What is the smallest number that is divisible by all numbers from 1 to 10?

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?

Least Common Multiple(LCM):

In arithmetic and number theory, the least common multiple, lowest common multiple, or smallest common multiple of two integers a and b, usually denoted by lcm, is the smallest positive integer that is divisible by both a and b.

Example:

LCM(6,7,21)

Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60

Multiples of 7: 7, 14, 21, 28, 35, 42, 56, 63

Multiples of 21: 21, 42, 63

Find the smallest number that is on all of the lists. We have it in bold above.

So LCM(6, 7, 21) is 42

The best solution with Euclid GCD algorithm

Euclid’s GCD algorithm:

1- The smallest positive number that is evenly divided (divided without remainder) by a > set of numbers is called the Least Common Multiple (LCM).

2- All we have to do to solve this problem is find the LCM for the integers {1, 2, 3, 4, > …, 20} using Euclid’s GCD algorithm.

3- After some reflection you might correctly realize that every integer is divisible by > > 1, so 1 can be removed from the list and use 2 through 20 instead.

4- we can eliminate other factors as well.

5- We leave 20 in the calculation but then remove its factors {2, 4, 5, 10}. Any number > > evenly divisible by 20 is also evenly divisible by these factors.

6- 19 is prime and has no factors— it stays.

7- 18 has factors {2, 3, 6, 9} and we already removed 2 but we can remove 3, 6, and 9. 17 > is prime — it stays.

8- We continue this numeric carnage until our original list of {1…20} becomes the much > > smaller {11…20}.

9- In general, all smaller factors of a bigger number in the list can be safely removed > > without changing the LCM.

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Figure out the prime factors of each of the numbers 1-10. 1 = 1*1. 2 = 2*1. 3 = 3*1. 4=2*2. 5 = 5*1. 6 = 2*3. 7= 7*1. 8 = 2*2*2. 9 = 3*3. 10 = 2*5. The smallest number that is divisible by all of these has to contain all of these factors.

Prime factors of: 12=2*2*3 11 is prime 10= 2*5 9 = 3*3 8 = 2*2*2 7 is prime 6= 2*3 5 is prime 4=2*2 3 is prime 2 is prime 1 divides into anything, although it is not prime.

2520.To evaluate this answer obtain the prime factors of the numbers 2 - 9 (as every integer is divisible by 1). Then take the product of the highest power of each prime factor.2 = 2 : 3 = 3 : 4 = 2 x 2 = 22 : 5 = 5 :6 = 2 x 3 :7 = 7 :8 = 2 x 2 x 2 = 23 :9 = 3 x 3 = 32The bold figures are the numbers that have to be multiplied to produce the Lowest Common Multiple : 5 x 7 x 23 x 32 = 2520.

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

Since all the four numbers 2, 3, 5, 7 are prime it means N will also be divisible by their product 2 × 3 × 5 × 7 = 210; For N < 3, no such number exists. So, print -1. For N = 3, the answer will be 210. For N > 3, the following computation needs to be done: Find Remainder R = 10 N-1 % N. Add 210 – R to 10 N-1.

The least common multiple of the values between 1 and 10 is what you are looking for. To do this, all you have to do is multiply each number by the next, so you would have 1*2*3*4*9*10. This will always give you value that is divisible by all of the numbers in the range.

However, you don't have to check all numbers 1..20. If a number is divisible by 20, it is divisible by 2, 5, 10. Extending this, it would be sufficient to check only divisors from 11..20. Another simple thing would be to increase the candidate solutions by 20 (number += 20) instead of 1, since any other number would not be divisible by 20.

Given a positive integers N, the task is to find the smallest N digit number divisible by N. Examples: Input: N = 2 Output: 10 Explanation: 10 is the smallest 2-digit number which is divisible by 2. Input: N = 3 Output: 102 Explanation: 102 is the smallest 3-digit number which is divisible by 3.

The prime factors of each number are1 : 1 (technically 1 is not a Prime number)2 : 23 : 34 : 2²5 : 56 : 2, 3The Lowest Common Multiple is the product of the highest power of each prime factor.This gives 1 × 2² × 3 × 5 = 60.which is the smallest whole number divisible by 1 to 6.

102 is the smallest 3-digit number which is divisible by 3. Recommended: Please try your approach on {IDE} first, before moving on to the solution. Naive Approach: The naive approach is to iterate from smallest N-digit number(say S ) to largest N-digit number(say L ).

