Номер строки Паскаля

Analyzing Pascal's triangle

I’m thinking through Pascal's triangle to identify the first row with a number divisible by 1210. Start with the prime factors of 1210: 2, 5, and 11. This requires a granular look at binomial coefficients.

Breaking down 1210

I’m working through the prime factors of 1210: 2, 5, and 11^2. For Pascal's triangle, I’m focusing on finding binomial coefficients divisible by these factors, aiming to pinpoint the smallest row.

Analyzing binomial coefficients

I’m curious about binomial coefficients divisible by 11^2 and how Legendre's formula helps identify higher multiplicities, particularly for binomial coefficients in Pascal's triangle, focusing on factors 2, 5, and 11.

Unpacking binomial coefficients

I’m exploring the divisibility of binomial coefficients by 11, specifically in Pascal's triangle. For large n, the largest binomial coefficient might need careful consideration to ensure it meets the divisibility requirement.

Assessing divisibility

I’m looking into smaller binomial coefficients divisible by 11 to identify the smallest row in Pascal's triangle where any C(n, k) is divisible by 121. Progress seems steady but slow.

Analyzing carries

I'm looking at Kummer's theorem to identify binomial coefficients in Pascal's triangle divisible by 11^2, aiming for rows where numbers represent at least two carries in base 11.

Applying Kummer's theorem

I’m piecing together that for a binomial coefficient to have a factor of 11^2, n must have at least two base 11 digits adding up to or exceeding 11.

Verifying smaller n values

Checking if n=22 or n=121 fits Pascal's triangle patterns. For n=121, need k to create two carries in base 11, ensuring divisibility by 11^2.

Crafting base 11 sums

I'm working with Pascal’s triangle coefficients, aiming for at least two carries with k=11 and n=121 in base 11, ensuring the sum is divisible by 11^2.

Piecing together base 11

I'm converting 110 decimal to base 11, aiming for a valid two-digit representation with 'A' for 10, ensuring clarity in Pascal's triangle coefficient calculations.

Adjusting my strategy

I’m working through adding k=11 and n-k=110 in base 11. The carry management aligns with Pascal's triangle and ensures the sum is divisible by 11^2.

Mapping out carries

I'm noting that I need to sum k=11 and n-k=110 in base 11, aiming for two carries, ensuring the sum of binomial coefficients is divisible by 11^2 in Pascal's triangle.

Evaluating base 11 sums

I'm thinking about how to generate two carries with k=11 and n=121, ensuring the sum in base 11 is divisible by 11^2. This method will confirm required divisibility.

Ensuring effective carries

To secure Pascal's triangle sums divisible by 11^2, I confirmed n should be at least 121, emphasizing the need for at least two carries when adding in base 11. progress is steady.

Ensuring strategic carries

To confirm the sum is divisible by 11^2, I systematically consider two-digit or three-digit base 11 numbers for k, ensuring their addition to n in base 11 consistently generates two carries.

Understanding base 11 carries

OK, let me see. To generate two carries in base 11, at least two digit columns of the sum must reach 11 or more. This ensures the sum is divisible by 11^2.

Evaluating carry possibilities

I'm mapping out how generating two carries in base 11 demands large digits in k, especially in the lower places, to achieve the sum "100"_11.

Ensuring accurate calculations

I'm mapping out the impact of carries when adding k and n−k in base 11, ensuring the total is divisible by 11^2. Progression to dividers is evident.

Mapping out carry details

Considering a smaller prime like 3, I’m examining n=9 in base 3 to generate two carries in base 3 addition. This approach clarifies the principle applied to Pascal’s triangle.

Confirming divisibility

OK, let me see: I verified C(9,4) in base 3 has two carries, ensuring divisibility by 3^2. This method's implications extend to base 11 for Pascal's triangle sums.

Crafting carry patterns

I'm focusing on forming k to generate two carries when adding in base 11, ensuring the sum in Pascal's triangle coefficients is divisible by 11^2. Progress is steady and methodical.

Managing carries

I'm figuring out how to generate two carries in base 11 addition, ensuring Pascal's triangle coefficients are divisible by 11^2, with the goal of achieving the sum "100"_11.

Ensuring the correct outcome

By setting k in base 11, I'm ensuring two carries align with Pascal's triangle coefficients, aiming for a sum divisible by 11^2. This approach clarifies the reasoning behind the coefficients.

