Author: ivasallay

853 You Can Do This Puzzle!

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This is a level 1 puzzle that is easier than even most other level 1 puzzles. You can do this puzzle! If mathematics makes you uncomfortable, you can still do this puzzle! Even if math class is your worse nightmare, you can complete this puzzle, and gain a little confidence. Go ahead, give it a try! Figure out where each number from one to ten goes in the top row and also in the first column so that the puzzle turns into a mixed-up multiplication table. It’s easier and far less time consuming than Sudoku. You CAN do this puzzle! Then, after you find all the factors, and are feeling really good about yourself, IF you want, you can fill in all the other cells of this mixed up multiplication table.

Print the puzzles or type the solution on this excel file: 10-factors-853-863

853 is a prime number that leaves a remainder of 1 when divided by 4, so 853 is the hypotenuse of a Pythagorean triple: 205-828-853.

23² + 18² = 853 so 205-828-853 can be calculated from 23² – 18², 2(23)(18), 23² + 18².

  • 853 is a prime number.
  • Prime factorization: 853 is prime.
  • The exponent of prime number 853 is 1. Adding 1 to that exponent we get (1 + 1) = 2. Therefore 853 has exactly 2 factors.
  • Factors of 853: 1, 853
  • Factor pairs: 853 = 1 × 853
  • 853 has no square factors that allow its square root to be simplified. √853 ≈ 29.20616

How do we know that 853 is a prime number? If 853 were not a prime number, then it would be divisible by at least one prime number less than or equal to √853 ≈ 29.2. Since 853 cannot be divided evenly by 2, 3, 5, 7, 11, 13, 17, 19, 23, or 29, we know that 853 is a prime number.

Here’s another way we know that 853 is a prime number: Since its last two digits divided by 4 leave a remainder of 1, and 23² + 18² = 853 with 23 and 18 having no common prime factors, 853 will be prime unless it is divisible by a prime number Pythagorean triple hypotenuse less than or equal to √853 ≈ 29.2. Since 853 is not divisible by 5, 13, 17, or 29, we know that 853 is a prime number.

 

852 and Level 6

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Print the puzzles or type the solution on this excel file: 12 factors 843-852

I knew that 852 was divisible by 3 as soon as I typed it in a straight line on the number pad. Any 3 digit number that lies on a straight line on a number pad or a phone dial pad is divisible by 3. And in case you’ve ever wondered why the numbers on a number pad or calculator and the numbers on a phone dial pad are reversed, ABC News has the answer.

852 is 705 in BASE 11, and it is 507 in BASE 13.

852 is palindrome 1E1 in BASE 23 (E is 14 base 10) because 1(23²) +14(23¹) + 1(23º) = 852.

852 is the sum of consecutive prime numbers 421 and 431.

852 is also the 24th pentagonal number because (3⋅24² – 24)/2 = 852

  • 852 is a composite number.
  • Prime factorization: 852 = 2 × 2 × 3 × 71, which can be written 852 = 2² × 3 × 71
  • 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 852 has exactly 12 factors.
  • Factors of 852: 1, 2, 3, 4, 6, 12, 71, 142, 213, 284, 426, 852
  • Factor pairs: 852 = 1 × 852, 2 × 426, 3 × 284, 4 × 213, 6 × 142, or 12 × 71,
  • Taking the factor pair with the largest square number factor, we get √852 = (√4)(√213) = 2√213 ≈ 29.189039

851 Give This Apple to Your Teacher This Year

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This puzzle looks a little like an apple. It’s a level 5 puzzle so it won’t be that easy. If you can solve the puzzle, give it to your teacher!


Print the puzzles or type the solution on this excel file: 12 factors 843-852

851 is the hypotenuse of a Pythagorean triple:

  • 276-805-851 which is 23 times (12-35-37)

851 is a palindrome in three other bases:

  • 353 BASE 16, because 3(16²) + 5(16¹) + 3(16º) = 851
  • 191 BASE 25, because 1(25²) + 9(25¹) + 1(25º) = 851
  • NN BASE 36 (N is 23 base 10) because 23(36¹) + 23(36º) = 23(37) = 851

Here is 851 factoring information:

  • 851 is a composite number.
  • Prime factorization: 851 = 23 × 37
  • 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 851 has exactly 4 factors.
  • Factors of 851: 1, 23, 37, 851
  • Factor pairs: 851 = 1 × 851 or 23 × 37
  • 851 has no square factors that allow its square root to be simplified. √851 ≈ 171904

851 is in this cool pattern:

 

How Often is 850 the Hypotenuse of a Pythagorean Triple?

