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.
Prime factorization: 279 = 3 x 3 x 31, which can be written (3^2) x 31
The exponents in the prime factorization are 2 and 1. Adding one to each and multiplying we get (2 + 1)(1 + 1) = 3 x 2 = 6. Therefore 279 has 6 factors.
Factors of 279: 1, 3, 9, 31, 93, 279
Factor pairs: 279 = 1 x 279, 3 x 93, or 9 x 31
Taking the factor pair with the largest square number factor, we get √279 = (√9)(√31) = 3√31 ≈ 16.703
Do you know who Superman is? I’ve asked many children in their first year of school that question. They know who he is, and they can tell me lots of things about him.
Superman’s secret identity is Clark Kent, but did you know that the numbers 1, 2, 3, 4, and 5 also have secret identities? Before today, hardly anybody has known what their secret identities are, but I will reveal them to you now!
Let’s start with five. Its secret identity is zero. Five can change into a zero by smoothing its top and curving it down. When you count by fives you can see it changing into zero and back into five again as fast as Superman can change into Clark Kent. 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100, 105, and so forth.
By using its secret identity , we can easily add or subtract five from a number ending in five or zero! 5 + 5 = 10, 10 + 5 = 15, and 15 – 5 = 10; 10 – 5 = 5.
Have you ever noticed how easy it is to turn a 4 into a 9?
That was as easy as Superman becoming Clark Kent!
When you count by 5’s starting at 4, these numbers turn into each other over and over again: 4, 9, 14, 19, 24, 29, and so forth. When you add 5 to either 4 or 9, it will turn into the other: 4 + 5 = 9, 9 + 5 = 14. The same thing happens when you subtract five: 14 – 5 = 9; 9 – 5 = 4. There isn’t any reason to count up or down; just remember how easy it is to turn a 4 into a 9 and vice versa.
A 3 can easily turn into an 8:
Counting by 5’s starting at 3 we get: 3, 8, 13, 18, 23, 28, and so forth.
If you cut off the tail at the bottom of a two, it can easily turn into its secret identity, seven:
Counting by 5’s starting at 2, we get: 2, 7, 12, 17, 22, 27, and so forth.
2 + 5 = 7; 7 + 5 = 12 and 12 – 5 = 7; 7 – 5 = 2.
Here’s how to discover the secret identity for the number 1. Take a strip of paper that looks like the number 1 and follow these SAFE directions for curling ribbon (or paper) with scissors. Curl the bottom half of the number 1 as well as the top fourth of that number 1 with scissors to make that number 1 look just like the number 6.
Counting by 5’s starting at 1, we get: 1, 6, 11, 16, 21, 26, and so forth.
1 + 5 = 6; 6 + 5 = 11 and 11 – 5 = 6; 6 – 5 = 1.
When Superman puts on glasses and a suit, we see his secret identity. Did you notice that odd digits have even secret identities and even digits have odd secret identities? Once a child memorizes these digits and their secret identities, he or she will be SUPER at adding or subtracting 5 and will never need to count up or down to get the answer.
Using these number transformations can help children memorize other addition facts. Once they know how to Add 1, 2, 3 and 4 and how to add 5, they can use that information to add 6, 7, 8, 9, or 10 by breaking up those numbers into a smaller number plus five. For example:
6 + 7 = 6 + (2 + 5) = (6 + 2) + 5 = 8 + 5 = 13.
7 + 9 = 7 + (4 + 5) = (7 + 4) + 5 = 11 + 5 = 16.
Learning facts for 6, 7, 8, 9, and 10 this way isn’t any more complicated than the common core approach[6 + 7 = 6 + (6 + 1) = (6 + 6) + 1 = 12 + 1 = 13], but is perhaps not as easy as just memorizing those facts and using flash cards or other means to drill them permanently into the brain. Secret identities and even “Flash” cards can be super fun.
Prime factorization: 275 = 5 x 5 x 11, which can be written (5^2) x 11
The exponents in the prime factorization are 2 and 1. Adding one to each and multiplying we get (2 + 1)(1 + 1) = 3 x 2 = 6. Therefore 275 has 6 factors.
Factors of 275: 1, 5, 11, 25, 55, 275
Factor pairs: 275 = 1 x 275, 5 x 55, or 11 x 25
Taking the factor pair with the largest square number factor, we get √275 = (√11)(√25) = 5√11 ≈ 16.583
Are six-year-olds too young to learn about odd and even numbers?
