Is the integer number 1,522 divisible by 3?

Approach 1. Dividing numbers. Approach 2. Integer numbers prime factorization.

Approach 1. Dividing numbers:

1,522 is divisible by 3, if there is an integer 'n' such that 1,522 = 'n' × 3.

Notice that dividing our numbers leaves a remainder:


1,522 ÷ 3 = 507 + 1;


There is no integer 'n' such that 1,522 = 'n' × 3.


1,522 is not divisible by 3;

1,522 is not divisible by 3
Dividing our numbers leaves a remainder.

Approach 2. Integer numbers prime factorization:

1,522 = 2 × 761;


3 is a prime number, it cannot be prime factorized into other prime factors;


1,522 does not have (all) the prime factors of the number 3;


1,522 is not divisible by 3.

1,522 is not divisible by 3.

Integer numbers prime factorization

Final answer:

1,522 is not divisible by 3.
Dividing our numbers leaves a remainder.
1,522 does not have (all) the prime factors of the number 3.

Is 5,072 divisible by 3?

Online calculator: numbers' divisibility check

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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.