Puzzles can be a great resource in teaching mathematics. I’ve made crossword puzzles to review vocabulary words before, but none of them looked as inviting to complete as the one Resourceaholic recommended in her 7 July 2015 post on end of term resources. What a difference a little clip art makes to a well-constructed crossword puzzle! The post also includes links to several other amazing mathematics-related puzzles including a polygon word search and a number sequence number search puzzle. Check it out!
A Logical Approach to solve a FIND THE FACTORS puzzle: Find the column or row with two clues and find their common factor. Write the corresponding factors in the factor column (1st column) and factor row (top row). Because this is a level three puzzle, you have now written a factor at the top of the factor column. Continue to work from the top of the factor column to the bottom, finding factors and filling in the factor column and the factor row one cell at a time as you go.
Factors of 546:
546 is a composite number.
Prime factorization: 546 = 2 x 3 x 7 x 13
The exponents in the prime factorization are 1, 1, 1, and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1)(1 + 1)(1 + 1) = 2 x 2 x 2 x 2 = 16. Therefore 546 has exactly 16 factors.
Factor pairs: 400 = 1 x 400, 2 x 200, 4 x 100, 5 x 80, 8 x 50, 10 x 40, 16 x 25, or 20 x 20
400 is a perfect square. √400 = 20
A few months ago I made a chart showing the number of factors for the first 300 counting numbers. Since this is my 400th post, I’d like to include a chart showing the number of factors for all the numbers from 301 to 400. I’m also interested in consecutive numbers with the same number of factors and whether or not the square root of a number can be reduced. The red numbers have square roots that can be reduced.
The longest streak of consecutive numbers with the same number of factors is only three. There are three sets of three consecutive numbers on this chart. (Between 200 and 300 there was a streak of four consecutive numbers with six factors each.)
How do the number of factors of these 100 numbers stack up against the previous 300? The following chart shows the number of integers with a specific number of factors and how many of those integers have reducible square roots:
39.5% or slightly less than 40% of the numbers up to 400 have reducible square roots.
Most of these numbers have 2, 4, or 8 factors. Numbers with two factors are prime numbers. Almost all numbers with four factors are the product of two different prime numbers, and nearly two-thirds of the numbers with eight factors are the product of three different prime numbers.
There isn’t much change between the percentages of reducible square roots from one list to the next.
The exponent of prime number 281 is 1. Adding 1 to that exponent we get (1 + 1) = 2. Therefore 281 has exactly 2 factors.
Factors of 281: 1, 281
Factor pairs: 281 = 1 x 281
281 has no square factors that allow its square root to be simplified. √281 ≈ 16.763
How do we know that 281 is a prime number? If 281 were not a prime number, then it would be divisible by at least one prime number less than or equal to √281 ≈ 16.763. Since 281 cannot be divided evenly by 2, 3, 5, 7, 11, or 13, we know that 281 is a prime number.
So far I have posted about one set of four and two sets of five consecutive reducible square roots. Sets of three consecutive reducible square roots are fairly common so I’ve ignored most of them. These consecutive square roots couldn’t be ignored:
The final square root features today’s prime number 281. Here are the prime factorizations and number of factors of each of these numbers:
Six is a popular number when counting the number of factors.
The exponent of prime number 269 is 1. Adding 1 to that exponent we get (1 + 1) = 2. Therefore 269 has exactly 2 factors.
Factors of 269: 1, 269
Factor pairs: 269 = 1 x 269
269 has no square factors that allow its square root to be simplified. √269 ≈ 16.401
How do we know that 269 is a prime number? If 269 were not a prime number, then it would be divisible by at least one prime number less than or equal to √269 ≈ 16.401. Since 269 cannot be divided evenly by 2, 3, 5, 7, 11, or 13, we know that 269 is a prime number.
As I have previously written, 844, 845, 846, 847, and 848 are the smallest FIVE consecutive numbers whose square roots can be simplified. Here are the second smallest FIVE with the same property.
The first number in the second set, 1680, equals 2 x 840 which is very close to the first number in the first set. Will strings of five consecutive numbers with reducible square roots occur about once every 850 numbers?
We can find the number of factors for these numbers by examining their prime factorizations.
The number of factors for each of the integers in this second set ranges from 3 to 40. Only two of the integers have the same number of factors. Finding another string of four or more numbers that have reducible square roots as well as the same number of factors may be difficult.
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 x 2 x 2 = 8. Therefore 266 has 8 factors.
Factors of 266: 1, 2, 7, 14, 19, 38, 133, 266
Factor pairs: 265 = 1 x 266, 2 x 133, 7 x 38, or 14 x 19
266 has no square factors that allow its square root to be simplified. √266 ≈ 16.3095
242, 243, 244, and 245 are the smallest four consecutive numbers that have the same number of factors. Each of them has exactly six factors, and as a result each one of their square roots can be simplified. Is it possible to have a longer string of consecutive numbers with exactly six factors? I don’t know yet. It seems reasonable that it could happen, so I am on the lookout for five, six, or seven consecutive numbers that have exactly six factors.
I won’t bother looking for a string of eight or more numbers with six factors. I already know that would be impossible. Here’s an example illustrating why:
846 ruins the run because it has an additional prime factor that doubles the number of factors that 846 has in all. Further down the number line that problem might be overcome in a different set of consecutive numbers.
However, the problem with 848 is a recurring problem that will never be overcome. 848 is divisible by 8, as is every eighth number. The prime factorization of numbers that are divisible by eight must contain a power of two that is greater than or equal to three. Its number of factors calculation would have to be at least (3 + 1)(1 + 1) = 4 x 2 = 8. (The ONLY number divisible by 8 that has exactly 6 factors is 32.)
Even though the numbers from 844 to 848 don’t have the same number of factors, they still have a distinction. They are the smallest five consecutive numbers whose square roots can be simplified!
Factor pairs: 260 = 1 x 260, 2 x 130, 4 x 65, 5 x 52, 10 x 26, or 13 x 20
Taking the factor pair with the largest square number factor, we get √260 = (√4)(√65) = 2√65 ≈ 16.125
Recently I wrote about the smallest four-consecutive-numbers whose square roots could all be simplified. The same numbers were also the smallest four consecutive numbers to have the same number of factors.
Each of those numbers had 6 factors, and guess what, ANY number with exactly 6 factors can have its square root simplified. The prime factorization of ANY number with exactly 6 factors can be expressed in one of the three following ways:
Since numbers with six factors always have a prime factor raised to a power greater than one, they can always have their square roots simplified. The fact that those four consecutive numbers have the same number of factors makes them extraordinary; that they all can have their square roots simplified is merely the natural consequence of that extraordinary fact.