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d() is the number of positive divisors of n, including 1 and n itself; σ() is the sum of the positive divisors of n, including 1 and n itselfs() is the sum of the proper divisors of n, including 1 but not n itself; that is, s(n) = σ(n) − n
364: 3 × 2 + 64 = 70, 1,764: 17 × 2 + 64 = 98. 15: It is divisible by 3 and by 5. [6] 390: it is divisible by 3 and by 5. 16: If the thousands digit is even, the number formed by the last three digits must be divisible by 16. 254,176: 176. If the thousands digit is odd, the number formed by the last three digits must be 8 times an odd number.
When a triple of numbers a, b and c forms a primitive Pythagorean triple, then (c minus the even leg) and one-half of (c minus the odd leg) are both perfect squares; however this is not a sufficient condition, as the numbers {1, 8, 9} pass the perfect squares test but are not a Pythagorean triple since 1 2 + 8 2 ≠ 9 2. At most one of a, b, c ...
A square has even multiplicity for all prime factors (it is of the form a 2 for some a). The first: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144 (sequence A000290 in the OEIS). A cube has all multiplicities divisible by 3 (it is of the form a 3 for some a). The first: 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728 (sequence A000578 ...
If none of its prime factors are repeated, it is called squarefree. (All prime numbers and 1 are squarefree.) For example, 72 = 2 3 × 3 2, all the prime factors are repeated, so 72 is a powerful number. 42 = 2 × 3 × 7, none of the prime factors are repeated, so 42 is squarefree. Euler diagram of numbers under 100:
Start with division by 2: the number is even, and n = 2 · 693. Continue with 693, and 2 as a first divisor candidate. 693 is odd (2 is not a divisor), but is a multiple of 3: one has 693 = 3 · 231 and n = 2 · 3 · 231. Continue with 231, and 3 as a first divisor candidate. 231 is also a multiple of 3: one has 231 = 3 · 77, and thus n = 2 ...
When Henderson broke a 66-yard TD run down the Oregon sideline for a 31-0 lead, both sides of the Rose Bowl stands rippled with disbelief. Oregon finally got moving on its final drive before halftime.
Here legal means that the coefficients are non-negative and sum to six, so that each die has six sides and every face has at least one spot. (That is, the generating function of each die must be a polynomial p(x) with positive coefficients, and with p(0) = 0 and p(1) = 6.)