Month: August 2017

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?

/ ivasallay / 1 Comment

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

 

 

843 is a Lucas Number

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

843 is the hypotenuse of Pythagorean triple 480-693-843 which is 3 times (160-231-281).

I think 843 looks interesting in a couple different bases:

  • It is palindrome 1101001011 in BASE 2 because 2⁹ + 2⁸ + 2⁶ + 2³ + 2¹ + 2⁰ = 843
  • It is 123 in BASE 28 because 1(28²) + 2(28¹) + 3(28º) = 843

From OEIS.org, I learned that 843 is the 14th Lucas number which is similar to being a Fibonacci number.

  • Fibonacci numbers start with 1, 1, and all the rest of the Fibonacci numbers are found by adding the previous two numbers in the series.
  • Lucas numbers start with 2, 1, and all the rest of the Lucas numbers are found by adding the previous two numbers in the series.

843 makes the list of Lucas numbers: 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, . . .

Even though it is the 15th number on the list, it is the 14th Lucas number. (The list starts with the zeroth number?)

Here is 843’s factoring information:

  • 843 is a composite number.
  • Prime factorization: 843 = 3 × 281
  • 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 843 has exactly 4 factors.
  • Factors of 843: 1, 3, 281, 843
  • Factor pairs: 843 = 1 × 843 or 3 × 281
  • 843 has no square factors that allow its square root to be simplified. √843 ≈ 03446228

 

 

842 and Level 6

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Print the puzzles or type the solution on this excel file: 10-factors-835-842

29² + 1² = 842

That means 842 is the hypotenuse of a Pythagorean triple:

  • 58-840-842, calculated from 2(29)(1), 29² – 1², 29² + 1²

842 is repdigit 222 in BASE 20 because 2(20²) + 2(20¹) + 2(20º) = 842

  • 842 is a composite number.
  • Prime factorization: 842 = 2 × 421
  • 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 842 has exactly 4 factors.
  • Factors of 842: 1, 2, 421, 842
  • Factor pairs: 842 = 1 × 842 or 2 × 421
  • 842 has no square factors that allow its square root to be simplified. √842 ≈ 29.017236

 

You’ll Be Impressed By 841’s Number Facts

/ ivasallay / 3 Comments

841 is a very cool square number. 29 × 29 = 841. You can write it more compactly: 29² = 841. Here are some facts about this square number:

841 is the sum of the 29th and the 28th triangular numbers because 841 = 435 + 406.

What’s that you say? Every other square number can make a similar claim? Oh. … Wait a minute…Not EVERY square number can do that. One is a square number, and it’s NOT the sum of two triangular numbers. So there…

How about this… 841 is the sum of the first 29 odd numbers. That makes it the sum of all the odd numbers from 1 to 57. That’s impressive!

What? All square numbers do that? Huh??? n² is always the sum of the first n odd numbers? Why’d you have to tell me that fact isn’t particularly unique either!

Okay…841 is an interesting number in a few other bases. Pay attention….  841 is

  • 100 in BASE 29
  • 121 in BASE 28
  • 144 in BASE 27
  • 169 in BASE 26
  • 441 in BASE 14

Ha! You didn’t know that one! … Now don’t go spoiling my glee by telling me that those same square numbers show up for a few other square numbers, too. Don’t tell me that!

Give me one more chance to impress you….This square number, 841, is the sum of two consecutive square numbers, so 20² + 21² = 841 = 29². That hasn’t happened to a square number since 3² + 4² = 25 = 5². That makes 841 the 21st Centered Square Number but only the SECOND square number that is both kinds of squares! There may be an infinite number of squares that do the same thing, but it is still a fairly unique characteristic.

There are 441 blue squares in that graphic. There are 400 squares that are orange, red, green, purple, or black. 441 + 400 = 841.