Start by thinking about the factorial n!. This is obviously divisible by all numbers less then n, but is not the smallest such number. For example, you could divide n! by 6, and the smaller result would still be divisible by 2 and by 3, and hence still divisible by 6.

2520 is: the smallest number divisible by all integers from 1 to 10, i.e., it is their least common multiple. half of 7! (5040), meaning 7 factorial, or 1×2×3×4×5×6×7. the product of five consecutive numbers, namely 3×4×5×6×7.

There is no smallest number since if any number is divisible by 8, then 8 less than that number is also divisible by 8 - and that argument continues all the way to minus infinity!The smallest positive number divisible by 8 is 8, itself.

And if not, we'll tell you -- the smallest number that can be divided by every number from 1 to 10 is 2,520. Read on for the solution. It's all about finding the unique factors among the numbers 1-10, which are 9, 8, 7, and 5. All numbers are divisible by 1, so we can ignore that.

Click here to see ALL problems on Divisibility and Prime Numbers Question 505530 : Find the smallest number evenly divisible by each counting number from 1 to 10. Explain in detail explanation.

All integers are divisible by 1. Even integers are divisible by 2, and 2, 4, 6, 8, and 10 are even digits divisible by 2. Integers ending in 0 are divisible by 10.

Remainder of "1*1000 + 3*100 + 3*10 + 2" divided by 9 can be written as : 1*1 + 3*1 + 3*1 + 2 = 9 The above expression is basically sum of all digits. Since 9 is divisible by 9, answer is yes. Below is the implementation of above idea.

If the difference of the sum of its digits at odd places and sum of its digits at even places is either 0 or a number divisible by 1 1. 2 8 7 1: Sum of the odd places = 2 + 7 = 9 Sum of the even places = 8 + 1 = 9 Difference = Sum of the odd places − Sum of the even places Difference = 9 − 9 = 0 So, 2 8 7 1 is divisible by 1 1.

This tutorial is to show you how we can find out the smallest divisor of a number in python. Number ‘y’ _is called divisor of number _‘x’ if ‘x/y’ is 0. Our program will ask the user to enter a no. It will then find out the lowest divisor of the number.

An efficient approach is to find smallest of all numbers, and check if it divides all the other numbers, if yes then the smallest number will be the required number.

Store all array elements into hash and find out the max element in array then up-to max element find out the multiples of a given number then if multiple of array element is in hash then that number is divisible by at-least one element of array.To remove duplicate values we store the value into set because if array has 2, 3 and 6 then only 6 is divisible by at-least one element of array, both 2 and 3 divide 6 so 6 will be stored only one time.

Consider we have an array A with few elements. We have to find an element from A, such that all elements can be divided by it. Suppose the A is like [15, 21, 69, 33, 3, 72, 81], then the element will be 3, as all numbers can be divisible by 3.

Program to find the Smallest Number in a List Example 3. The Python sort function sort List elements in ascending order. Next, we are using Index position 0 to print the first element in a List.

Time Complexity: O(L – S), where L and S is the largest and smallest N-digit number respectively. Efficient Approach: If the number divisible by N, then the number will be of the form N * X for some positive integer X.

Approach: Smallest number which is divisible by all digits from 1 to 9 is equal to the LCM of (1, 2, 3, 4, 5, 6, 7, 8, 9) = 2520. Therefore, the multiples of 2520 are also divisible by all digits from 1 to 9 implying that (N + 2520) will always satisfy the condition.

36 is the smallest number which is divisible by both the digits 2 and 3. Approach: The idea is to find the LCM of all the non-zero digits of X and then just find the next greater multiple of that LCM value which is greater than N .

Don't worry about how you know that. Given the fact that 840 is the smallest number that is divisible by 1, 2, 3, 4, 5, 6, 7 and 8, and that 9 is 3 times 3, how do you find a number that is divisible by 1, 2, 3, 4, 5, 6, 7, 8 and 9? Plainly multiplying 840 by 9 gives a number that is too large, because 9 is already 3 x 3.

This is the smallest number divisible by 1, 2, 3, , 20 _____ Note that number 9,699,690 only has 1 factor of 2, while numbers 4, 12, and 20 have 2 factors of 2, number 8 has 3 factors of 2, and number 16 has 4 factors of 2. For this reason, 4, 8, 12, 16, and 20 cannot divide 9,699,690.