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We can tell if a number is the hypotenuse of a Pythagorean triple by looking at its prime factorization.

  • If NONE of its prime factors leave a remainder of 1 when divided by 4, then it will NOT be the hypotenuse of Pythagorean triple.
  • If at least one of its prime factors leave a remainder of 1 when divided by 4, then it WILL be the hypotenuse of Pythagorean triple.
  • If ALL of its prime factors leave a remainder of 1 when divided by 4, then it will also be the hypotenuse of at least one PRIMITIVE Pythagorean triple.

850 = 2 × 5² × 17¹. Its factor, 2, prevents 850 from being the hypotenuse of a primitive Pythagorean triple, but 5² × 17¹ will actually make it the hypotenuse of SEVEN Pythagorean triples. Some of those we can find by looking at the ways we can make 850 from the sum of two squares:

29² + 3² = 850, 27² + 11² = 850, and 25² + 15² = 850

  • 29² + 3² gives us 174-832-850, calculated from 2(29)(3), 29² – 3², 29² + 3², and is 2 times (87-416-425)
  • 27² + 11² gives us 594-608-850, calculated from 27² – 11², 2(27)(11), 27² + 11², and is 2 times (297-304-425)
  • 25² + 15² gives us 400-750-850 calculated from 2(25)(15), 25² – 15², 25² + 15², and is (8-15-17) times 50.

Let’s look a little closer at 850’s factoring information:

  • 850 is a composite number.
  • Prime factorization: 850 = 2 × 5 × 5 × 17, which can be written 850 = 2 × 5² × 17
  • The exponents in the prime factorization are 1, 2, and 1. Adding one to each and multiplying we get (1 + 1)(2 + 1)(1 + 1) = 2 × 3 × 2 = 12. Therefore 850 has exactly 12 factors.
  • Factors of 850: 1, 2, 5, 10, 17, 25, 34, 50, 85, 170, 425, 850
  • Factor pairs: 850 = 1 × 850, 2 × 425, 5 × 170, 10 × 85, 17 × 50, or 25 × 34
  • Taking the factor pair with the largest square number factor, we get √850 = (√25)(√34) = 5√34 ≈ 29.154759.

Here are the SEVEN ways 850 is the hypotenuse of a Pythagorean triple with five of 850’s factor pairs in bold print:

  • 174-832-850 which is 2 times (87-416-425)
  • 594-608-850 which is 2 times (297-304-425)
  • 510-680-850 which is (3-4-5) times 170.
  • 130-840-850 which is 10 times (13-84-85).
  • 360-770-850 which is 10 times (36-77-85).
  • 400-750-850 which is (8-15-17) times 50
  • 238-816-850 which is(7-24-25) times 34

When I wrote about 845, I said I would explore a conjecture a little more:

My conjecture: If prime numbers x and y are Pythagorean triple hypotenuses, and A and B are integers with B ≥ A and A ≥ 1, then xᴬ × yᴮ will have two primitive triples. The total number of triples xᴬ × yᴮ will have will be A + B + 2Bᴬ

So…how many Pythagorean triples does 2 × 5³ × 17¹ = 4250 have? It will have the same number as 5³ × 17¹ = 2125.

From the conjecture I figure that 2125 and 4250 will each have 1 + 3 + 2(3¹) = 10 total triples. Let’s see if I’m right …

Besides 1 and itself, the factors of 2125 are 5, 17, 25, 85, 125, and 425, all Pythagorean triple hypotenuses. Each of their respective primitive Pythagorean triples has a multiple with 4250 as the hypotenuse:

  1. 850 times 5’s primitive
  2. 250 times 17’s primitive
  3. 175 times 25’s primitive
  4. 34 times 125’s primitive
  5. 50 times 85’s two primitives
  6. 10 times 425’s two primitives

That’s a total of 8 Pythagorean triples from that list. We will also have triples that are 2 times 2125’s primitives. We can find those triples by looking at the sums of two squares that equal 4250.

  1. 65² + 5² = 4250; but 5 is a factor of both 65 and 5, so this will produce a duplicate of one of the triples already given.
  2. 61² + 23² = 4250; gives us 2806-3192-4250, calculated from 2(61)(23), 61² – 23², 61² + 23²
  3. 55² + 35² = 4250; but 5 is a factor of both 55 and 35, so this will produce a duplicate of one of the triples already given.
  4. 49² + 43² = 4250; gives us 552-4214-4250, calculated from 49² – 43², 2(49)(43), 49² + 43²

That gives us 2 more triples to add to the previous 8 for a total of 10 Pythagorean triples, and my conjecture still holds true.

