Nei quadrati dei numeri dispari la cifra delle decine non è mai dispari?

Question

Sapreste spiegare se è un colpo d;=)B< fortuna o qualcosa che ha un motivo specifico?
(E poi, succede anche con altre basi di numerazione, e quali?)

Excuse me, I was sure to be in the italian Yahoo!Answers. This is the translation:

"In all squares od odd numbers is the "second last" digit (how to say, the second from the right) never odd? Is it by chance or there is a good reason for it? And in other bases is it the same? In which ones?"

*In all squares of odd numbers...

Answer 1

First off, I apologize for not speaking Italian.

Any odd number can be expressed as 10a + 2b + 1,
- where "b" is 0, 1, 2, 3, or 4
- where "a" is any whole number

(10a + 2b + 1)^2
= 100a^2 + 20ab + 10a + 20ab + 4b^2 + 2b + 10a + 2b + 1
= 100a^2 + 40ab + 20a + 4b^2 + 4b + 1

100a^2 + 40ab + 20a always has an even second-to-last digit and a zero last digit, therefore 100a^2 + 40ab + 20a + 4b^2 + 4b + 1 has an even second-to-last digit if 4b^2 + 4b + 1 has an even second-to-last digit.

If b = 0 then 4b^2 + 4b + 1 = 01
If b = 1 then 4b^2 + 4b + 1 = 09
If b = 2 then 4b^2 + 4b + 1 = 25
If b = 3 then 4b^2 + 4b + 1 = 49
If b = 4 then 4b^2 + 4b + 1 = 81

Since for all possible values of "b" there is an even second-to-last digit, for all odd numbers, the square has an even second-to-last digit in base 10.

Thus the answer to your first question is "yes".

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The answer to your second question can be found be contemplating my demonstration of the answer to your first question.

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Any odd number in an even base can be expressed as ca + 2b + 1,
- where "c" is even
- where "b" is any whole number 0 thru c/2-1
- where "a" is any whole number

(ca + 2b + 1)^2
= c^2*a^2 + 2cab + ca + 2cab + 4b^2 + 2b + ca + 2b + 1
= c^2*a^2 + 4cab + 2ca + 4b^2 + 4b + 1

c^2*a^2 + 4cab + 2ca always has an even second-to-last digit and a zero last digit, therefore c^2*a^2 + 4cab + 2ca + 4b^2 + 4b + 1 has an even second-to-last digit if 4b^2 + 4b + 1 has an even second-to-last digit.

If c = 2 then
- If b = 0 then 4b^2 + 4b + 1 = 01 base 2
If c = 4 then
- If b = 0 then 4b^2 + 4b + 1 = 01 base 4
- If b = 1 then 4b^2 + 4b + 1 = 21 base 4
If c = 6 then
- If b = 0 then 4b^2 + 4b + 1 = 01 base 6
- If b = 1 then 4b^2 + 4b + 1 = 13 base 6
- If b = 2 then 4b^2 + 4b + 1 = 41 base 6
If c = 8 then
- If b = 0 then 4b^2 + 4b + 1 = 01 base 8
- If b = 1 then 4b^2 + 4b + 1 = 11 base 8
- If b = 2 then 4b^2 + 4b + 1 = 31 base 8
- If b = 3 then 4b^2 + 4b + 1 = 61 base 8
If c=12 then
- If b = 0 then 4b^2 + 4b + 1 = 01 base 12
- If b = 1 then 4b^2 + 4b + 1 = 09 base 12
- If b = 2 then 4b^2 + 4b + 1 = 21 base 12
- If b = 3 then 4b^2 + 4b + 1 = 41 base 12
- If b = 4 then 4b^2 + 4b + 1 = 69 base 12
- If b = 5 then 4b^2 + 4b + 1 = A1 base 12
And so on...

Since for all possible values of "b" in bases 2, 4, and 12 there is an even second-to-last digit, for all odd numbers in these bases, the square has an even second-to-last digit in these bases.

Thus, the answer to your third question is "It is the same in some, but not all".

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From the answers to your first and third questions we see that it is the same for bases 2, 4, 10, and 12, and not for bases 6 and 8.

In base 3, 21^2=1211
In base 5, 3^2=14
In base 7, 3^2=12
In base 9, 3^2=10
In base 11, 5^2=23

Therefore, it is not the same in bases 3, 5, 7, 9, and 11.

Consider that ( x + 2 )^2 < 2x^2 for x=5

(x + 2)^2 < 2x^2
(x + 2)^2 + 2x < 2x^2 + 3x + 2
(x + 2)^2 + 2x + 5 < 2x^2 + 3x + 2 + x
x^2 + 4x + 4 + 2x + 5 < 2x^2 + 4x + 2
x^2 + 6x + 9 < 2(x^2 + 2x + 1)
(x + 3)^2 < 2(x + 1)^2
[(x + 1) + 2]^2 < 2(x + 1)^2

Thus, for all odd whole numbers x equals 5 or greater, the square of the next odd number is less than half the square of the current odd number. This means that if there is no second-to-last digit in a square of an odd number, then the second-to-last digit of the square of the next odd number can be no larger than 1 in any base greater than 12.

Therefore, bases 2, 4, 10, and 12 are the only bases for which the squares of odd numbers always have an even second-to-last digit.

Answer 2

1^2=01
3^2=09
5^2=25
7^2=49
9^2=81
11^2=121
13^2=169
15^2=225
17^2=289
19^2=361
so whenever you find the square of an odd no the digit in the 10s place will always be an even number

Answer 3

Translate in English First.

Answer 4

please write your question in english when you ask it in World Wide Web for World Wide Web expans World Wide you see and english is the World Wide language.