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Number Play - Checking Divisibility Quickly

Grade 8CBSE

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

🔑Concepts

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A number is divisible by 1010 if its units digit is 00. If the units digit is any other number, the number is not divisible by 1010.

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A number is divisible by 55 if its units digit is either 00 or 55.

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A number is divisible by 22 if its units digit is an even number (0,2,4,6,0, 2, 4, 6, or 88).

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A number is divisible by 99 if the sum of its digits is divisible by 99. For a number abcabc, it is divisible if (a+b+c)(a + b + c) is a multiple of 99.

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A number is divisible by 33 if the sum of its digits is divisible by 33. For a number abcabc, it is divisible if (a+b+c)(a + b + c) is a multiple of 33.

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Generalized form: Any two-digit number abab can be written as 10a+b10a + b, and any three-digit number abcabc can be written as 100a+10b+c100a + 10b + c.

📐Formulae

N=100a+10b+cN = 100a + 10b + c

Divisibility by 9: (a+b+c)÷9=k, where k∈Z\text{Divisibility by 9: } (a + b + c) \div 9 = k, \text{ where } k \in \mathbb{Z}

Divisibility by 3: (a+b+c)÷3=k, where k∈Z\text{Divisibility by 3: } (a + b + c) \div 3 = k, \text{ where } k \in \mathbb{Z}

General Form: 10nan+10n−1an−1+⋯+101a1+a0\text{General Form: } 10^n a_n + 10^{n-1} a_{n-1} + \dots + 10^1 a_1 + a_0

💡Examples

Problem 1:

Check if the number 214365214365 is divisible by 99.

Solution:

Sum of digits =2+1+4+3+6+5=21= 2 + 1 + 4 + 3 + 6 + 5 = 21. Since 2121 is not divisible by 99, the number 214365214365 is not divisible by 99.

Explanation:

To check divisibility by 99, we add all the digits and see if the resulting sum is a multiple of 99.

Problem 2:

If 21y521y5 is a multiple of 99, where yy is a digit, what is the value of yy?

Solution:

Sum of digits =2+1+y+5=8+y= 2 + 1 + y + 5 = 8 + y. For the number to be divisible by 99, 8+y8 + y must be a multiple of 99. The nearest multiple is 99, so 8+y=9  ⟹  y=18 + y = 9 \implies y = 1.

Explanation:

Since yy is a single digit (00 to 99), the only value that makes the sum a multiple of 99 is 11.

Problem 3:

Find the value of kk if 31k231k2 is divisible by 33.

Solution:

Sum of digits =3+1+k+2=6+k= 3 + 1 + k + 2 = 6 + k. For the number to be divisible by 33, 6+k6 + k must be 6,9,12,…6, 9, 12, \dots. Thus, kk can be 0,3,6,0, 3, 6, or 99.

Explanation:

The sum of digits must be a multiple of 33. We test all single-digit values of kk from 00 to 99 that satisfy this condition.

Problem 4:

Determine if the number 12301230 is divisible by 2,5,2, 5, and 1010.

Solution:

The units digit is 00.

  1. Divisible by 22: Yes (units digit is 00).
  2. Divisible by 55: Yes (units digit is 00).
  3. Divisible by 1010: Yes (units digit is 00).

Explanation:

A units digit of 00 satisfies the requirements for all three divisibility tests simultaneously.