Now one more thing about the number 850, here’s how to write it in a couple other bases:

  • 505 BASE 13, because 5(13²) + 5(1) = 5(170) = 850.
  • PP BASE 33 (P is 25 base 10) because 25(33) + 25(1) = 25(34) = 850

 

 

849 and Level 4

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Print the puzzles or type the solution on this excel file: 12 factors 843-852

8 + 4 + 9 = 21, a number divisible by 3 so odd number 849 can be evenly divided by 3, AND 849 can be written as the sum of three consecutive numbers and six consecutive numbers.

  • 282 + 283 + 284 = 849.
  • 139 + 140 + 141 + 142 + 143 + 144 = 849

849 can also be written as the sum of three odd numbers: 281 + 283 + 285 = 849.

849 can be written as the difference of two squares two different ways:

  • 143² – 140² = 849
  • 425² – 424² = 849

True, similar things can be written about every other number that is divisible by 3, but they are still fun facts about the number 849.

  • 849 is a composite number.
  • Prime factorization: 849 = 3 × 283
  • 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 849 has exactly 4 factors.
  • Factors of 849: 1, 3, 283, 849
  • Factor pairs: 849 = 1 × 849 or 3 × 283
  • 849 has no square factors that allow its square root to be simplified. √849 ≈ 29.1376

848 and Level 3

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Print the puzzles or type the solution on this excel file: 12 factors 843-852

848 is a palindrome, and all but three of its factors are palindromes, too. (Single digit numbers are also palindromes.)

848 is the sum of two squares: 28² + 8² = 848

848 is the hypotenuse of a Pythagorean triple:

  • 448-720-848, calculated from 2(28)(8), 28² – 8², 28² + 8²

844, 845, 846, 847, and 848 are the smallest five consecutive numbers whose square roots can be simplified.

  • 848 is a composite number.
  • Prime factorization: 848 = 2 × 2 × 2 × 2 × 53, which can be written 848 = 2⁴ × 53
  • The exponents in the prime factorization are 4 and 1. Adding one to each and multiplying we get (4 + 1)(1 + 1) = 5 × 2 = 10. Therefore 848 has exactly 10 factors.
  • Factors of 848: 1, 2, 4, 8, 16, 53, 106, 212, 424, 848
  • Factor pairs: 848 = 1 × 848, 2 × 424, 4 × 212, 8 × 106, or 16 × 53
  • Taking the factor pair with the largest square number factor, we get √848 = (√16)(√53) = 4√53 ≈ 29.1204396

847 Sending Love to My Sister in Louisianna

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Print the puzzles or type the solution on this excel file: 12 factors 843-852

My sister, Sue, lives in Louisiana. Several years ago Katrina upset her life, and now Harvey is pounding at her door. I have not heard from her since yesterday when she posted this dreary picture on facebook with the caption, “Flooded at my street.”

Sue, I hope you are okay. If you need a diversion, I hope this puzzle helps at least a tiny bit. I made it just for you. If you need someplace to stay, you can stay with me and my family. We send lots of love and prayers your way.

We also have a son, daughter-in-law, and two grandchildren who live in the Houston area. They are doing okay, but many of their friends are struggling. We pray for them as well.

After the freightening wind died down some, my daughter-in-law posted this picture with the caption, “Day 2 of Hurricane Harvey: We found a Craw-Dad in the back yard!”

My daughter-in-law later posted, “For those of you who are not in Houston I wanted to give you an update. We are located in Kingwood which is northeast of Houston. We have had rain since last Friday and many of our lakes, rivers and bayous are flowing out of their banks. Our home has been very blessed to be in a neighborhood where the rain water is draining nicely, so far. But many of our friends are not as lucky and have had to evacuate due to high water in their homes. We had one small leak in our kitchen, but were able to cover it and stop the dripping. We feel very blessed, but also very concerned for our friends and neighbors. Houston could use your prayers.”

I would like to add that Louisiana and several other towns and cities in Texas could use our prayers, help, and donations.

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Now I’ll write a little about the number 847:

844, 845, 846, 847, and 848 are the smallest five consecutive numbers whose square roots can be simplified.

847 is palindrome 1011101 in BASE 3 because 3⁶ + 3⁴ + 3³ + 3² + 3º = 847.

847 is also 700 in BASE 11 because 7(11²) = 847.

