How many two digit numbers are divisible by the sum of their digits?

How many two digit numbers are divisible by the sum of their digits?

We have now considered all the potential values of m. Hence we conclude that the only 2-digit positive integers which are divisible by both the sum and product of their digits are 12, 24, and 36.

How do you find a number which is divisible by all of its digits?

Approach:

1. Get all the digits of the given number.
2. For each digit check if number%digit==0 (means divisible).
3. If any of them is non zero return false.
4. If for all digits number%digit==0, return true.

What is the sum of the digits in each product?

Say we have a number greater than or equal to 5 and less than 10, like 5. If we multiply it by 2, we get 10, and if we take the sum of the digits of 10, we get 1. If instead of adding up the digits of 10, we subtract 9, we also get 1.

Is a number always divisible by the sum of its digits?

Since any power of 10 is 999.. 9 + 1 (with the appropriate amount of 9s), and the 999.. 9 part is always divisible by 3, what’s left is the sum of the digits, which implies the divisibility by 3 of the number. Another rule says that if the sum of the digits is divisible by 9, the number is divisible by 9.

How many 2 digit numbers have the sum of their digits divisible by 4?

In the given AP i.e., 12,16,20,……,96, first term is a1=12, common difference is d=4 and nth term is an=96. So, the series of 2-digit numbers which are divisible by 4 consists of a total 22 terms. Therefore, there are a total 22 two-digit numbers which are divisible by 4.

What is the smallest four digit number divisible by 12 18 and 21?

Thus, the smallest 4 digit number such that when it is divided by 12, 18, 21 and 28 is 1008.

What is the largest multiple of 3?

For example, let us consider 8760, it is a multiple of 3 because sum of digits is 8 + 7+ 6+ 0 = 21, which is a multiple of 3. 2) If a number is multiple of 3, then all permutations of it are also multiple of 3. For example, since 6078 is a multiple of 3, the numbers 8760, 7608, 7068, ….. are also multiples of 3.

How do you find the highest divisible numbers?

1 Answer

1. Generate biggest number from digits (it means sort them in desc order)
2. check if it is divisible by B-1.
3. If yes sucess just return n-th digit if no go to step 1. but generate next number (by swaping different digits)

What is the sum of the first 100 whole numbers?

5050
Therefore, the sum of first 100 natural numbers = 5050.

Why is a number divisible by 3 if its digits sum is divisible by 3?

A number is divisible by 3 if the sum of its digits is divisible by 3. For large numbers this rule can be applied again to the result. A.) 504: 5 + 0 + 4 = 9, so it is divisible by 3.

Which is a number that is divisible by the sum of its digits?

Take a positive integer X. This number is part of the sequence we are interested in if the sum of all digits of X is a divisor of X, and if the product of all digits of X is a divisor of X. For example, 135 is such a number because 1 + 3 + 5 = 9 which divides 135 = 9 * 15 and 1 * 3 * 5 = 15 which also divides 135.

How to calculate sum and product of digits?

Input: N = 12 Output: Sum = 3, product = 2 1 and 2 divide 12. So, their sum is 3 and product is 2. Input: N = 1012 Output: Sum = 4, product = 2 1, 1 and 2 divide 1012.

Is the sum of all digits of X a divisor?

This number is part of the sequence we are interested in if the sum of all digits of X is a divisor of X, and if the product of all digits of X is a divisor of X. For example, 135 is such a number because 1 + 3 + 5 = 9 which divides 135 = 9 * 15 and 1 * 3 * 5 = 15 which also divides 135.

Which is the sum of the digits of M?

Let m be an n -digit number, and its digits in reverse order be d0, d1, up to dn-1, so each di is a whole number between 0 and 9 inclusive. so obviously, if m is divisible by 9, so is the sum of m’s digits, and vice versa. Also, since 9 is divisible by 3, the same occurs for 3. QED. So why 9 and 3?