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Playing with Numbers - Tests for Divisibility of Numbers

Grade 6CBSE

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

🔑Concepts

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Divisibility by 2, 5, and 10 depends only on the last digit of the number. A number is divisible by 22 if its unit digit is 0,2,4,6,0, 2, 4, 6, or 88. It is divisible by 55 if the unit digit is 00 or 55. It is divisible by 1010 if the unit digit is strictly 00. Visualizing a number line, these are the patterns where multiples land on specific ending markers.

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Divisibility by 3 and 9 depends on the sum of all digits in the number. If the sum of the digits is a multiple of 33, the number is divisible by 33. Similarly, if the sum of the digits is a multiple of 99, the number is divisible by 99. For example, in the number 729729, the sum is 7+2+9=187 + 2 + 9 = 18, which is divisible by both 33 and 99.

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Divisibility by 4 and 8 depends on the last few digits. A number with 33 or more digits is divisible by 44 if the number formed by its last two digits (tens and ones) is divisible by 44. A number with 44 or more digits is divisible by 88 if the number formed by the last three digits is divisible by 88. Think of this as 'zooming in' on the right-hand tail of the number.

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Divisibility by 6 is a composite test. A number is divisible by 66 if it satisfies the divisibility rules for both 22 and 33 simultaneously. This means the number must be an even number and the sum of its digits must be divisible by 33. If it fails either one, it is not divisible by 66.

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Divisibility by 11 involves an alternating sum pattern. Find the difference between the sum of the digits at odd places (starting from the right) and the sum of the digits at even places. If the difference is either 00 or divisible by 1111, then the number is divisible by 1111. You can visualize this by placing the digits in two separate columns (Odd Position vs. Even Position) and comparing their totals.

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General Divisibility Properties: If a number is divisible by another number, then it is divisible by each of the factors of that number. For example, if a number is divisible by 1212, it is also divisible by 2,3,4,2, 3, 4, and 66. Also, if a number is divisible by two co-prime numbers, it is divisible by their product.

📐Formulae

Divisibility by 2: Last digit∈{0,2,4,6,8}\text{Last digit} \in \{0, 2, 4, 6, 8\}

Divisibility by 3: (∑digits)÷3=integer(\sum \text{digits}) \div 3 = \text{integer}

Divisibility by 4: (Last 2 digits)÷4=integer(\text{Last 2 digits}) \div 4 = \text{integer}

Divisibility by 5: Last digit∈{0,5}\text{Last digit} \in \{0, 5\}

Divisibilityby6:Divisible by 2 AND Divisible by 3Divisibility by 6: \text{Divisible by 2} \text{ AND } \text{Divisible by 3}

Divisibility by 8: (Last 3 digits)÷8=integer(\text{Last 3 digits}) \div 8 = \text{integer}

Divisibility by 9: (∑digits)÷9=integer(\sum \text{digits}) \div 9 = \text{integer}

Divisibility by 11: ∣(Sum of digits at odd places)−(Sum of digits at even places)∣=11n, where n=0,1,2...|(\text{Sum of digits at odd places}) - (\text{Sum of digits at even places})| = 11n, \text{ where } n = 0, 1, 2...

💡Examples

Problem 1:

Check if the number 297144297144 is divisible by 66.

Solution:

Step 1: Test for divisibility by 22. The last digit is 44, which is even. So, 297144297144 is divisible by 22. \ Step 2: Test for divisibility by 33. Find the sum of the digits: 2+9+7+1+4+4=272 + 9 + 7 + 1 + 4 + 4 = 27. Since 2727 is divisible by 33 (3×9=273 \times 9 = 27), the number is divisible by 33. \ Conclusion: Since the number is divisible by both 22 and 33, it is divisible by 66.

Explanation:

To check divisibility by 66, we must verify two conditions: the number must be even and the sum of its digits must be a multiple of 33.

Problem 2:

Is the number 71389657138965 divisible by 1111?

Solution:

Step 1: Identify digits at odd positions (from right): 5,9,3,75, 9, 3, 7. Sum =5+9+3+7=24= 5 + 9 + 3 + 7 = 24. \ Step 2: Identify digits at even positions (from right): 6,8,16, 8, 1. Sum =6+8+1=15= 6 + 8 + 1 = 15. \ Step 3: Find the difference: 24−15=924 - 15 = 9. \ Conclusion: Since 99 is not 00 and not divisible by 1111, the number 71389657138965 is not divisible by 1111.

Explanation:

The rule for 1111 requires calculating the difference between the sums of digits in alternating positions. If the result is not 00 or a multiple of 1111, the number fails the test.