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..

문제가 짧군요.How many 18-digit numbers n (without leading zeros) are there such that no digit occurs more than three times in n?18자리수중에 어느 한 숫자가 3번넘게 반복되지 않는 경우의 수를 구하랍니다.난이도가 55%라(푼 문제중 60%가 최고난도) 쫄아가지고 보류해놨는데 알고보니까 쉬운 문제네요 ㅇㅅㅇ;;일단 1)쓰이는 서로 다른 숫자의 갯수를 정했습니다. 모든 수가 3번씩 쓰일 경우, 6개의 서로 다른 숫자가 쓰일거고, 10개의 모든 한자리수가 쓰일수가 있겠죠.2)각 숫자가 몇번 등장하는지를 구했습니다. n번(1,2,3)쓰이는 숫자의 갯수를 xn이라 하면 x1+x2+x3=k(k는 6부터 10), x1+2*x..

The smallest number m such that 10 divides m! is m=5. The smallest number m such that 25 divides m! is m=10. Let s(n) be the smallest number m such that n divides m!. So s(10)=5 and s(25)=10. Let S(n) be ∑s(i) for 2 ≤ i ≤ n. S(100)=2012.Find S(108). s(n)은 n이 나눌 수 있는 최소의 factorial인 m!을 찾아서, 그 m을 값으로 합니다. n=25를 예로 들면, 10!에 5라는 소인수가 두개 있으므로 s(25)=10이 되고, n=10=2*5인데 5!에 소인수 2, 5 둘다 잇으니까 s(10)=5가 되고...

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!..

106개 풀었네요. 처음엔 열문제정도만 풀다가 방치했는데 군대와서 심심해가지고 반년동안 끄적되니까 어느새 이렇게 되네요. 덕분에 matlab도 많이 익숙해지고 python도 도전해보고, 알고리즘에도 관심가지게되고, 덕분에 좋은 경험하고 있습니다. 군대에서 여가시간쓰기도 좋고..

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..

문제를 살펴봅시다. The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9 in some order, but it also has a rather interesting sub-string divisibility property. Let d1 be the 1st digit, d2 be the 2nd digit, and so on. In this way, we note the following: d2d3d4=406 is divisible by 2 d3d4d5=063 is divisible by 3 d4d5d6=635 is divisible by 5 d5d6d7=357 is di..

문제 The largest integer ≤ 100 that is only divisible by both the primes 2 and 3 is 96, as 96=32*3=25*3. For two distinct primes p and q let M(p,q,N) be the largest positive integer ≤N only divisible by both p and q and M(p,q,N)=0 if such a positive integer does not exist. E.g. M(2,3,100)=96. M(3,5,100)=75 and not 90 because 90 is divisible by 2 ,3 and 5. Also M(2,73,100)=0 because there does not ..