The smallest number that is divisible by all the numbers 2 through 12 is 27,720.

Smallest Number Divisible by 1 -100 Numbers There are two types of numbers, Prime Numbers and Composite Numbers. The Prime Numbers between 1 - 100 are 1,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97.

das Nerd: you are almost right, but the number you suggest is not divisible by 4,8,16,32,64,9,27,81,25, or 49. To find the correct result, you must multiply all the prime numbers to the largest power they appear in any number from 1-100.

Yes. You should remember that a number [math]a[/math] is divisible by another number [math]b[/math] if there exists a number [math]c[/math] such that [math]a = b \times c[/math].

Find the smallest number that is made up of each of the digits 1 through 9 exactly once and is divisible by 99. Solution The first thing we note is that 1 + 2 + + 9 = 45 and as the sum of the digits is divisible by 9 then any arrangement of those digits will produce a number that is divisible by 9.

The Smallest number exactly divisible by 6, 8 and 12 = LCM of 6, 8, 1 2 = 2 4. We have to find the smallest 3 -digit multiple of 24. It can be seen that 24 × 4 = 96 and 24 ×5 = 120. The smallest 3-digit number which is exactly divisible by 6,8 and 12 is 120.

If any number ends in a two digit number that you know is divisible by 4 (e.g. 24, 04, 08, etc.), then the whole number will be divisible by 4 regardless of what is before the last two digits. Alternatively, one can simply divide the number by 2, and then check the result to find if it is divisible by 2.

The smallest k digit number divisible by x is 152 A class named Demo contains a function anmed ‘smallest_k’. This function returns the minimum number of digits of ‘k’ that completely divide the number ‘x’. In the main function, a value for ‘x’ and ‘k’ is defined.

Project Euler 5 relates to the divisibility of numbers and is an almost trivial problem that is included here only for completeness. Project Euler 5 2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

The number in question must be divisible by all of the numbers between 1 and 20 (assume inclusive for the ranges). As a result, it must be divisible by all of the numbers between 1 and 10. The smallest number that is divisible by all the numbers between 1 and 10 (given by Project Euler) is 2520.

Numbers Divisible by 7 from 0 to 100 in Java. That program will print the output on screen and shows only those numbers which are divisible by 7.

Write a Java Program to Check whether Number is Divisible by 5 and 11 using If Else statement and the Conditional Operator with example. This Java program helps the user to enter any number. Next, it checks whether the given number is divisible by both 5 and 11 using If Else Statement .

What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20? Solution The solution will also be divisible by the numbers 1 to 10, so we can start at 2520 and increment by 2520.

Below, we list what numbers can be divided by 30 and what the answer will be for each number. 30 / 1 = 30 30 / 2 = 15 30 / 3 = 10 30 / 5 = 6 30 / 6 = 5 30 / 10 = 3 30 / 15 = 2 30 / 30 = 1 What is 31 divisible by? Now you know what 30 is divisible by. You may also be interested in the answer to the next number on our list.

The smallest five digit number = 10,000 In order to find the smallest five digit number Divide the number 10,000 by 35 We get quotient = 285 and remainder = 25 So 10,000 = 285 x 35 + 25 Adding the number 10 on both sider ⇒ 10,000 +10 = 285 x 35 + 25+10 ⇒ 10,010 = 285 x 35 + 35 Notice that RHS is can be exactly divisible by 35.

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

C Exercises: Smallest positive number divisible by 1-20 2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

Which one is divisible by 10?

Divisibility rule for 10 states that any number whose last digit is 0, is divisible by 10. Example: 10, 20, 30, 1000, 5000, 60000, etc.

How to get 2520?

The factors of 2520 in pairs are:.

1 × 2520 = (1, 2520).

2 × 1260 = (2, 1260).

3 × 840 = (3, 840).

4 × 630 = (4, 630).

5 × 504 = (5, 504).

6 × 420 = (6, 420).

7 × 360 = (7, 360).

8 × 315 = (8, 315).

How many numbers from 1 to N are divisible by all numbers from 2 to 10?

So any number divisible by 2520 is divisible by all the numbers from 2 to 10. So, the count of such numbers is N / 2520.