A STORAGE FOR THOUGHT

드디어 한 문제를 풀었습니다... 이게 다 오버워치때문이야 A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k; for example, R(6) = 111111.Given that n is a positive integer and GCD(n, 10) = 1, it can be shown that there always exists a value, k, for which R(k) is divisible byn, and let A(n) be the least such value of k; for example, A(7) = 6 and A(41) = 5.You are gi..

오늘은 피타고라스로 달립니다It is easily proved that no equilateral triangle exists with integral length sides and integral area. However, the almost equilateral triangle 5-5-6 has an area of 12 square units.We shall define an almost equilateral triangle to be a triangle for which two sides are equal and the third differs by no more than one unit.Find the sum of the perimeters of all almost equilateral trian..

It turns out that 12 cm is the smallest length of wire that can be bent to form an integer sided right angle triangle in exactly one way, but there are many more examples.12 cm: (3,4,5) 24 cm: (6,8,10) 30 cm: (5,12,13) 36 cm: (9,12,15) 40 cm: (8,15,17) 48 cm: (12,16,20)In contrast, some lengths of wire, like 20 cm, cannot be bent to form an integer sided right angle triangle, and other lengths a..

문제를 볼까?For a prime p let S(p) = (∑(p-k)!) mod(p) for 1 ≤ k ≤ 5.For example, if p=7, (7-1)! + (7-2)! + (7-3)! + (7-4)! + (7-5)! = 6! + 5! + 4! + 3! + 2! = 720+120+24+6+2 = 872. As 872 mod(7) = 4, S(7) = 4.It can be verified that ∑S(p) = 480 for 5 ≤ p < 100.Find ∑S(p) for 5 ≤ p < 108.소수 p에 대해서 S(p) = (∑(p-k)!) mod(p) (1 ≤ k ≤ 5)라 정의하자. 예를 들자면 p가 7일 때 (7-1)! + (7-2)! + (7-3)! + (7-4)! + (7-5)! = 6!..

Let r be the remainder when (a−1)n + (a+1)n is divided by a2.For example, if a = 7 and n = 3, then r = 42: 63 + 83 = 728 ≡ 42 mod 49. And as n varies, so too will r, but for a = 7 it turns out that rmax = 42.For 3 ≤ a ≤ 1000, find ∑ rmax. r을 (a−1)n + (a+1)n 을 a2로 나눴을때의 나머지라 하자; 예를 들면 a = 7이고 n = 3이면, r = 42(63 + 83 = 728 ≡ 42 mod 49)이다. 그리고 n이 달라지면, r,도 달라질테지만, 어쨌건간에 a = 7이 경우는 r의 최댓값은 42이다. 3부터..