A Multiplication Based Logic Puzzle

Posts tagged ‘Puzzle’

834 and Level 6

834 is the sum of consecutive prime numbers two different ways:

  • 127 + 131 + 137 + 139 + 149 + 151 = 834; that’s six consecutive primes
  • 31 + 37 + 41 + 43 + 47 + 53 + 59 + 61 + 67 + 71 + 73 + 79 + 83 + 89 = 834; that’s fourteen consecutive primes

The ONLY Pythagorean triple that contains the number 834 is 834 – 173888 – 173890.

  • 834 is a composite number.
  • Prime factorization: 834 = 2 × 3 × 139
  • 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 834 has exactly 8 factors.
  • Factors of 834: 1, 2, 3, 6, 139, 278, 417, 834
  • Factor pairs: 834 = 1 × 834, 2 × 417, 3 × 278, or 6 × 139
  • 834 has no square factors that allow its square root to be simplified. √834 ≈ 28.879058

There was a solar eclipse in the United States today. People where I lived were able to experience 91.32% obstruction of the sun. I love this interactive map of today’s eclipse and past and future ones as well.

Here are a few tweets I saw about eclipses on twitter:

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Category:

Level 6 Puzzle, Mathematics

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833 and Level 5

Print the puzzles or type the solution on this excel file: 12 factors 829-834

Wrap Your Mind Around This: 7² × 17 = 833 = 28² + 7²

833 is the hypotenuse of a Pythagorean triple:

  • 392-735-833 calculated from 2(28)(7), 28² – 7², 28² + 7²

833 can be written as the sum of consecutive prime numbers two ways. One of those ways starts with one of its prime factors, 17:

  • 833 is the sum of the seventeen prime numbers from 17 to 83
  • 833 is also the sum of the eleven prime numbers from 53 to 101

Since 833 has three factor pairs where both factors are odd, it can be written as the difference of two squares three different ways:

  • 833 × 1 = 833 means 417² – 416² = 833
  • 119 × 7 = 833 means 63² – 56² = 833
  • 49 × 17 = 833 means 33² – 16² = 833

The 41st triangular number will be 861. We must use less than 41 consecutive numbers if we want to express 833 as the sum of consecutive numbers. 833 has 3 odd factors (1, 7, 17) less than 41. Thus 833 can be written as the sum of 7 consecutive numbers and as the sum of 17 consecutive numbers. Notice 833’s factor pairs below highlighted in red.

  • 833 = 116 + 117 + 118 + 119 + 120 + 121 + 122; that’s 7 consecutive numbers
  • 833 = 41 + 42 + 43 + 44 + 45 + 46 + 47 + 48 + 49 + 50 + 51 + 52 + 53 + 54 + 55 + 56 + 57; that’s 17 consecutive numbers

The factor of 833 that is the highest power of 2 is 1 because 2º=1. Each of those odd factors, (1, 7, 17), times 2 × 1 is still less than 40, so 833 can also be written as the sum of 2 consecutive numbers, the sum of 14 consecutive numbers, and the sum of 34 consecutive numbers:

  • 833 = 416 + 417; that’s 2 consecutive numbers
  • 833 = 53 + 54 + 55 + 56 + 57 + 58 + 59 + 60 + 61 + 62 + 63 + 64 + 65 + 66; that’s 14 consecutive numbers
  • 833 = 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 + 21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30 + 31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 + 40 + 41; that’s 34 consecutive numbers

Here is 833’s factoring information:

  • 833 is a composite number.
  • Prime factorization: 833 = 7 × 7 × 17, which can be written 833 = 7² × 17
  • 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 833 has exactly 6 factors.
  • Factors of 833: 1, 7, 17, 49, 119, 833
  • Factor pairs: 833 = 1 × 833, 7 × 119, or 17 × 49
  • Taking the factor pair with the largest square number factor, we get √833 = (√49)(√17) = 7√17 ≈ 28.861739

Category:

Level 5 Puzzle

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832 and Level 4

Print the puzzles or type the solution on this excel file: 12 factors 829-834

832 has many factors, but it can be written as the sum of consecutive numbers only one way:

  • 58 + 59 + 60 + 61 + 62 + 63 + 64 + 65 + 66 + 67 + 68 + 69 + 70 = 832; that’s thirteen consecutive numbers.

