Talk:241 (number)

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Digits to show that 241 is not a perfect square[edit]

Any way to show that 241 is not a perfect square using digital rules??

We know that:

Final digit[edit]

All squares end in 0, 1, 4, 5, 6, or 9. This rules out both 2 and 3. The smallest number that cannot be ruled out is 5.

Final 2 digits[edit]

Using E for even and Q for odd (an O cannot be told from a 0,) squares must end in 00, E1, E4, 25, Q6, or E9. This rules out 5, which ends in 05. The smallest number that cannot be ruled out as a perfect square using this rule is 21.

Final 3 digits[edit]

How about 3-digit endings?? Squares that end in 00 must end in 000, 100, 400, 500, 600, or 900. For 5, there's 025, 225, and 625. No 3-digit endings ending in 4 or 6 besides those that have already been ruled out by 2-digit endings can be ruled out. For 1 and 9, the rule is that the last 2 digits before the final digit must be a multiple of 4. This rules out 21, becuase the 3-digit endings allowed are 121, 321, 521, 721, and 921. The smallest number that cannot be ruled out using the rules above is now 24.

Final 4+ digits[edit]

A theorem used for finding groups of 4 or more digits squares cannot end in is:

If n can be ruled out as a perfect square by its final d digits, then 4n and 25n can be ruled out by their final n+2 digits.

This can rule out 24. The 4-digit endings for a square that ends in 024 are 1024, 3024, 5024, 7024, and 9024. The smallest number that cannot be ruled out is 41.

Digital root[edit]

Now, there don't appear to be any additional rules for groups of final digits that can be used to rule out squares ending in 1 or 9. Thus showing 41 is not a perfect square is like showing 10 is not a multiple of 3 in that you can't use groups of final digits. You can, however, use the digital root. The digital root of a square must be 1, 4, 7, or 9, and the digital root of 41 is 5. All numbers that are not perfect squares from 2 to 240 can be ruled out using one of these methods. However, 241 still cannot be ruled out. Any additional digit-based rule anyone has?? Georgia guy 20:46, 24 February 2006 (UTC)

Is this going to rule out 241 as a perfect square??[edit]

It says at Divisor that you can decide a number's divisiblity by 7 by looking at the last digit and adding it to 3 times the number made out of all the other digits. Thus, 72 + 1 = 73, then 21 + 3 = 24, then 6 + 4 = 10, then 3 + 0 = 3. This results in 3 for 241. Any number the final digit has to be for a perfect square?? Georgia guy 22:31, 24 February 2006 (UTC)

The answer appears to be yes. The final digit for 241 is 3, but the final digit for a perfect square has to be 1, 2, 4, 7, 8, or 9. It finally worked! Georgia guy 22:33, 24 February 2006 (UTC)


I believe we should use the same format for the infobox for all prime numbers, please comment if you don't feel this should be the case. (talk) 04:35, 19 May 2013 (UTC)