Here are a few other tidbits about 841:

841 is the sum of the nine prime numbers from 73 to 109. It is also the sum of three consecutive primes: 277 +  281 + 283 = 841

841 is the hypotenuse of TWO Pythagorean triples:

  • 580-609-841 which is 29 times (20-21-29)
  • 41-840-841 calculated from 21² – 20², 2(21)(20), 21² + 20²

841 Pythagorean triple Recursion: Hmm…How Quickly Can I Give You a Headache? Warning! Reading this set of bullet points might overload your brain:

  • 5² + 2² = 29, so (2∙5∙2)² + (5² – 2²)² = (5² + 2²)² which means 20² + 21² = 29² = 841.
  • 21² + 20² = 841, so (2∙21∙20)² + (21² – 20²)² = (21² + 20²)² which means 840² + 41² = 841² = (29²)² = 707281.
  • 840² + 41² = 707281, so (2∙840∙41)² + (840² – 41²)² = (840² + 41²)² which means 68880² + 703919² = 707281² = ((29²)²)² = 500,246,412,961
  • We could go on forever with even bigger powers of 29 …

We’ll finish with just some simple, easy-on-the-brain facts about 841:

29 is the tenth prime number. Its square, 841, is only the tenth number to have exactly three factors.

  • 841 is a composite number.
  • Prime factorization: 841 = 29²
  • The exponent in the prime factorization is 2. Adding one we get (2 + 1) = 3. Therefore 841 has exactly 3 factors.
  • Factors of 841: 1, 29, 841
  • Factor pairs: 841 = 1 × 841 or 29 × 29
  • 841 is a perfect square. √841 = 29


I’m impressed by all this, even if you aren’t.

 

840 is the Smallest Number that Is the Top Number in THREE Sum-Difference Puzzles

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Factor Trees for 840:

840 has more factors than any previous number. Those factors will help us write 840 as the sum of consecutive counting numbers, consecutive even numbers, and consecutive odd numbers. What are the factors of 840? Here are a couple of the many possible factor trees for 840:

The red leaves on the tree are prime numbers. Gathering the six red leaves from either factor tree above gives us 840’s prime factorization: 840 = 2³ × 3 × 5 × 7.  Now 840 is not the smallest number to have six red leaves. In fact, there are smaller numbers with as many as nine leaves, but 840’s six innocent-looking red leaves will turn into a whopping 32 factors!

Factors of 840:

  • 840 is a composite number.
  • Prime factorization: 840 = 2 × 2 × 2 × 3 × 5 × 7, which can be written 840 = 2³ × 3 × 5 × 7.
  • 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 840 has exactly 32 factors.
  • Factors of 840: 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 15, 20, 21, 24, 28, 30, 35, 40, 42, 56, 60, 70, 84, 105, 120, 140, 168, 210, 280, 420, 840
  • Factor pairs: 840 = 1 × 840, 2 × 420, 3 × 280, 4 × 210, 5 × 168, 6 × 140, 7 × 120, 8 × 105, 10 × 84, 12 × 70, 14 × 60, 15 × 56, 20 × 42, 21 × 40, 24 × 35 or 28 × 30
  • Taking the factor pair with the largest square number factor, we get √840 = (√4)(√210) = 2√210 ≈ 28.98275.

840 is also the smallest number that can be evenly divided by the first eight counting numbers!

Sum-Difference Puzzles:

210 has eight factor pairs. One of those factor pairs adds up to 29, and another one subtracts to 29. Another factor pair adds up to 37, and a different one subtracts to 37. Thus, 210 produces two different Sum-Difference Puzzles shown in the first two graphics below!

840 is a multiple of 210 and  has sixteen factor pairs! One of the factor pairs adds up to ­58 and another to 74. A different factor pair subtracts to 58, and still a different one subtracts to 74. If you can identify those factor pairs, then you can solve the two puzzles that are next to the 210 puzzles below!

840 has yet another factor pair that adds up to 113 and a different one that subtracts to 113. If you can find those factor pairs, then you can solve this next primitive puzzle:

If you need help with any of those 840 Sum-Difference puzzles, the chart below shows the sums and differences of all of 840’s factor pairs. See which sums also appear in the difference column.