832 can be written as the difference of two squares five different ways because it has five factor pairs in which both numbers are even:

  • 416 × 2 = 832 means 209² – 207² = 832
  • 208 × 4 = 832 means 106² – 102² = 832
  • 104 × 8 = 832 means 56² – 48² = 832
  • 52 × 16 = 832 means 34² – 18² = 832
  • 32 × 26 = 832 means 29² – 3² = 832

832 is also the sum of two squares:

  • 24² + 16² = 832

832 is the hypotenuse of a Pythagorean triple:

  • 320-768-832 calculated from 24² – 16², 2(24)(16), 24² + 16²
  • 320-768-832 is also 64 times (5-12-13)

832 is repdigit QQ in BASE 31 (Q is 26 base 10). That’s because 26(31) + 26(1) = 832, which is the same as saying 26 × 32 = 832.

  • 832 is a composite number.
  • Prime factorization: 832 = 2 × 2 × 2 × 2 × 2 × 2 × 7, which can be written 832 = 2⁶ × 7
  • The exponents in the prime factorization are 6, and 1. Adding one to each and multiplying we get (6 + 1)(1 + 1) = 7 × 2 = 14. Therefore 832 has exactly 14 factors.
  • Factors of 832: 1, 2, 4, 8, 13, 16, 26, 32, 52, 64, 104, 208, 416, 832
  • Factor pairs: 832 = 1 × 832, 2 × 416, 4 × 208, 8 × 104, 13 × 64, 16 × 52, or 26 × 32
  • Taking the factor pair with the largest square number factor, we get √832 = (√64)(√13) = 8√13 ≈ 28.8444102

Category:

Level 4 Puzzle

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831 and Level 3

Print the puzzles or type the solution on this excel file: 12 factors 829-834

Before I write a blog post, I look to see how the post number is expressed in different bases. Today I noticed that 831 is 30333 in BASE 4. I was intrigued by all those 3’s because I knew that 831 = 3 × 277. It seems logical that 277 would be 10111 in BASE 4, it turns out that it is! I looked at 831 in all the bases up to BASE 36. Did any others have only multiples of 3 as its digits? Yes, a few did, so I’ve made a chart of 277 and 831 in those five bases to make comparing them easy. I also used only base 10 numbers and not letters of the alphabet to represent the digits in the other bases. As you look at this chart, remember 3 × 277 = 831.

Why are those the ONLY bases for which 3 times the digits of 277 equals the digits for 831? Because in every other base, at least one of the digits times 3 will be greater than or equal to the base and some complicated carrying will have to take place to determine the digits for 831 in that base.

For example, 277 is palindrome 1 11 1 in BASE 12. Obviously 3 times 1 11 1 is 3 33 3. Since 33 is bigger than 12, we somehow end up with non-palindrome 5 9 3 in BASE 12 for 831. This is how that somehow happened: 33÷12 = 2R9. The 9 becomes the middle digit while the 2 is added to the original 3 to make the first digit, 5.

Here’s a little more about the number 831:

Because 277 is one of its factors, 831 is the hypotenuse of a Pythagorean triple: 345-756-831, which is 3 times primitive (115-252-277).

  • 831 is a composite number.
  • Prime factorization: 831 = 3 × 277
  • 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 831 has exactly 4 factors.
  • Factors of 831: 1, 3, 277, 831
  • Factor pairs: 831 = 1 × 831 or 3 × 277
  • 831 has no square factors that allow its square root to be simplified. √831 ≈ 8270706

 

 

Category:

Level 3 Puzzle, Mathematics

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830 I Can Divide These Polynomials By (x – 2) Without Even Looking at Them

Print the puzzles or type the solution on this excel file: 12 factors 829-834

(x – 2) is a factor of an infinite number of polynomials. I am listing only a small, but very special subset of them here. First look for the pattern that allows us to generate a polynomial from a given number in base 2. Then look for another pattern when the polynomial is divided by (x – 2).

