Is the integer number 113 divisible by 23?

Is 113 divisible by 23?

Approach 1. Dividing numbers:

One integer A is divisible by another integer B, if after dividing them, A ÷ B, the remainder is zero.


113 is divisible by 23, if there is an integer 'n' such that:
113 = 'n' × 23.


Notice that dividing our numbers leaves a remainder:


113 ÷ 23 = 4 + 21;


There is no integer 'n' such that 113 = 'n' × 23.


113 is not divisible by 23.


Note:

1) If you subtract the remainder of the above operation, 21, from the initial number, 113, you'll get as a result a number that is evenly divisible by the second number, 23:


113 - 21 = 92;


92 = 4 × 23.


2) If you subtract the remainder of the above operation, 21, from the second number, 23, and then add the result to the initial number, 113, you'll get as a final result a number that is 'evenly divisible' by the second number, 23:

23 - 21 = 2;


113 + 2 = 115;


115 = 5 × 23.


113 is not divisible by 23
Dividing our numbers leaves a remainder.

Approach 2. Integer numbers prime factorization:

When are two numbers divisible?

Number 113 is divisible by number 23 if it has as factors all the prime numbers that occur in the prime factorization of 23.


Integer numbers prime factorization:

Prime Factorization of a number: finding the prime numbers that multiply together to make that number.


113 is a prime number, it cannot be broken down to other prime factors;


23 is a prime number, it cannot be broken down to other prime factors;



* Positive integers that are only dividing by themselves and 1 are called prime numbers. A prime number has only two factors: 1 and itself.
* A composite number is a positive integer that has at least one factor (divisor) other than 1 and itself.


113 does not have (all) the prime factors of the number 23;


113 is not divisible by 23.


113 is not divisible by 23.

Final answer:
113 is not divisible by 23.
Dividing our numbers leaves a remainder.
113 does not have (all) the prime factors of the number 23.
Note:
92 is divisible by 23
115 is divisible by 23

More operations of this kind:

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Tutoring: What is the integer numbers divisibility? Divisibility rules.

Divisibility of integer numbers

Let's divide two different numbers, 12 and 15, by 4. When we divide 12 by 4, the quotient is 3 and the operation leaves no remainder. But when we divide 15 by 4, the quotient is 3 and the operation leaves a remainder of 3. We say that the number 12 is divisible by 4 and 15 is not. We also say that 4 is a divisor of 12, but is not a divisor of 15.

In general, we say that "a" is divisible by "b", if there is an integer number "n", so that: a = n × b. Number "b" is called the divisor of "a" ("n" is also a divisor of "a").

0 is divisible by any number other than zero itself.

Any number "a", different of zero, is divisible at least by 1 and itself, which are called improper divisors.

Some divisibility rules

The number 84 is divisible by 4 and 3 and is also divisible by 4 × 3 = 12. This is not true unless the two divisors are coprime.

In general, if "a" is divisible by both "m" and "n" and greatest common factor (m; n) = 1 (coprime numbers), then "a" it is also divisible by their product, (m × n).

Calculating divisors (factors) is very useful when simplifying fractions (reducing fractions to lower terms). The established rules for finding factors (divisors) are based on the fact that the numbers are written in the decimal system. Mutiples of 10 are divisible by 2 and 5, because 10 is divisible by 2 and 5; multiples of 100 are divisible by 4 and 25, because 100 is divisible by 4 and 25; multiples of 1000 are divisible by 8, because 1000 is divisible with 8. All the powers of 10, when divided by 3, or 9, have a remainder equal to 1.

Due to the rules of operation with remainders, we have the following remainders when dividing numbers by 3 or 9: 600 leaves a remainder equal to 6 = 1 × 6; 240 = 2 × 100 + 4 × 10, then the remainder will be equal to 2 × 1 + 4 × 1 = 6. On dividing a number by 3 or 9 the remainder will be equal to that left from dividing the sum of digits of that number by 3 or 9; 7,309 has the sum of the numbers 7 + 3 + 0 + 9 = 19, which is divided without a remainder to neither 3 nor 9. So 7,309 is not divisible by 3 or 9.

All even powers of 10, 100, 10,000, 1,000,000, etc., when divided by 11 left a remainder of 1, and the odd powers of 10, when divided by 11 left a remainder equal to 10 or 10 - 11 = -1. In this case, the alternating sum of the digits bears the same remainder as when dividing by 11, as if the whole number were being divided by 11. How to calculate the alternating sum is shown in the example below.

For instance, for the number: 85,976: 8 + 9 + 6 = 23, 5 + 7 = 12, the alternating sum of the digits: 23 - 12 = 11. So 85,976 is divisible by 11.

A number is divisible by:
  • 2 if the last digit is divisible by 2
  • 4 if the last two digits form a number divisible by 4;
  • 8, if the last three digits form a number divisible by 8;
  • 5 if the last digit is divisible by 5 (5 and 0)
  • 25, if the last two digits form a number divisible by 25
  • 3, if the sum of digits is divisible by 3;
  • 9, if the sum of digits is divisible by 9;
  • 11 if the alternating sum of digits is divisible by 11.

What is a prime number?

What is a composite number?

Prime numbers up to 1,000

Prime numbers up to 10,000

Sieve of Eratosthenes

Euclid's algorithm

Simplifying ordinary (common) math fractions (reducing to lower terms): steps to follow and examples