Crafting the carry

I’m honing Pascal's triangle additions in base 11 to ensure two carries in each column. This technique guarantees coefficients are divisible by 11^2, showcasing a structured approach.

Aligning final digits

I'm confirming the process in base 11 to ensure Pascal's triangle coefficients are divisible by 11^2, with k digits adjusted to align with the final digit constraints.

Confirming factorization

I’m validating that when n=121 and k=120 in base 11, the binomial coefficient C(121,1) equals 121, which is divisible by 11^2. This dish ensures that Pascal's triangle coefficients are accurate.

Ensuring divisibility

I'm making sure Pascal's triangle coefficients at n=121 also include factors of 2 and 5 to ensure divisibility by 1210. This is a crucial step in validating the coefficients.

Confirming divisibility

OK, let me see. Ensuring Pascal's triangle coefficients are divisible by 1210 involves systematically verifying coefficients for factors of 2, 5, and 11^2, particularly focusing on large coefficients like C(121,2).

Confirming divisibility

I’m verifying that no earlier Pascal's triangle row has coefficients divisible by 11^2. For instance, in row 22, no carries lead to divisibility by 11^2.

Evaluating base 11 carries

I'm working through base 11 digit pairs to produce two carries for Pascal's triangle coefficients. Interestingly, no smaller row precisely meets this condition.

Verifying binomial divisibility

I confirmed that the smallest row in Pascal's triangle with a coefficient divisible by 1210 is n=121, thanks to Wolstenholme's theorem and the combinational properties of binomial coefficients.

Piecing together factors

Hmm, I’m checking Wolstenholme's theorem and its implications for prime factors in binomial coefficients. For instance, C(21,10) and C(22,1), to see if 11^2 divides them.

Rechecking factorial factors

I’m recalculating the prime factors in the factorials for C(33,11). Despite the earlier confusion, it turns out there’s no factor of 11. This confirms that larger multiples are needed for higher prime factors.

Mapping out higher powers

I’m sorting through larger Pascal's triangle rows to ensure coefficients are divisible by 11^2. Notably, factors of 2 and 5 are abundant and typically expected.

Ensuring divisibility for larger rows

I’m confirming that for row 121 in Pascal's triangle, the binomial coefficients are divisible by 11^2. This ensures any coefficient you're working with is accurately divisible by 1210.

Confirming divisibility

OK, I’m verifying that row 121 in Pascal's triangle contains binomial coefficients divisible by 11^2. This ensures coefficients are divisible by 1210 by verifying prime factors.

Mapping base 11

OK, let’s see: to ensure two carries in base 11, digit pairs must sum to at least 11. This process validates Pascal's triangle coefficients divisible by 11^2.

Calculating factors

I’ve figured out the factors in 33! and 22! to check divisibility by 11^2 in binomial coefficients. No smaller row has this divisor, ensuring large primes are considered.

Mapping out carries

I'm analyzing Pascal's triangle coefficients to ensure binomial coefficients are divisible by 11^2. This involves ensuring at least two carries in base 11 addition for precise verification.

Ensuring binomial coefficients

I’m focusing on Pascal's triangle coefficients, ensuring they’re divisible by 11^2 by managing two carries in base 11. This method is critical for maintaining accurate divisibility.

Ensuring accurate sums

To achieve robust Pascal's triangle coefficients divisible by 11^2, I’m concentrating on summing digits to 11. This method ensures precise divisibility.

Evaluating digit pairs

Ensuring Pascal's triangle coefficients divisible by 11^2 involves pinpointing digit pairs summing to 11, ensuring accurate divisibility in the ones and tens columns.

Verifying carries

I’m mapping out Pascal's triangle detail, ensuring that coefficients are divisible by 11^2. This involves verifying that two carries align correctly in the ones and tens columns by summing digits to 11.

Verifying digit alignment

I'm confirming that Pascal's triangle coefficients divisible by 11^2 ensure two carries align correctly in the ones and tens columns, with the final sum clearly matching n without overwriting or rounding.

Ensuring the correct sum

OK, let me see. I’m carefully selecting digits to ensure the final sum matches "A0"_11 without causing overflow or exceeding n, maintaining precise divisibility by 11^2 in Pascal's triangle.

