245 – The Last of Four Consecutive Numbers

  • 245 is a composite number.
  • 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

Square roots 242 - 245

 

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.

3 thoughts on “245 – The Last of Four Consecutive Numbers

  1. Wow! I’ve never seen this before. I don’t know enough number theory to know if this is a sensible question or not, but are longer sequences possible?

  2. I’m surprised too. That’s a neat string of numbers.

    I would imagine longer sequences should be possible, but I don’t know that they are and I’m too weak on number theory to do much one way or another with the question.

  3. I wish I knew more number theory, too. It seems reasonable that longer sequences would be possible.

    Here’s a string of five consecutive numbers whose square roots can all be simplified, but the numbers in this string do not have the same number of factors:
    √844 = 2√211
    √845 = 13√5
    √846 = 3√94
    √847 = 11√7
    √848 = 4√53

    http://www.mathwarehouse.com/arithmetic/square-root-calculator.php
    helped me find that string of five consecutive numbers in about an hour.

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