How to find consecutive counting numbers that add up to 840:

840 is more than the 40th triangular number (820) and less than the 41st triangular number (861). We can also arrive at the number 40 by using √(1 + 840×2) – 1 = 40, no rounding necessary. 840 has six odd factors that are not more than 40, namely 1, 3, 5, 7, 15, 21 and 35.

I’ll describe the ways we can write 840 as the sum of consecutive numbers. Can you write out the sums? I’ve done one of them for you:

  • using 3 numbers with 280 as the middle number,
  • using 5 numbers with 168 as the middle number,
  • using 7 numbers with 120 as the middle number; 117 + 118 + 119 + 120 + 121 + 122 + 123 = 840
  • using 15 numbers with 56 as the middle number,
  • using 21 numbers with 40 as the middle number
  • using 35 numbers with 24 as the middle number.

Notice each of those ways has a factor pair of 840 in the description.

The largest power of 2 that is a factor of 840 is 8, which doubled becomes 16. Which of 840’s odd factors multiplied by 16 are not more than 40? 1 × 16 = 16, and 3 × 16 = 48. Oops, that’s too much. The rest of its odd factors times 16 will be too much as well. Nevertheless, we can write 840 as the sum of 16 counting numbers. 840÷16 = 52.5 so 52 and 53 will be the 8th and 9th numbers in the sum.

  • 45 + 46 + 47 + 48 + 49 + 50 + 51 + 52 + 53 + 54 + 55 + 56 + 57 + 58 + 59 + 60 = 840

How do we find consecutive EVEN numbers that add up to an even number?

Only even numbers can be the sum of consecutive even numbers. Let’s use 840 as an example again. 840÷2 = 420.

First, we will find all the ways to write 420 as the sum of consecutive numbers. Then we will simply double the middle number and surround it with the appropriate number of even numbers to get a sum of even numbers that add up to 840:

√(1 + 420×2) – 1 = 28, no rounding necessary, so we will make a list of the odd factors that are not more than 28. They are 1, 3, 5, 7, 15, and 21. We also note that the largest power of 2 that is a factor of 420 is 4. Doubling 4, we get 8. Which of 420’s odd factors multiplied by 8 are not more than 28? 1 × 8 = 8, and 3 × 8 = 24. All of the rest will be too much.

I’ll describe the ways we can write 840 as the sum of consecutive even numbers. You can see 420’s factor pairs and 840’s factor pairs in the descriptions. Can you write out the sums? I’ve done one of them for you:

  • using 3 even numbers with 2 × 140 = 280 as the middle number
  • using 5 even numbers with 2 × 84 = 168 as the middle number
  • using 7 even numbers with 2 × 60 = 120 as the middle number
  • using 15 even numbers with 2 × 28 = 56 as the middle number
  • using 21 even numbers with 2 × 20 = 40 as the middle number; 20 + 22 + 24 + 26 + 28 + 30 + 32 + 34 + 36 + 38 + 40 + 42 + 44 + 46 + 48 + 50 + 52 + 54 + 56 + 58 + 60 = 840

We can also write 840 as the sum of an even amount of consecutive even numbers.

  • using 8 even numbers: 2 times (49 + 50 + 51 + 52 + 53 + 54 + 55 + 56 = 420) becomes 98 + 100 + 102 + 104 + 106 + 108 + 110 + 112 = 840. Notice that 840÷8 = 105, the odd number that is between the two numbers in the exact middle of the sum.
  • Likewise, using 24 even numbers: 12 + 14 + 16 + 18 + 20 + 22 + 24 + 26 + 28 + 30 + 32 + 34 + 36 + 38 + 40 + 42 + 44 + 46 + 48 + 50 + 52 + 54 + 56 + 58 = 840. Notice that 840÷ 24 = 35, the odd number that is between the two numbers in the exact middle of the sum.

How to find consecutive ODD numbers that add up to an even number like 841: 

Let me first point out that every square number, n², is the sum of the first n numbers.

