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Number - Divisibility Rules and Tests

Grade 6IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A number nn is said to be divisible by dd if the quotient n÷dn \div d results in an integer with a remainder of 00.

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Divisibility by 22: A number is divisible by 22 if its last digit is even (0,2,4,6,80, 2, 4, 6, 8).

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Divisibility by 33: A number is divisible by 33 if the sum of its digits is divisible by 33.

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Divisibility by 44: A number is divisible by 44 if the last two digits form a number that is divisible by 44.

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Divisibility by 55: A number is divisible by 55 if its last digit is either 00 or 55.

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Divisibility by 66: A number is divisible by 66 if it is divisible by both 22 and 33 (it must be even and the sum of its digits must be a multiple of 33).

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Divisibility by 88: A number is divisible by 88 if the last three digits form a number divisible by 88.

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Divisibility by 99: A number is divisible by 99 if the sum of its digits is divisible by 99.

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Divisibility by 1010: A number is divisible by 1010 if its last digit is 00.

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Divisibility by 1111: Find the sum of digits in odd positions and the sum of digits in even positions. If the difference between these sums is 00 or a multiple of 1111, the number is divisible by 1111.

📐Formulae

Sum of Digits=∑i=0kdi\text{Sum of Digits} = \sum_{i=0}^{k} d_i

Rule for 11: ∣(Sum of odd-place digits)−(Sum of even-place digits)∣=11k, where k∈Z\text{Rule for 11: } |(\text{Sum of odd-place digits}) - (\text{Sum of even-place digits})| = 11k, \text{ where } k \in \mathbb{Z}

n≡0(modd)n \equiv 0 \pmod{d}

💡Examples

Problem 1:

Check if the number 4,7524,752 is divisible by 99.

Solution:

Sum of digits: 4+7+5+2=184 + 7 + 5 + 2 = 18. Since 1818 is divisible by 99 (18÷9=218 \div 9 = 2), the number 4,7524,752 is divisible by 99.

Explanation:

We use the divisibility rule for 99 which states that the sum of the digits must be a multiple of 99.

Problem 2:

Is the number 1,3311,331 divisible by 1111?

Solution:

Sum of digits in odd positions: 1+3=41 + 3 = 4. Sum of digits in even positions: 3+1=43 + 1 = 4. Difference: 4−4=04 - 4 = 0. Since the difference is 00, 1,3311,331 is divisible by 1111.

Explanation:

The rule for 1111 involves the alternating sum of digits. If the result is 00 or a multiple of 1111, the rule is satisfied.

Problem 3:

Determine if 1,5241,524 is divisible by 66.

Solution:

  1. Check divisibility by 22: The last digit is 44 (even), so it is divisible by 22.
  2. Check divisibility by 33: Sum the digits: 152+412\begin{array}{r} 1 \\ 5 \\ 2 \\ + 4 \\ \hline 12 \end{array} Since 1212 is divisible by 33, the number 1,5241,524 is divisible by 33. Because it is divisible by both 22 and 33, it is divisible by 66.

Explanation:

Divisibility by 66 is a composite rule requiring both the rules for 22 and 33 to be true.