Again, 1320 has sixteen factor pairs. One of the factor pairs adds up to 73, and a different one subtracts to 73. If you can identify those factor pairs, then you can solve this puzzle!
If you would like a little help finding those factor pairs that make sum-difference, the chart below lists all of 1320’s factor pairs with their sums and their differences.
More about the Number 1320:
1320 is the sum of consecutive primes FOUR different ways: It is the sum of the eighteen primes from 37 to 109, 107 + 109 + 113 + 127 + 131 + 137 + 139 + 149 + 151 + 157 = 1320, 149 + 151 + 157 + 163 + 167 + 173 + 179 + 181 = 1320, and 659 + 661 = 1320.
1320 is the hypotenuse of a Pythagorean triple: 792-1056-1320 which is (3-4-5) times 264.
Can you find the factors that will turn this puzzle into a multiplication table? It’s a level 6 so you might find it to be a challenge. Use logic. Don’t guess and check.
Here are a few facts about the puzzle number, 1308:
1308 is a composite number.
Prime factorization: 1308 = 2 × 2 × 3 × 109, which can be written 1308 = 2² × 3 × 109
The exponents in the prime factorization are 2, 1, and 1. Adding one to each and multiplying we get (2 + 1)(1 + 1)(1 + 1) = 3 × 2 × 2 = 12. Therefore 1308 has exactly 12 factors.
Taking the factor pair with the largest square number factor, we get √1308 = (√4)(√327) = 2√327 ≈ 36.16628
1308 is the hypotenuse of a Pythagorean triple:
720-1092-1308 which is 12 times (60-91-109)
As shown in their factor trees below, 1308, 1309, 1310, and 1311 each have three distinct prime numbers in their prime factorizations. They are the smallest set of four consecutive numbers with the same number of prime factors. Thank you OEIS.org for alerting me to that fact.
The exponents in the prime factorization are 1 and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1) = 2 × 2 = 4. Therefore 1294 has exactly 4 factors.
Factors of 1294: 1, 2, 647, 1294
Factor pairs: 1294 = 1 × 1294 or 2 × 647
1294 has no square factors that allow its square root to be simplified. √1294 ≈ 35.97221
1294 is also the sum of the twenty prime numbers from 23 to 107.
The exponents in the prime factorization are 1 and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1) = 2 × 2 = 4. Therefore 1286 has exactly 4 factors.
Factors of 1286: 1, 2, 643, 1286
Factor pairs: 1286 = 1 × 1286 or 2 × 643
1286 has no square factors that allow its square root to be simplified. √1286 ≈ 35.86084
1286 is also the sum of six consecutive prime numbers:
197 + 199 + 211 + 223 + 227 + 229 = 1286
To me, today’s level 6 puzzle looks a little like a puppy dog. If you know or use a multiplication table, then with proper training, finding the factors of this puzzle will be no problem.
I’d like to tell you a little about the number 1280:
1280 is a composite number.
Prime factorization: 1280 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5, which can be written 1280 = 2⁸ × 5
The exponents in the prime factorization are 8 and 1. Adding one to each and multiplying we get (8 + 1)(1 + 1) = 9 × 2 = 18. Therefore 1280 has exactly 18 factors.
Taking the factor pair with the largest square number factor, we get √1280 = (√256)(√5) = 16√5 ≈ 35.77709.
1280 is the sum of the fourteen prime numbers from 61 to 127. Do you know what those prime numbers are?
32² + 16² = 1280
1280 is the hypotenuse of a Pythagorean triple:
768-1024-1280 which is (3-4-5) times 256 That triple can also be calculated from 32² – 16², 2(32)(16), 32² + 16²
Since 1280 is the 5th multiple of 256, I would expect that a number close to 1280 would be the 500th number whose square root could be simplified. That number was 1275, just five numbers ago.
Both 6 and 12 are allowable common factors of 60 and 12. Likewise, both 8 and 12 are allowable common factors of 96 and 72. In each case, only one of those common factors will work with this puzzle. Don’t guess and check each one. Study the other clues and at least one wrong common factor will be eliminated. Have fun solving it!
Now I’ll write a few things about the number 1257:
1257 is a composite number.
Prime factorization: 1257 = 3 × 419
The exponents in the prime factorization are 1 and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1) = 2 × 2 = 4. Therefore 1257 has exactly 4 factors.
Factors of 1257: 1, 3, 419, 1257
Factor pairs: 1257 = 1 × 1257 or 3 × 419
1257 has no square factors that allow its square root to be simplified. √1257 ≈ 35.4542
1257 is the difference of two squares two different ways:
211² – 208² = 1257
629² – 628² = 1257
The clues in one of the columns for this puzzle as well as one of the rows are 9 and 3. You will need to figure out where to put the factors 1, 3, 3, and 9 to make those clues work. You might think it doesn’t matter where you write those factors, but believe me, it does matter. My advice: Don’t start with those clues. Find another logical place to start. Good luck with this one!
Prime factorization: 1250 = 2 × 5 × 5 × 5 × 5, which can be written 1250 = 2 × 5⁴
The exponents in the prime factorization are 1 and 5. Adding one to each and multiplying we get (1 + 1)(4 + 1) = 2 × 5 = 10. Therefore 1250 has exactly 10 factors.
Taking the factor pair with the largest square number factor, we get √1250 = (√625)(√2) = 25√2 ≈ 35.35534
1250 is the sum of consecutive squares two different ways:
193 + 197 + 199 + 211 + 223 + 227 = 1250
619 + 631 = 1250
1250 is the sum of two squares THREE different ways:
31² + 17² = 1250
25² + 25² = 1250
35² + 5² = 1250
1250 is the hypotenuse of FOUR Pythagorean triples:
750-1000-1250 which is (3-4-5) times 250,
672-1054-1250 which is 2 times (336-527-625) and
can also be calculated from 31² – 17², 2(31)(17), 31² + 17²,
440-1170-1250 which is 10 times (44-117-125), and
350-1200-1250 which is (7-24-25) times 50 and
can also be calculated from 2(35)(5), 35² – 5², 35² + 5²
If you use logic, you can figure out the solution to this puzzle. You will have to study all the clues just to know where to start, but I think you’ll find a lot of satisfaction in finding the solution.
The exponents in the prime factorization are 1 and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1) = 2 × 2 = 4. Therefore 1238 has exactly 4 factors.
Factors of 1238: 1, 2, 619, 1238
Factor pairs: 1238 = 1 × 1238 or 2 × 619
1238 has no square factors that allow its square root to be simplified. √1238 ≈ 35.18522
Because of its prime factors, I know that 1238 is part of only one Pythagorean triple:
1238-383160-383162
1238 is a palindrome in two other bases:
It’s 646 in BASE 14, and
it’s 383 in BASE 19.