Shifting my strategy

To ensure two carries align with n=121 in Pascal's triangle, I’m contemplating a three-digit base 11 representation, preventing overflow and maintaining precise divisibility by 11^2.

Ensuring precision

I’m working on aligning the final sum with "A0"_11, avoiding overflow with too few digits, and leveraging a three-digit base 11 representation for clearer division by 11^2.

Shifting my strategy

I’m piecing together that ensuring the final sum matches "A0"_11 requires at least a three-digit base 11 representation to prevent overflow and maintain exact divisibility by 11^2 in Pascal's triangle.

Ensuring proper representation

Aligning the final sum with "A0"_11 necessitates a three-digit base 11 representation, skillfully managing carries to ensure divisibility by 11^2 and meet Pascal's triangle coefficient requirements.

Thought about строке треугольника Паскаля für 1m 34s

Краткий ответ: Впервые число, кратное 1210, появляется в 121-й строке треугольника Паскаля.

Подробный разбор:

Для удобства нумеруем строки треугольника Паскаля начиная с нулевой (верхняя строка, содержащая единственную единицу, имеет номер 0). Тогда n-я строка содержит коэффициенты $\displaystyle C(n,0), C(n,1), \dots, C(n,n)$.

Нам нужно найти первую строку, в которой встречается число, кратное 1210. Разложим 1210 на простые множители:

$$1210 = 2 \times 5 \times 11^2.$$

Таким образом, мы ищем первую строку, в которой присутствует биномиальный коэффициент, делящийся на $2 \times 5 \times 11^2 = 2 \times 5 \times 121 = 1210.$

Из этих трёх простых факторов (2, 5 и $11^2$) наиболее "сложным" условием является наличие фактора $11^2 = 121$ в коэффициенте. Мелкие простые числа 2 и 5 практически всегда встречаются в изобилии в факториалах (а значит и в биномиальных коэффициентах) уже на сравнительно малых значениях $n$. Основным ограничивающим фактором является именно $11^2$.

Почему трудно получить $11^2$?
Известно, что для простого числа p все внутренние биномиальные коэффициенты строки с номером p (то есть $C(p,k)$ при $1 \le k \le p-1$) делятся на p. Так, для p=11 в строке n=11 все коэффициенты, кроме крайних единиц, делятся на 11. Однако они имеют лишь одну степень простого числа 11, а не $11^2$.

Чтобы добиться делимости на $11^2$, нужно рассмотреть более глубокую структуру делимости биномиальных коэффициентов. Здесь на помощь приходит теорема Куммера: показатель степени простого числа p в разложении $\displaystyle C(n,k)$ равен количеству "переносов" при сложении k и n–k в системе счисления с основанием p.

Для того чтобы $\displaystyle 11^2$ делил $C(n,k)$, необходимо как минимум два переноса при сложении k + (n–k) = n в 11-ричной системе счисления. Один перенос гарантирует наличие фактора 11, а два переноса – наличие фактора $11^2$.

Рассмотрим представление числа n в системе счисления по основанию 11. Чтобы обеспечить два переноса, n должно быть достаточно большим, чтобы в разряде 11^2 (то есть в третьем разряде 11-ричной записи) можно было "погасить" эти переносы. Проще говоря, если n меньше $11^2 = 121$, то его запись в 11-ричной системе имеет не более 2 разрядов, и добиться двух переносов, которые в итоге дадут ровно n, чрезвычайно сложно или невозможно. Первый момент, когда у нас появляется "трёхразрядное" число в базе 11 – это $n = 121$, так как $121_{10} = 100_{11}$.

И действительно, при $n=121$ уже можно подобрать такой k, что в сложении в системе с основанием 11 будет два переноса, следовательно $C(121,k)$ будет делиться на $11^2$.

Например, рассмотрим коэффициент $C(121,2)$:

$$C(121,2) = \frac{121 \times 120}{2} = 121 \times 60 = 7260.$$

Проверим делимость 7260 на 1210:

$$\frac{7260}{1210} = 6.$$

Таким образом, $C(121,2)$ кратно 1210. А поскольку 121 – это первое число, при котором появляется возможность получить фактор $11^2$ в коэффициентах (и, следовательно, 1210 тоже), никакая более ранняя строка не даст такой результат.

Итог: Первая строка, в которой встречается число, кратное 1210, – это строка с номером 121.