For example: 100 = 10², and 100 is also the sum of the first 10 odd numbers as illustrated below:

A similar statement is true for every other square number:

Yes, every square number, n², is the sum of the first n odd numbers.

To write a non-square even number as the sum of consecutive odd numbers, the first thing we must do is determine if the number can be written as the difference of two squares. If an even number has a factor pair, b × a, in which b > a, and BOTH a and b are even, then that even number can be expressed as the difference of two squares by using [(b + a)/2]² – [(b – a)/2]².

Now as long as an even number can be expressed as the difference of two squares, B² – A², then that number can be written as the sum of all the odd numbers from 2A+1 to 2B-1.

840 is an even number with eight factor pairs in which both numbers are even. Let’s use those even factor pairs to find the ways to write 840 as the sum of consecutive ODD numbers:

  • 420 × 2 = 840 means 211² – 209² = 840, and that means the sum of the 2 consecutive odd numbers from 419 to 421 = 840
  • 210 × 4 = 840 means 107² – 103² = 840, and that means the sum of the 4 consecutive odd numbers from 207 to 213 = 840
  • 140 × 6 = 840 means 73² – 67² = 840, and that means the sum of the 6 consecutive odd numbers from 135 to 145 = 840
  • 84 × 10 = 840 means 47² – 37² = 840, and that means the sum of the 10 consecutive odd numbers from 75 to 93 = 840
  • 70 × 12 = 840 means 41² – 29² = 840, and that means the sum of the 12 consecutive odd numbers from 59 to 81 = 840
  • 60 × 14 = 840 means 37² – 23² = 840, and that means the sum of the 14 consecutive odd numbers from 47 to 73 = 840
  • 42 × 20 = 840 means 31² – 11² = 840, and that means the sum of the 20 consecutive odd numbers from 23 to 61 = 840
  • 30 × 28 = 840 means 29² – 1² = 840, and that means the sum of the 28 consecutive odd numbers from 3 to 29 = 840

Thus, 840 with its record setting 32 factors, can be written as the sum of 7 consecutive numbers, 7 consecutive even numbers, and 8 consecutive odd numbers!

More about the Number 840:

Incidentally, being able to write 840 as the difference of two squares, eight different ways also makes 840 a leg in at least eight different Pythagorean triples. Those Pythagorean triples can be calculated using the numbers from the difference of two squares. For example, 682-840-1082 can be calculated from 2(31)(11), 31² – 11², 31² + 11².

840 was the leg for those eight triples. It is possible that looking at 2(b)(a), where b × a = 420, will produce some more Pythagorean triples with 840 as the leg.

840 is also the hypotenuse of one Pythagorean triple, 504-672-840.

On a note totally unrelated to anything I’ve written above, 840 is a repdigit in two bases:

  • SS BASE 29 (S is 28 base 10) Note that 28(29) + 28(1) = 28(30) = 840
  • 00 BASE 34 (0 is 24 base 10) Note that 24(34) + 24(1) = 24(35) = 840

840 is also the sum of twin prime numbers 419 and 421.

839 and Level 5

/ ivasallay / Leave a comment

839 is the sum of the five prime numbers from 157 to 179:

  • 157 + 163 + 167 + 173 + 179 = 839

Print the puzzles or type the solution on this excel file: 10-factors-835-842

  • 839 is a prime number.
  • Prime factorization: 839 is prime.
  • The exponent of prime number 839 is 1. Adding 1 to that exponent we get (1 + 1) = 2. Therefore 839 has exactly 2 factors.
  • Factors of 839: 1, 839
  • Factor pairs: 839 = 1 × 839
  • 839 has no square factors that allow its square root to be simplified. √839 ≈ 28.9654967

How do we know that 839 is a prime number? If 839 were not a prime number, then it would be divisible by at least one prime number less than or equal to √839 ≈ 28.97. Since 839 cannot be divided evenly by 2, 3, 5, 7, 11, 13, 17, 19, or 23, we know that 839 is a prime number.