Mathematics

864 Factor Trees

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Factor Trees for 864:

The prime factorization of 864 is 2⁵ × 3³. The sum of the exponents is 5 + 3 = 8. Since 8 is a power of 2,  a couple of 864’s factor trees are full and well-balanced:

All of those prime factors lined up in numerical order. That didn’t happen for the next one, but it still makes a good looking tree, and all the prime factors are easy to find.

Is it possible to make a factor tree for 864 that hardly looks like a tree and isn’t as easy to find all the prime factors? Yes, it is. Here’s an example:

Factors of 864:

  • 864 is a composite number.
  • Prime factorization: 864 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3, which can be written 864 = 2⁵ × 3³
  • The exponents in the prime factorization are 5 and 3. Adding one to each and multiplying we get (5 + 1)(3 + 1) = 6 × 4 = 24. Therefore 864 has exactly 24 factors.
  • Factors of 864: 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 72, 96, 108, 144, 216, 288, 432, 864
  • Factor pairs: 864 = 1 × 864, 2 × 432, 3 × 288, 4 × 216, 6 × 144, 8 × 108, 9 × 96, 12 × 72, 16 × 54, 18 × 48, 24 × 36, or 27 × 32
  • Taking the factor pair with the largest square number factor, we get √864 = (√144)(√6) = 12√6 ≈ 29.3938769

Sum-Difference Puzzles:

6 has two factor pairs. One of those pairs adds up to 5, and the other one subtracts to 5. Put the factors in the appropriate boxes in the first puzzle.

864 has twelve factor pairs. One of the factor pairs adds up to ­60, and a different one subtracts to 60. If you can identify those factor pairs, then you can solve the second puzzle!

More Facts about the Number 864:

864 looked interesting to me in a few other bases:

  • 4000 BASE 6 because 4(6³) = 864
  • 600 BASE 12 because 6(12²) = 864
  • RR BASE 31 (R is 27 base 10) because 27(31) + 27(1) = 27(32) = 864
  • OO BASE 35 (O is 24 base 10) because 24(35) + 24(1) = 24(36) = 864
  • O0 BASE 36 (Oh zero) because 24(36) + 0(1) = 864

864 is the sum of the 20 prime numbers from 7 to 83.

131 + 137 + 139 + 149 + 151 + 157 = 864; that’s six consecutive primes.

431 + 433 = 864; that’s the sum of twin primes.

864 is in this cool pattern:

 

863 These Primitive Pythagorean Triples Have Some Terrific Legs!

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We’ve seen that the sum of the hypotenuse of a primitive Pythagorean triple and either of its legs is either a perfect square or two times a perfect square. What about just the legs themselves? Is there anything special about their sums? I think so. I think primitive Pythagorean triples have some terrific legs! Look at this chart and see if you agree.

If a sum is a prime number, there is only one way to get that sum. If the sum is repeated, then its prime factorization only uses prime numbers already on the list.

I’m not as impressed with the differences of the legs, but it is interesting that the same prime numbers show up:

863 is the sum of consecutive primes several ways:

  • 863 is the sum of the fifteen prime numbers from 29 to 89.
  • 863 is the sum of the thirteen prime numbers from 41 to 97.
  • 107 + 109 + 113 + 127 + 131 + 137 + 139 = 863; that’s seven consecutive primes.
  • 163 + 167+ 173 + 179 + 181 = 863; that’s five consecutive primes.

863(863 + 6) = 749,947, a palindrome. Thank you OEIS.org for that fun fact.

  • 863 is a prime number.
  • Prime factorization: 863 is prime.
  • The exponent of prime number 863 is 1. Adding 1 to that exponent we get (1 + 1) = 2. Therefore 863 has exactly 2 factors.
  • Factors of 863: 1, 863
  • Factor pairs: 863 = 1 × 863
  • 863 has no square factors that allow its square root to be simplified. √863 ≈ 29.37686

How do we know that 863 is a prime number? If 863 were not a prime number, then it would be divisible by at least one prime number less than or equal to √863 ≈ 29.4. Since 863 cannot be divided evenly by 2, 3, 5, 7, 11, 13, 17, 19, 23, or 29, we know that 863 is a prime number.

 

857 Why do Primitive Pythagorean Triples Do That?

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29² + 4² = 857, so 857 is the hypotenuse of a Pythagorean triple that just happens to be primitive:

  • 232-825-857 which was calculated from 2(29)(4), 29² – 4², 29² + 4².

