Can you answer those questions of the day (Not easy-but-possible questions)?

Question

Here are challenging questions that I can answer:

Pythagoras: There are infinitely many integer solutions (a,b,c) to the equation associated with Pythagoras' Theorem: c² = a² + b²

Consider the special case for this equation given by (m + n)² = (m + 1)² + (n + 1)². Determine all integer values for m and n that fit the equation.

Odometer Palindromes: Suppose an automobile's odometer can represent mileage up to 999,999, and that leading 0s are not shown. For example, 2511 miles would not be shown as 002511.

How many mileage readings from 1 to 999,999 are palindromes?

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Pitágoras: Hay una infinidad de soluciones muchos enteros (a, b, c) a la ecuación asociada con el teorema de Pitágoras: c ² = a ² + b ²

Consideremos el caso especial de esta ecuación por (m + n) ² = (m + 1) ² + (n + 1) ². Determinar todos los valores enteros m y n para que se ajusten a la ecuación.

Palindromes Millaje: Supongamos que un automóvil puede representar kilometraje del odómetro de hasta 999.999, y que 0s líderes no se muestran. Por ejemplo, 2511 millas no se muestra como 002511.

Como las lecturas de muchos kilómetros de 1 a 999.999 son palíndromos?

Pythagore: Il existe une infinité de nombreuses solutions entières (a, b, c) à l'équation associée avec le théorème de Pythagore: C ² = a ² + B ²

Prenons le cas particulier de cette équation donnée par (m + n) ² = (m + 1) ² + (n + 1) ². Déterminer toutes les valeurs entières de m et n qui correspondent à l'équation.

Palindromes Odomètre: Supposons un compteur kilométrique automobile peut représenter jusqu'à kilométrage à 999.999, et que 0s leader sont pas présentés. Par exemple, 2511 miles ne serait pas démontré que 002511.

Comment kilométrage de nombreuses lectures de 1 à 999.999 sont des palindromes?

畢達哥拉斯：有無窮多整數解（甲，乙，丙）方程與畢達哥拉斯定理：ç ² = 1 ² + b ²

特殊情況考慮對這個方程給出（m + n條）² =（米+ 1）2 +（n +1個）2。確定所有的整數值為 m和n適合公式。

里程表回文：假設汽車的里程表可以代表里程達 999,999，和領導 0的不顯示。例如，二千五百十一英里不會顯示為 002511。

多少里程讀數從 1到999999頃回文？

피타고라스 :이 방정식은 피타고라스, 피타고라스 '정리와 관련된 C)에 무한히 많은 정수 솔루션 (A와 B, : C ² = ² + B를 ²있습니다

이 방정식 (㎡ + n)이 ² = (난 + 1) + 회 (N + 1) ² ²에 의해 주어진 특별한 경우를 고려합니다. M 및 n은 방정식을 맞는 대한 모든 정수 값을 확인합니다.

주행 Palindromes : 최대 999,999 위해 마일리지를 나타내는 수 있으며, 자동차의 주행 거리계 가정은 최고의 좋아한다는 표시되지 않습니다. 예를 들어, 2천5백11마일 002,511으로 표시되지 않을 것이다.

999999에 1부터 얼마나 많은 마일리지를 판독 palindromes입니까?

Answer 1

(m + n)² = (m + 1)² + (n + 1)²

m² + 2mn + n² = m² + 2m + 1 + n² + 2n + 1

m² + 2mn + n² = m² + n² + 2(m + n + 1)

2mn = 2(m + n + 1)

mn = m + n + 1

mn - m = n + 1

m(n - 1) = n + 1

m = (n + 1)/(n - 1) assuming n =/= 1

n = (m + 1)/(m - 1) assuming m =/= 1



Since m = (n + 1)/(n - 1) and n = (m + 1)/(m - 1), neither m nor n can be greater than 3 and still make an integer pair:

m > 3

m + 1 > 4

m - 1 > 2

(m + 1)/(m - 1) > 2



Now assume n = (m + 1)/(m - 1) is greater than 3:

(m + 1)/(m - 1) > 3

m + 1 > 3m - 3

2m < 4

m < 2

Therefore, if n > 3, m < 2.



So for all m > 3, 2 < n < 3, which means that n is not an integer. The same process can be used to conclude that n must be less than 3 for m to be an integer, so m < 3 and n < 3.





(0,-1), (-1,0), (3,2), and (2,3) are all integer values for (m,n) such that (m + n)² = (m + 1)² + (n + 1)².







Palindromes: For the case with x digits shown, let (a_1, a_2, ..., a_x) be the set of digits from greatest to least.



You can organize this into pairs which must be equal for the number to be a palindrome: a_1 = a_x, a_2 = a_(x-1), etc.



For all odd numbers of digits shown, because the middle number can be any digit from 0 to 9 and not change the numbers status as a palindrome, the number of sets of digits that form palindromes equals the number of digits for x - 1 digits that form palindromes multiplied by 10.



So, for x = 2n | n is a natural number, the number of possible palindromes is the number of possible values for each pair of digits, which, because leading 0's are ignored, is 9*10^(n-1).



For x = 2n + 1 | n is a natural number, the number of possible palindromes is 9*10^n.



Now, summing the numbers, you get:



9*10^0 + 9*10^0 + 9*10^1 + 9*10^1 + 9*10^2 + 9*10^2

9 + 9 + 90 + 90 + 900 + 900

18 + 180 + 1800

1800 + 198

1998



There are 1998 possible palindromes.







As a result of this, the number of palindromes for any x digits where x >= 1 will be sum(n=0,x-1)[9*10^floor(n/2)].

Answer 2

The palindromes can be n = 1, 2, 3, 4, 5 or 6 digits.



n = 1: all 9 can be considered palindromes



n = 2: aa

There are 9 ways of choosing the first number, and 1 way of choosing the second.

Number = 9



n = 3: aba

There are 9 ways of choosing hte first, 10 ways for the second, 1 way for the last

Number = 90



n = 4: abba

Number = 90



n = 5: abcba

Number = 9*10*10*1*1 = 900



n = 6: abccba

Number = 900



Total = (9 + 90 + 900)*2 = 1998



I guess the pattern would continue, so that if the odometer had 2n digits, the total number of palindromes would be

2*(10^n - 1)



****

For the pythagoras's theorem question, building on what gago... has already done.

m = (n+1) / (n-1)

If you're only interested in integer solutions, then there cannot exist any for n>3 since for n > 3, n-1 is more than half of n+1, and thus cannot be divided.

The only non-trivial pair is (2,3).

Answer 3

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