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The first: 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600 (sequence A000142 in the OEIS). 0! = 1 is sometimes included. A k-smooth number (for a natural number k) has its prime factors ≤ k (so it is also j-smooth for any j > k). m is smoother than n if the largest prime factor of m is below the largest of n.
The table below shows all 72 divisors of 10080 by writing it as a product of two numbers in 36 different ways. The highly composite number: 10080 10080 = (2 × 2 × 2 × 2 × 2) × (3 × 3) × 5 × 7
Continuing this process until every factor is prime is called prime factorization; the result is always unique up to the order of the factors by the prime factorization theorem. To factorize a small integer n using mental or pen-and-paper arithmetic, the simplest method is trial division : checking if the number is divisible by prime numbers 2 ...
A sentence diagram is a pictorial representation of the grammatical structure of a sentence. The term "sentence diagram" is used more when teaching written language, where sentences are diagrammed. The model shows the relations between words and the nature of sentence structure and can be used as a tool to help recognize which potential ...
The sum of its factors (including one and itself) sum to 360, exactly three times 120. Perfect numbers are order two ( 2-perfect ) by the same definition. 120 is the sum of a twin prime pair (59 + 61) and the sum of four consecutive prime numbers (23 + 29 + 31 + 37), four consecutive powers of two (8 + 16 + 32 + 64), and four consecutive powers ...
If two primes which end in 3 or 7 and surpass by 3 a multiple of 4 are multiplied, then their product will be composed of a square and the quintuple of another square. In other words, if p, q are of the form 20k + 3 or 20k + 7, then pq = x 2 + 5y 2. Euler later extended this to the conjecture that
Suppose, to the contrary, there is an integer that has two distinct prime factorizations. Let n be the least such integer and write n = p 1 p 2... p j = q 1 q 2... q k, where each p i and q i is prime. We see that p 1 divides q 1 q 2... q k, so p 1 divides some q i by Euclid's lemma. Without loss of generality, say p 1 divides q 1.
77 is not a multiple of 3, since the sum of its digits is 14, not a multiple of 3. It is also not a multiple of 5 because its last digit is 7. The next odd divisor to be tested is 7. One has 77 = 7 · 11, and thus n = 2 · 3 2 · 7 · 11. This shows that 7 is prime (easy to test directly). Continue with 11, and 7 as a first divisor candidate.