OEIS.org informs us that 847 is the sum of the digits of 2¹⁴ – 1, the 14th Mersenne prime. Since the sum of its digits is 847, that prime number has to be at least 95 digits long!

  • 847 is a composite number.
  • Prime factorization: 847 = 7× 11 × 11, which can be written 847 = 7 × 11²
  • The exponents in the prime factorization are 2 and 1. Adding one to each and multiplying we get (1 + 1)(2 + 1) = 2 × 3  = 6. Therefore 847 has exactly 6 factors.
  • Factors of 847: 1, 7, 11, 77, 121, 847
  • Factor pairs: 847 = 1 × 847, 7 × 121, or 11 × 77
  • Taking the factor pair with the largest square number factor, we get √847 = (√121)(√7) = 11√7 ≈ 29.1032644

846 and Level 2

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Print the puzzles or type the solution on this excel file: 12 factors 843-852

844, 845, 846, 847, and 848 are the smallest five consecutive numbers whose square roots can be simplified.

846 can be written as the sum of consecutive prime numbers two different ways. Together they use ALL the prime numbers from 13 to 127 exactly one time.

  • 13 + 17 + 19 + 23 + 29 + 31 + 37 + 41 + 43 + 47 + 53 + 59 + 61 + 67 + 71 + 73 + 79 + 83 = 846; that’s eighteen consecutive primes.
  • 89 + 97 + 101 + 103 + 107 + 109 + 113 + 127 = 846; that’s eight consecutive primes.

OEIS.org informs us that 846² = 715,716. Not quite a Ruth Aaron pair, but still quite impressive.

  • 846 is a composite number.
  • Prime factorization: 846 = 2 × 3 × 3 × 47, which can be written 846 = 2 × 3² × 47
  • The exponents in the prime factorization are 2, 1, and 1. Adding one to each and multiplying we get (1 + 1)(2 + 1)(1 + 1) = 2 × 3 × 2 = 12. Therefore 846 has exactly 12 factors.
  • Factors of 846: 1, 2, 3, 6, 9, 18, 47, 94, 141, 282, 423, 846
  • Factor pairs: 846 = 1 × 846, 2 × 423, 3 × 282, 6 × 141, 9 × 94, or 18 × 47
  • Taking the factor pair with the largest square number factor, we get √846 = (√9)(√94) = 2√209 ≈ 29.086079

How Often Is 845 the Hypotenuse in a Pythagorean Triple?

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Mathematics is a wonderful collection of patterns.

If we look at the patterns of Pythagorean triples, can we predict how often 845 is the hypotenuse of a Pythagorean triple? Yes, we can.

The prime factorization of 845 is 5 × 13². Now 5, 13, 169 each has one primitive triple while 65 has two primitive triples. That’s a total of five primitive triples. Each of those primitives has a multiple with 845 as the hypotenuse.

  1. 5’s primitive (3-4-5) times 169 is 507-676-845
  2. 13’s primitive (5-12-13) times 65 is 325-780-845
  3. 169’s primitive (119-120-169) times 5 is 595-600-845
  4. 65’s first primitive (16-63-65) times 13 is 208-819-845
  5. 65’s second primitive (33-56-65) times 13 is 429-728-845

Because ALL of its prime factors are primitive Pythagorean triple hypotenuses, 845 will also have some primitives of its own. How many?

Now 325, which is 5² × 13, has five non-primitive triples and two primitive triples, 36-323-325 and 204-253-325 . Since 325 and 845 have the same number of non-primitive triples, and similar looking prime factorizations, is it reasonable to assume that 845 will have two primitive triples as well? Yes, it is!

We can find 845’s primitive Pythagorean triples by looking at the ways to write 845 as the sum of two squares:

29² + 2² = 845, 26² + 13² = 845, and 22² + 19² = 845

26 is a multiple of 13 so it will duplicate one of the triples given by one of 845’s factors. The other two equations give us these two primitive Pythagorean triples:

  1. 116-837-845, which was calculated from 2(29)(2), 29² – 2², 29² + 2²
  2. 123-836-845, which was calculated from 22² – 19², 2(22)(19), 22² + 19²

Thus 845 has a total of 5 + 2 = 7 Pythagorean triples in which it is the hypotenuse!

What if we wanted to know how many times 5² × 13² = 4225 is the hypotenuse of a Pythagorean triple. Could we predict how many times that would be? Yes, we can!

