...s whose coordinates are no larger than 40. Find the length of the longest proper sequence of dominos that can be formed using the dominos of <math>D_{40}.</
...ected segment from one vertex of the <math>n</math> -gon to another, and a proper sequence is represented as a path that retraces no segment. Each time that
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9 KB (1,659 words) - 12:54, 25 May 2025
...</math> be the sum of the base <math>10</math> [[logarithm]]s of all the [[proper divisor]]s (all [[divisor]]s of a number excluding itself) of <math>1000000
...re <math>(6 + 1)(6 + 1) = 49</math> divisors, of which <math>48</math> are proper. The sum of multiple logarithms of the same base is equal to the logarithm
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3 KB (487 words) - 19:52, 16 September 2020
...an 1 will be called ''nice'' if it is equal to the product of its distinct proper divisors. What is the sum of the first ten nice numbers?
Let <math>p(n)</math> denote the product of the distinct proper divisors of <math>n</math>. A number <math>n</math> is ''nice'' in one of t
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3 KB (511 words) - 08:29, 9 January 2023
...actly three proper divisors, each of which is less than <math>30</math> (a proper divisor of <math>n</math> is a positive integer that divides but is not equ
...qr</math>. In other words, only perfect squares of primes have exactly two proper divisors (hence <math>2397 = 2^2 + 3^2 + 5^2 + 7^2 + \cdots + 29^2</math>),
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1 KB (226 words) - 18:58, 29 May 2008
...<math>n</math>. For example, 22 is deficient because its [[proper divisor|proper factors]] sum to 14 < 22. The smallest deficient numbers are 1, 2, 3, 4, 5,
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719 bytes (88 words) - 11:39, 9 February 2018
How many [[positive integer]]s have exactly three [[proper divisor]]s (positive integral [[divisor]]s excluding itself), each of which
...n</math> is such an [[integer]]. Because <math>n</math> has <math>3</math> proper divisors, it must have <math>4</math> divisors,, so <math>n</math> must be
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2 KB (249 words) - 08:37, 23 January 2024
...divisor <math>1</math>, and all other positive integers have at least two proper divisors.
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436 bytes (63 words) - 15:27, 10 May 2021
...actly three proper divisors, each of which is less than <math>30</math> (a proper divisor of <math>n</math> is a positive integer that divides but is not equ
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4 KB (582 words) - 20:57, 8 May 2019
...<math>n</math> itself. For example, 12 is abundant because the sum of its proper divisors is 1 + 2 + 3 + 4 + 6 = 16 > 12.
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934 bytes (91 words) - 11:27, 9 February 2018
...green, or blue so that each number has a different color from each of its proper divisors?
...no restrictions because the set of integers does not contain a multiple or proper factor of <math>5</math> or <math>7</math>. There are 3 ways to paint each,
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4 KB (623 words) - 00:31, 3 November 2024
A <i>proper divisor</i> of a positive integer <math>N</math> is a positive divisor of <
...1</math>, the integer <math>a_{n+1}</math> is the sum of the three largest proper divisors of <math>a_n</math>.
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669 bytes (110 words) - 18:52, 19 July 2025
A <i>proper divisor</i> of a positive integer <math>N</math> is a positive divisor of <
...1</math>, the integer <math>a_{n+1}</math> is the sum of the three largest proper divisors of <math>a_n</math>.
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4 KB (775 words) - 15:56, 4 September 2025
...cobian is everywhere invertible (already established), and that \( f \) is proper, meaning the preimage of any compact set is compact.
...compact sets are closed, so they are compact. This proves that \( f \) is proper.
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5 KB (839 words) - 17:23, 18 August 2025
'''Lemma 2:''' For any proper integral ideal <math>J</math>, there is some <math>\gamma\in K\setminus R</
...ath>R\not\subseteq (a)</math>, <math>n\ge 1</math>. As <math>J</math> is a proper ideal, it must be contained in some maximal ideal, <math>P</math>. Since ma
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9 KB (1,648 words) - 15:36, 14 October 2017
...ectful to such a superior power. Therefore, it is best to let Gmaas make a proper sculpture of himself, so he is respected. If you want to respect him the mo
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69 KB (11,844 words) - 21:54, 20 October 2025
...math>H</math> be a proper subgroup of <math>H</math>. Then there exists a proper normal subgroup <math>A</math> of <math>G</math> such that <math>H \subsete
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9 KB (1,768 words) - 16:55, 5 June 2008
...2.''' If <math>G</math> is a <math>p</math>-group, and <math>H</math> is a proper [[subgroup]] of <math>G</math>, then the normalizer of <math>H</math> is di
'''Proposition.''' Let <math>H</math> be a proper subgroup of a <math>p</math>-group <math>G</math>. Then there exists a nor
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4 KB (816 words) - 12:51, 25 February 2025
...ose <math>I(PQ) \neq (1)</math>. Then by [[Krull's Theorem]], there is a (proper) maximal ideal <math>\mathfrak{m}</math> that contains <math>I(PQ)</math>.
is a proper subset of <math>(a,b)(c,d)</math>.
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3 KB (488 words) - 19:37, 28 September 2024
...a proper [[ideal]] <math>I\le R</math> which is not contained in any other proper ideal of <math>R</math>. (That is, <math>I\neq R</math>, and there is no id
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968 bytes (183 words) - 18:50, 23 August 2009
maximal proper submodule.
be a generator for <math>M</math>. Let <math>N</math> be a maximal proper submodule of <math>M</math>
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6 KB (1,183 words) - 14:02, 18 August 2009