Paula Beardell Krieg gave me permission to use the pictures of this flexible number line she designed in this post:
I recently read a post at mathfour.com that discussed the “basic” concept of odd and even numbers and children’s ability to understand the difference. The article made me very curious so I talked briefly to 45 first grade students about even and odd numbers. What did I find out?
Almost all of them had been introduced to the concept in kindergarten and knew that 1, 3, 5, 7, 9 are odd numbers while 2, 4, 6, 8, 10 are even.
A few accelerated learning students were able to explain to me that the one’s digit of a number determines if the number is even or odd,
But most of these first graders did not understand that fact because about a third of the students thought that 32 is odd!
One little girl explained to me how odd and even numbers alternate. She said, “If 99 is even, then 100 will be odd.” She remembered that concept but didn’t understand it well enough to apply it to the example she gave!
Even though odd and even numbers may be a difficult concept to learn, teach the concept and use it anyway. In fact, talk about it to preschoolers while you put on their socks, shoes, or mittens. One,_Two,_Buckle_My_Shoe.
Children learn to recite numbers in order before they learn how to count, and that helps them learn how to count and later how to add or subtract 1 from a number. I have tutored bewildered looking students who weren’t sure what to do with 8 + 1 = until I told them that 8 + 1 = means “what number comes right after 8 when you count?” Likewise, 8 – 1 = means “what number comes right before 8 when you count?” After hearing those questions, these students immediately knew the answer, and they didn’t count to find it.
Children who can quickly recite the odd numbers to 11 and the even numbers to 10 will have an easier time adding or subtracting two from a number. When they see 3 + 2 =, they can remember that 3 is odd and then ask themselves what odd number comes after 3. Likewise when they see 8 – 2, they can remember that 8 is even and recall that 6 is the even number right before 8.
The way I remember it, I was in second grade when I first was told that an even number plus an even number is even, an odd number plus an odd number is even, while an even number plus an odd number is odd. Any student learning to add or subtract would benefit from that tip.
Adding 3 to an odd number gives an even number, in fact, it’s the second even number after the original number. Adding 3 to an even number gives an odd number which is the second odd number after the original number. Subtracting 3 has the same rule, but substitute the word “before” for the word “after.”
Adding 4 to an odd number gives the second odd number after it while adding 4 to an even number gives the second even number after it. Subtracting 4 has a similar rule.
Adding 3 or 4 will mean additionally memorizing that 12 and 14 are even and 13 is odd, but that will be all a first grader needs to know about odd and even numbers. Later these two categories of numbers will be useful throughout their lives for many, many reasons.
What are some ways to help children to memorize odd and even numbers? Paula Beardell Krieg has designed the most captivating number line in the world.
The transformation can be done by a child or an adult. This number line that is made with envelopes is pretty enough to hang on a classroom wall, but it can fold up like a book, or be played with and changed so that real learning can take place. Paula Beardell Krieg shows several uses of it in her post, the-flux-capacity-of-an-artful-number-line, and promises to give directions on how to make one soon!
Try these rhymes: 0, 2, 4, 6, 8; Being EVEN is just great! 1, 3, 5, 7, 9; Being ODD is just fine!
Smartfirstgraders.com has several activities and rhymes to help students memorize the odd and even numbers.
Finally, if you clap when you say ODD, you will clap one time. 1 is an odd number.
If you clap when you say EVEN, you will clap two times, 2 is even.
And as mathfour.com pointed out in more detail then I’m showing here: ODD has 3 letters, and 3 is odd.
Also EVEN has 4 letters to help us remember that 4 is even.
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!
The exponent of prime number 263 is 1. Adding 1 to that exponent we get (1 + 1) = 2. Therefore 263 has exactly 2 factors.
Factors of 263: 1, 263
Factor pairs: 263 = 1 x 263
263 has no square factors that allow its square root to be simplified. √263 ≈ 16.217
How do we know that 263 is a prime number? If 263 were not a prime number, then it would be divisible by at least one prime number less than or equal to √263 ≈ 16.217. Since 263 cannot be divided evenly by 2, 3, 5, 7, 11, or 13, we know that 263 is a prime number.
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Joseph Nebus reads hundreds of comics every day as he looks for ones with a mathematical theme. He regularly shares these finds on his blog and writes a short explanation about the mathematics mentioned in the comics. I especially loved his reading-the-comics-october-14-2014 edition.
Why mathematics students should show their work is clearly explained under Jeff Mallet’s Frazz (October 12) comic strip. Joseph Nebus basically gives two reasons to show work. The first reason has probably been stated by teachers thousands of times, but the second is truly an inspiration, and I highly recommend teachers and students alike read it!
How should the work be shown? This You-tube video does a very good job showing how to show work and make that work as readable as possible.