Besides 1 and itself, the factors of 4225 are 5, 13, 25, 65, 169, 325, and 845, all Pythagorean triple hypotenuses. Each of their respective primitive Pythagorean triples has a multiple with 4225 as the hypotenuse:

  1. 845 times 5’s primitive
  2. 325 times 13’s primitive
  3. 169 times 25’s primitive
  4. 25 times 169’s primitive
  5. 65 times 65’s two primitives
  6. 13 times 325’s two primitives
  7. 5 times 845’s two primitives

That’s a total of 10 non-primitive Pythagorean triples for 4225.

We know that 4225 will have some primitives of its own. If the pattern holds true, I hypothesize that 4225 will also have two primitives. So I ask, “What sums of two squares equal 4225?”

  • 63² + 16² = 4225; 63 and 16 have no common prime factors so this will produce a primitive triple.
  • 60² + 25² = 4225; 60 and 25 have 5 as a common factor so this will duplicate one of 4225’s other non-primitives.
  • 56² + 33² = 4225; 56 and 33 have no common prime factors so this will produce a primitive triple.
  • 52² + 39² = 4225; 52 and 39 have 13 as a common factor so this will duplicate one of 4225’s other non-primitives.

Thus the pattern held true and my hypothesis was correct! 4225 has a total of 10 + 2 = 12 Pythagorean triples! All of this has led me to some conclusions:

If x is a prime number with a remainder of 1 when divided by 4, and a is an integer that is greater than or equal to one, then xᴬ will have one primitive triple while the total number of Pythagorean triples will be A.

If x and y are prime numbers each with a remainder of 1 when divided by 4, and if A and B are integers with B ≥ A and A ≥ 1, then xᴬ × yᴮ will have two primitive triples. It appears that the total number of triples xᴬ × yᴮ will have will be A + B + 2Bᴬ. That is my conjecture.

Let’s check my conjecture for the numbers discussed in this post:

  • 5¹ and 13¹ each have 1 triple
  • 5² and 13² each have 2 triples
  • 5² × 13¹ and 5¹ × 13² each have 1 + 2 + 2(2¹) = 7 triples
  • 5² × 13² has 2 + 2 + 2(2²) = 12 triples

So far the conjecture appears to ring true. Perhaps I’ll explore it a little more when I write post 2 × 5² × 17¹ later this week.

Now here’s a little more about the number 845:

844, 845, 846, 847, and 848 are the smallest five consecutive numbers whose square roots can be simplified.

845 is palindrome 5A5 in BASE 12 (A is 10 base 10) because 5(12²) + 10(12¹) + 5(12º) = 845

845 is 500 in BASE 13 because 5(13²) = 845

  • 845 is a composite number.
  • Prime factorization: 845 = 5 × 13 × 13, which can be written 845 = 5 × 13²
  • The exponents in the prime factorization are 1 and 2. Adding one to each and multiplying we get (1 + 1)(2 + 1) = 2 × 3  = 6. Therefore 845 has exactly 6 factors.
  • Factors of 845: 1, 5, 13, 65, 169, 845
  • Factor pairs: 845 = 1 × 845, 5 × 169, or 13 × 65
  • Taking the factor pair with the largest square number factor, we get √845 = (√169)(√5) = 13√5 ≈ 29.06888

Let’s revisit √844

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844 is repdigit 444 in BASE 14 because 4(14²) + 4(14¹) + 4(14⁰) = 844.

None of the following should surprise us:

  • Since 844 is 444 in BASE 14, we know 4 is a factor of 844.
  • (844)/2 = 422 which is 222 in BASE 14.
  • (844)/4 = 211 which is 111 in BASE 14

Back when I wrote Post 266 I made some observations about the numbers 844, 845, 846, 847, and 848. Those are the smallest five consecutive numbers whose square roots can be simplified. Now that I’m writing Post 844, I am just as amazed by those five consecutive numbers and their square roots.

844, 845, 846, 847, 848

Although I included that graphic and the next one in that previous post. Both of them are worth a second look.

844 - 848 prime factorization

  • 844 is a composite number.
  • Prime factorization: 844 = 2 × 2 × 211, which can be written 844 = 2² × 211
  • The exponents in the prime factorization are 2 and 1. Adding one to each and multiplying we get (2 + 1)(1 + 1) = 3 × 2  = 6. Therefore 844 has exactly 6 factors.
  • Factors of 844: 1, 2, 4, 211, 422, 844
  • Factor pairs: 844 = 1 × 844, 2 × 422, or 4 × 211
  • Taking the factor pair with the largest square number factor, we get √844 = (√4)(√211) = 2√211 ≈ 29.051678