Now, one of the legs of a primitive Pythagorean triple will be divisible by 4. In this case that number is 232. As I indicated when I wrote 821 Why Do Primitive Pythagorean Triples Do That? The sum of the number divisible by 4 in a primitive Pythagorean triple and the hypotenuse always equals a perfect square. So 232 + 857 = 1089 = 33².

This week I noticed something about the OTHER leg that might be even more amazing! The other leg in the triple, 825, is not divisible by 4, but get this: One half of the sum of a primitive’s hypotenuse and that leg that is NOT divisible by 4 also equals a square number: For example, (857 + 825)/2 = 1681/2 = 841 = 29².

But not only that, one half of the DIFFERENCE of a primitive’s hypotenuse and that leg that is NOT divisible by 4 equals a perfect square, too: (857 – 825)/2 = 32/2 = 16 = 4².

Now 29² and 4² look a little familiar to me today. Hmm… We’ve seen those two numbers together earlier in this post!

It seems that for Primitive Pythagorean Triples, what goes around comes around!

Check out the sums and difference of these other primitive Pythagorean triples:

Go ahead, try it with any other primitive triple on this list or otherwise!

If we let s = √(Sum/2) and let d = √(Difference/2), then we can rewrite some Pythagorean triple like this:

  • s² – d², 2sd, s² + d², so s² and d² are REALLY significant perfect squares!

Besides being the hypotenuse of a primitive Pythagorean triple, here are some other facts about the number 857:

857 is the sum of three consecutive prime numbers: 281 + 283 + 293.

857 is also the sum of the nineteen prime numbers from 11 to 83.

  • 857 is a prime number.
  • Prime factorization: 857 is prime.
  • The exponent of prime number 857 is 1. Adding 1 to that exponent we get (1 + 1) = 2. Therefore 857 has exactly 2 factors.
  • Factors of 857: 1, 857
  • Factor pairs: 857 = 1 × 857
  • 857 has no square factors that allow its square root to be simplified. √857 ≈ 29.27456

How do we know that 857 is a prime number? If 857 were not a prime number, then it would be divisible by at least one prime number less than or equal to √857 ≈ 29.3. Since 857 cannot be divided evenly by 2, 3, 5, 7, 11, 13, 17, 19, 23, or 29, we know that 857 is a prime number.

 

Here’s another way we know that 857 is a prime number: Since its last two digits divided by 4 leave a remainder of 1, and 29² + 4² = 857 with 29 and 4 having no common prime factors, 857 will be prime unless it is divisible by a prime number Pythagorean triple hypotenuse less than or equal to √857 ≈ 29.3. Since 853 is not divisible by 5, 13, 17, or 29, we know that 857 is a prime number.

Something Cool about 854 and Its Square

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OEIS.org informs us that you can find all the digits 1 to 9 exactly one time when 854 is combined with its square. I thought that was pretty cool so I made this gif:

854 and Its Square

make science GIFs like this at MakeaGif

854 can be written as the sum of consecutive numbers in several different ways:

  • as the sum of 4 consecutive numbers: 212 + 213 + 214 + 215 = 854
  • as the sum of the 7 consecutive numbers from 119 to 125 with 122 as the middle number.
  • as the sum of the 28 consecutive numbers from 17 to 44 with 30 and 31 as the middle numbers.
  • as the sum of 2 consecutive even numbers: 426 + 428 = 854
  • as the sum of the 7 consecutive even numbers from 116 to 128 with 122 as the middle number.
  • as the sum of the 14 consecutive even numbers from 48 to 74 with 60 and 62 as the middle numbers.
  • Even number 854 does not have any factor pairs in which both numbers are even, so it cannot be written as the sum of consecutive odd numbers.
854 is the hypotenuse of a Pythagorean triple: 154-840-854, which is 14 times (11-60-61).
  • 854 is a composite number.
  • Prime factorization: 854 = 2 × 7 × 61
  • The exponents in the prime factorization are 1, 1, and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1)(1 + 1) = 2 × 2 × 2 = 8. Therefore 854 has exactly 8 factors.
  • Factors of 854: 1, 2, 7, 14, 61, 122, 427, 854
  • Factor pairs: 854 = 1 × 854, 2 × 427, 7 × 122, or 14 × 61
  • 854 has no square factors that allow its square root to be simplified. √854 ≈ 29.223278.

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

 

 

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

 

 

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

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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.