factor

261 and Level 2

/ ivasallay / Leave a comment

  • 261 is a composite number.
  • Prime factorization: 261 = 3 x 3 x 29
  • 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 261 has 6 factors.
  • Factors of 261: 1, 3, 9, 29, 87, 261
  • Factor pairs: 261 = 1 x 261, 3 x 87, 9 x 29
  • Taking the factor pair with the largest square number factor, we get √261 = (√9)(√29) = 3√29 ≈ 16.155

This puzzle will look like a multiplication table after you find the factors and then fill in the rest of the table.

2014-41 Level 2

Print the puzzles or type the factors on this excel file: 10 Factors 2014-10-13

2014-41 Level 2 Factors

260 Some Thoughts on Those Four Consecutive Numbers

/ ivasallay / 3 Comments

  • 260 is a composite number.
  • 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.
  • Factors of 260: 1, 2, 4, 5, 10, 13, 20, 26, 52, 65, 130, 260
  • 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.

4 numbers; 6 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:

6 factors

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.

259 and Level 1

/ ivasallay / Leave a comment

  • 259 is a composite number.
  • Prime factorization: 259 = 7 x 37
  • The exponents in the prime factorization are 1 and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1) = 2 x 2 = 4. Therefore 259 has 4 factors.
  • Factors of 259: 1, 7, 37, 259
  • Factor pairs: 259 = 1 x 259 or 7 x 37
  • 259 has no square factors that allow its square root to be simplified. √259 ≈ 16.093

Here’s a level 1 Find the Factors puzzle:

2014-41 Level 1

Print the puzzles or type the factors on this excel file: 10 Factors 2014-10-13

2014-41 Level 1 Factors

258 and Level 6

/ ivasallay / Leave a comment

  • 258 is a composite number.
  • Prime factorization: 258 = 2 x 3 x 43
  • 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 258 has 8 factors.
  • Factors of 258: 1, 2, 3, 6, 43, 86, 129, 258
  • Factor pairs: 258 = 1 x 258, 2 x 129, 3 x 86, or 6 x 43
  • 258 has no square factors that allow its square root to be simplified. √258 ≈ 16.062

Level 6 puzzles aren’t easy. Can you solve this one?

2014-40 Level 6

Print the puzzles or type the factors on this excel file: 12 Factors 2014-10-06

2014-40 Level 6 Logic

257 and Level 5

/ ivasallay / Leave a comment

  • 257 is a prime number.
  • Prime factorization: 257 is prime.
  • The exponent of prime number 257 is 1. Adding 1 to that exponent we get (1 + 1) = 2. Therefore 257 has exactly 2 factors.
  • Factors of 257: 1, 257
  • Factor pairs: 257 = 1 x 257
  • 257 has no square factors that allow its square root to be simplified. √257 ≈ 16.031

How do we know that 257 is a prime number? If 257 were not a prime number, then it would be divisible by at least one prime number less than or equal to √257 ≈ 16.031. Since 257 cannot be divided evenly by 2, 3, 5, 7, 11, or 13, we know that 257 is a prime number.

257 is 2⁸ + 1 and is the second largest known Fermat Prime.  Click that link to see how unusual and cool Fermat Primes are!

Can you find the factors of the clues in this Level 5 puzzle?

2014-40 Level 5

Print the puzzles or type the factors on this excel file: 12 Factors 2014-10-06

2014-40 Level 5 Logic

256 and Level 4

/ ivasallay / Leave a comment

  • 256 is a composite number.
  • Prime factorization: 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2, which can be written 256 = 2⁸
  • The exponent in the prime factorization is 8. Adding one we get (8 + 1) = 9. Therefore 256 has 9 factors.
  • Factors of 256: 1, 2, 4, 8, 16, 32, 64, 128, 256
  • Factor pairs: 256 = 1 x 256, 2 x 128, 4 x 64, 8 x 32, or 16 x 16
  • 256 is a perfect square. √256 = 16.

((2²)²)² = 256

1 + 3 + 5 + 7 + 9 + 11 + … + 29 + 31 = 256; that’s the sum of the first 16 odd numbers.

Because it is a power of 2, it is impossible to write 256 as the sum of consecutive whole numbers.

ALL of the factors of 256 are powers of 2: 2⁰=1, 2¹=2, 2²=4, 2³=8, 2⁴=16, 2⁵=32, 2⁶=64, 2⁷=128, 2⁸=256

NONE of the clues in this level 4 puzzle are powers of 2, but can you still solve it and figure out where the powers of 2 belong in the puzzle?

2014-40 Level 4

Print the puzzles or type the factors on this excel file: 12 Factors 2014-10-06

2014-40 Level 4 Logic

255 and Level 3

/ ivasallay / Leave a comment

  • 255 is a composite number.
  • Prime factorization: 255 = 3 x 5 x 17
  • 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 255 has 8 factors.
  • Factors of 255: 1, 3, 5, 15, 17, 51, 85, 255
  • Factor pairs: 255 = 1 x 255, 3 x 85, 5 x 51, or 15 x 17
  • 255 has no square factors that allow its square root to be simplified. √255 ≈ 15.969

Here’s a factoring puzzle for you to solve:

2014-40 Level 3

Print the puzzles or type the factors on this excel file: 12 Factors 2014-10-06

A Logical Approach to FIND THE FACTORS: 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.

2014-40 Level 3 Factors

254 and Level 2

/ ivasallay / Leave a comment

  • 254 is a composite number.
  • Prime factorization: 254 = 2 x 127
  • The exponents in the prime factorization are 1 and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1) = 2 x 2 = 4. Therefore 254 has 4 factors.
  • Factors of 254: 1, 2, 127, 254
  • Factor pairs: 254 = 1 x 254 or 2 x 127
  • 254 has no square factors that allow its square root to be simplified. √254 ≈ 15.937

2014-40 Level 2

Print the puzzles or type the factors on this excel file: 12 Factors 2014-10-06

2014-40 Level 2 Factors

253 and Level 1

/ ivasallay / Leave a comment

  • 253 is a composite number.
  • Prime factorization: 253 = 11 x 23
  • The exponents in the prime factorization are 1 and 1. Adding one to each and multiplying we get (1 + 1)(1 + 1) = 2 x 2 = 4. Therefore 253 has 4 factors.
  • Factors of 253: 1, 11, 23, 253
  • Factor pairs: 253 = 1 x 253 or 11 x 23
  • 253 has no square factors that allow its square root to be simplified. √253 ≈ 15.906

2014-40 Level 1

Print the puzzles or type the factors on this excel file: 12 Factors 2014-10-06

2014-40 Level 1 Factors

252 How likely can this square root be simplified?

/ ivasallay / 2 Comments

  • 252 is a composite number.
  • 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.
  • Factors of 252: 1, 2, 3, 4, 6, 7, 9, 12, 14, 18, 21, 28, 36, 42, 63, 84, 126, 252
  • 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.)

 

1 third

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 - 2 thirds

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.

9 twenty-fifths

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:

nearing 40%

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%.