Do you see the patterns? I do.

From the first pattern I know there is a similar special polynomial that ends with -830. AND I know from the second pattern what I will get if I divide THAT polynomial by (x – 2). Now get this: Even though I haven’t seen the polynomial yet, I know what the quotient will be! When the polynomial ending in -830 is divided by (x – 2), it will be. . . .

  • x⁸ + 3x⁷ + 6x⁶ + 12x⁵ + 25x⁴ + 51x³ + 103x² + 207x + 415

And guess what, I’m right! I found the quotient without showing any steps or even looking at what I was dividing.

How did I know what that quotient would be without writing down the problem and doing some division first? Well, not only is this polynomial special, but the quotient is special, too!

All I needed to know was that the last term was -830. I then divided 830 repeatedly by 2. Any time my quotient was an odd number, I subtracted one from it before I divided it again by 2. I repeated the process until I reached zero. That is how I got all my coefficients. Even though I could do this problem without showing any work, I made a gif so you and anyone else can quickly see how I did it, but you’ll have to look sideways at it to see it. Showing steps is ALWAYS a good thing.
Find 830 in BASE 2

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As a side benefit, this is another way to find out what 830 is in BASE 2.

Here’s a little more about the number 830:

  • 830 is a composite number.
  • Prime factorization: 830 = 2 × 5 × 83
  • 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 830 has exactly 8 factors.
  • Factors of 830: 1, 2, 5, 10, 83, 166, 415, 830
  • Factor pairs: 830 = 1 × 830, 2 × 415, 5 × 166, or 10 × 83
  • 830 has no square factors that allow its square root to be simplified. √830 ≈ 28.80972

830 is the sum of four consecutive prime numbers:

  • 197 + 199 + 211 + 223 = 830

Because 5 is one of its factors, 830 is the hypotenuse of a Pythagorean triple:

  • 498-664-830; that’s 166 times (3-4-5)

 

Category:

Level 2 Puzzle

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828 Try Synthetic Division on These Special Polynomials

Print the puzzles or type the solution on this excel file: 10-factors-822-828

828 has a lot of factors so I decided to use it in my examples of synthetic division. What are the factors of 828?

  • 828 is a composite number.
  • Prime factorization: 828 = 2 × 2 × 3 × 3 × 23, which can be written 828 = 2² × 3² × 23
  • The exponents in the prime factorization are 2, 2 and 1. Adding one to each and multiplying we get (2 + 1)(2 + 1)(1 + 1) = 3 × 3 × 2 = 18. Therefore 828 has exactly 18 factors.
  • Factors of 828: 1, 2, 3, 4, 6, 9, 12, 18, 23, 36, 46, 69, 92, 138, 207, 276, 414, 828
  • Factor pairs: 828 = 1 × 828, 2 × 414, 3 × 276, 4 × 207, 6 × 138, 9 × 92, 12 × 69, 18 × 46 or 23 × 36
  • Taking the factor pair with the largest square number factor, we get √828 = (√36)(√23) = 6√23 ≈ 28.774989.

Synthetic division is taught in many schools in the United States, but in other places in the world it typically is not taught at all. I like synthetic division. I disagree with those few people who describe it as a mostly useless trick that isn’t worth learning. Yes, its usefulness is limited, but when it can be used, it can be absolutely wonderful. Personally, I almost always use synthetic division when dividing polynomials by (x-a) or (x+a) where a is any whole number. (If a is a fraction, synthetic division can still be done, but it might not be much fun.)

What are some of the advantages of using synthetic division?

  • If you had a polynomial where x is raised to several different powers, such as x⁹ + x⁸ + x⁷ + x⁶ + x⁵ + x⁴ + x³ + x² + x – 8, you would only have to write 1 1 1 1 1 1 1 1 1 -8 to perform the algorithm. That could prevent writer’s cramp if the polynomial is quite long. Also less writing means fewer chances for mistakes.
  • Instead of needing 9×2 lines to do long division for the problem, only three total lines are needed. That saves paper.
  • Using a instead of (x-a) or -a instead of (x+a) in the algorithm means we use addition instead of subtraction to find the quotient. Most people make fewer mistakes adding numbers than they do subtracting. Fewer mistakes means less frustration and less erasing.