Prime factorization: 260 = 2 x 2 x 5 x 13, which can be written 260 = (2^2) x 5 x 13
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 x 2 x 2 = 12. Therefore 260 has 12 factors.
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.
Prime factorization: 252 = 2 x 2 x 3 x 3 x 7, which can be written 252 = (2^2) x (3^2) x 7
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 x 3 x 2 = 18. Therefore 252 has 18 factors.
Factor pairs: 252 = 1 x 252, 2 x 126, 3 x 84, 4 x 63, 6 x 42, 7 x 36, 9 x 28, 12 x 21, or 14 x 18
Taking the factor pair with the largest square number factor, we get √252 = (√7)(√36) = 6√7 ≈ 15.875
The square root of a whole number can be simplified if it has a square number factor. How likely is that condition met by any random whole number?
4 is 2 x 2 and therefore a square number. 1 out of every four whole numbers (or 25%) is divisible by 4
3^2 = 9. Likewise 1 out of every nine whole numbers is divisible by square number 9 (about 11.1%).
Some numbers, like 252, are divisible by both 4 and 9. (1 out of every 36 numbers are divisible by both 4 and 9.)
Thus 1/3 of all whole numbers are divisible by 4, 9 or both.
That means that 2/3 of the numbers in the set of all whole numbers are NOT divisible by 4, 9 or both. It is often easier to compute the probability of something NOT happening and then subtract that fraction from 1 to determine the probability of something actually happening. The probability a number is NOT divisible by 4 is 3/4 while the probability a whole number is NOT divisible by 9 is 8/9. We get the same result either way.
1/3 of all whole numbers (about 33.3%) are divisible by either 4 or 9! That fact is very cool because it is so easy to tell if a number is divisible by 4 or 9: If the last 2 digits of a number is divisible by 4, the entire number is divisible by 4 and if the sum of the digits of a whole number is divisible by 9, that whole number is divisible by 9.
It is also very easy to tell if a number is divisible by 5 x 5 or 25. If the last two digits of the number are 25, 50, 75 or 00, then it is divisible by 25. Let’s compute how likely it is that the square root of a number can be simplified because that number is divisible by 4, 9, or 25.
Thus 36% of all whole numbers are divisible by 4, 9, or 25 and therefore have square roots that can be simplified! It is not as easy to tell if a number is divisible by 49, 121, 169, or any other number that is the perfect square of a prime number. The percentage of numbers that are divisible by these other perfect squares doesn’t go up much more either. Consider this infinite product subtracted from 1:
When I’ve computed the partial product up to 3480/(59 x 59) and subtracted it from 1, the probability only increased to 39.010%. I used excel to compute the probability of a number being divisible by a square factor up to 1,495,729 (which is 1223^2) and it is only 39.201%. There isn’t much change in the percentage between the 17th prime number (59) and the 200th prime number (1223).
As n gets larger (n^2 -1)/(n^2) gets closer and closer to 1. I conclude that the probability that a random whole number can have its square root simplified is about 40%.
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 246 has 8 factors.
Factors of 246: 1, 2, 3, 6, 41, 82, 123, 246
Factor pairs: 246 = 1 x 246, 2 x 123, 3 x 82, or 6 x 41
246 has no square factors so its square root cannot be simplified. √246 ≈ 15.684
Prime factorization: 245 = 5 x 7 x 7, which can be written 245 = 5 x (7^2)
The exponents in the prime factorization are 1 and 2. Adding one to each and multiplying we get (1 + 1)(2 + 1) = 2 x 3 = 6. Therefore 245 has 6 factors.
Factors of 245: 1, 5, 7, 35, 49, 245
Factor pairs: 245 = 1 x 245, 5 x 49, or 7 x 35
Taking the factor pair with the largest square number factor, we get √245 = (√5)(√49) = 7√5 ≈ 15.652
I was surprised when I noticed that the square roots of these 4 consecutive numbers – 242, 243, 244, and 245 could all be simplified.
The square root of a whole number can only be simplified if that whole number has a square number as one of its factors. All four of these numbers meet that condition, and they are the first four consecutive numbers to do so.
For numbers less than or equal to 240, there are only 3 sets of 3 consecutive square roots that can be simplified.
√48 = 4√3
√49 = 7
√50 = 5√2
√98 = 7√2
√99 = 3√11
√100 = 10
√124 = 2√31
√125 = 5√5
√126 = 3√14
242, 243, 244, and 245 also have another distinction. They each have exactly 6 factors and are the smallest consecutive four numbers to have the same number of factors.
