## Smallest multiple

### Problem 5

2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?

Ans:-**232792560**

Code

```
import fractions
def compute():
ans = 1
for i in range(1, 21):
ans *= i // fractions.gcd(i, ans)
return str(ans)
print(compute())
```

Programming problem from Hackerrank.com

**2520 **is the smallest number that can be divided by each of the numbers from **1** to **10** without any remainder.

What is the smallest positive number that is evenly divisible(divisible with no remainder) by all of the numbers from **1** to **N** ?

**Input Format**

First line contains **T** that denotes the number of test cases. This is followed by **T **lines, each containing an integer, **N **.

**Constraints**

**1⩽T⩽10****1⩽N⩽40**

**Output Format**

Print the required answer for each test case.

**Sample Input **

```
2
3
10
```

**Sample Output **

```
6
2520
```

**Explanation **

- You can check
**6**is divisible by each of**{1,2,3}**, giving quotient of**{6,3,2}**respectively. - You can check
**2520**is divisible by each of**{1,2,3,4,5,6,7,8,9,10}**giving quotient of**{2520,1260,840,630,504,420,360,315,280,252}**respectively.

**Solution:-**

- The smallest number n that is evenly divisible by every number in a set {k1, k2, …, k_m}

is also known as the lowest common multiple (LCM) of the set of numbers. - The LCM of two natural numbers x and y is given by LCM(x, y) = x * y / GCD(x, y).
- When LCM is applied to a collection of numbers, it is commutative, associative, and idempotent.
- Hence LCM(k1, k2, …, k_m) = LCM(…(LCM(LCM(k1, k2), k3)…), k_m).

Code:-

```
import math
def compute(n):
ans = 1
for i in range(1, n+1):
ans *=i// math.gcd(i, ans)
return str(ans)
t = int(input().strip())
for a0 in range(t):
n = int(input().strip())
print(compute(n))
```