Before we can do synthetic division we need to write some polynomials. Since this is my 828th post, I will write some polynomials based on the following chart, and they will be very special polynomials!

The numbers in bold print end in a zero because the corresponding base number is a factor of 828. For base 11 or greater, sometimes a digit is represented by a letter of the alphabet. The key to translating those letters to the corresponding number in base 10 is A = 10, B = 11, C = 12, D = 13, E = 14, F = 15, G = 16, H = 17, I = 18, J = 19, K = 20, L = 21, M = 22. This chart goes to BASE 28 because √828 ≈ 28.77.

We can write a polynomial for any of those bases using the digits given. The last digit for these special polynomials will be replaced with -828, but as you will see, that original last digit will not be forgotten.

Because 828 is 30330 in BASE 4, let’s use that information as our first example:

  • The digits 30330 make the polynomial 3x⁴ + 0x³ + 3x² + 3x -828.
  • The digits 3 0 3 3 -828 will be used as the coefficients in our synthetic division algorithm.
  • BASE 4 will be seen in the divisor (x – 4) and as “4” in the algorithm.

Now watch as this gif uses synthetic division to find the quotient.

 

828 Synthetic Division

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The remainder is zero, and the last digit of 30330 is zero. From the remainder theorem we also know that 3(4⁴) + 3(4²) + 3(4) -828 = 0.

It turns out we can know what the remainder is for each of these special polynomials BEFORE we do any dividing! The remainder will be the last digit times negative one. That does not usually happen when we use synthetic division on a polynomial, but it will always happen on these special polynomials!

Here are a four more examples of writing one of these special polynomials and dividing it using synthetic division. Try writing the rest of the problems using some of the other bases and doing the division yourself, too.

Here are a few notes that wouldn’t fit in the table giving a logical solution to Find the Factors #828:

  1. Clue 27 will use a 3, so clue 9 cannot be 3×3. Thus, clues  9 and 18 will put 9 in the first column and 1 and 2 in the top row.
  2. Can both 40’s be 4×10? No, because that would use both 10’s, and make the 8 and the 18 use both 2’s. That would mean that clue 10 could not be 10×1 or 2×5.
  3. So 56 and one of the 40’s will use both 8’s. That means 24 has to use 4 and 6. Thus 24 and 42 will use both 6’s, so 30 will be 10×3.
  4. We know one of the 40’s is 4×10, but we don’t know which one. Nevertheless, we know that its 4 will be in the first column because its 10 cannot be. Since 24 must use 4 and 6, its 4 must be in the top row above the 24.

That was pretty complicated, so here’s where all the factors go, too. 🙂

 

 

Category:

Level 6 Puzzle, Mathematics

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827 and Level 5

827 is one of the prime numbers in the fourth prime decade, (821, 823, 827, 829).

827 = 103 + 107+ 109 + 113+ 127 + 131 + 137, that’s the sum of 7 consecutive prime numbers.

Here’s today’s puzzle:

Print the puzzles or type the solution on this excel file: 10-factors-822-828

  • 827 is a prime number.
  • Prime factorization: 827 is prime.
  • The exponent of prime number 827 is 1. Adding 1 to that exponent we get (1 + 1) = 2. Therefore 827 has exactly 2 factors.
  • Factors of 827: 1, 827
  • Factor pairs: 827 = 1 × 827
  • 827 has no square factors that allow its square root to be simplified. √827 ≈ 28.7576

How do we know that 827 is a prime number? If 827 were not a prime number, then it would be divisible by at least one prime number less than or equal to √827 ≈ 28.8. Since 827 cannot be divided evenly by 2, 3, 5, 7, 11, 13, 17, 19, or 23, we know that 827 is a prime number.

Category:

Level 5 Puzzle

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