Project Euler #5: Smallest multiple Discussions |

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3 days ago+ 0 comments

All tests cleared, for possible test case(0) {1,1} fails without exception handling for this solution in java

Here is a solution solution for Java includes recursive gcd, lcm and LCM for arrays and exception handling https://github.com/a-dilmen/Hacker_Rank_Project_Euler/blob/main/SmallestMultiple.java

1 week ago+ 0 comments

my code but biving one test case error.

import sys
import math

test_case = int(input().strip())
for a0 in range(test_case):
    rang = int(input().strip())
    num = []
    for i in range(1,rang+1):
        num.append(i)
    num.sort()
    lcm = math.lcm(num[0],num[1])
    for i in range(2,len(num)):
        lcm = math.lcm(lcm,num[i]) 
    print(lcm)

1 month ago+ 0 comments

Python Solutions: https://github.com/Siliquastrum/ProjectEuler/blob/main/1-to-100/1-to-10/5-SmallestMultiple.py

Other ProjectEuler Problems Python Solutions: https://github.com/Siliquastrum/ProjectEuler

1 month ago+ 0 comments

Python:

from functools import reduce
gcd = lambda a, b: a if not b else gcd(b, a % b)
lcm = lambda a, b: (a * b) // gcd(a, b)
[print(reduce(lcm, range(1, int(input())+1))) for _ in range(int(input()))]

1 month ago+ 0 comments

My Solution

Python 3

t = int(input().strip())
for a0 in range(t):
    n=int(input())

    lst=[i for i in range(1,n+1)]
    k=1
    for i in range(n):
        if lst[0]==1:
            del lst[0]
        if not len(lst):
            break
        num=lst[0]
        for j in range(len(lst)):
            if lst[j]%num==0:
                lst[j]=lst[j]//num
        k*=num
    print(k)

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