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Algebra Play - Decoding Divisibility Tricks

Grade 8CBSE

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

🔑Concepts

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A two-digit number abab can be written in generalised form as 10a+b10a + b. If the digits are reversed to baba, the number becomes 10b+a10b + a.

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A three-digit number abcabc is written as 100a+10b+c100a + 10b + c.

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Sum of a 2-digit number abab and the number obtained by reversing its digits baba is always divisible by 1111 and the sum of the digits (a+b)(a+b).

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Difference of a 2-digit number abab and its reverse baba (where a>ba > b) is always divisible by 99 and the difference of the digits (a−b)(a-b).

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Difference between a 3-digit number abcabc and its reverse cbacba (where a>ca > c) is always divisible by 9999 and (a−c)(a-c).

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Divisibility by 10: A number is divisible by 10 if its units digit is 00.

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Divisibility by 5: A number is divisible by 5 if its units digit is either 00 or 55.

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Divisibility by 2: A number is divisible by 2 if its units digit is 0,2,4,6,0, 2, 4, 6, or 88.

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Divisibility by 3 and 9: A number is divisible by 3 (or 9) if the sum of its digits is divisible by 3 (or 9).

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Cryptarithmetic: In puzzles where letters replace digits, each letter must represent only one digit, and the first digit of a number cannot be 00.

📐Formulae

10a+b+10b+a=11(a+b)10a + b + 10b + a = 11(a + b)

(10a+b)−(10b+a)=9(a−b)(10a + b) - (10b + a) = 9(a - b)

(100a+10b+c)−(100c+10b+a)=99(a−c)(100a + 10b + c) - (100c + 10b + a) = 99(a - c)

Sum of digits of abc=a+b+c\text{Sum of digits of } abc = a + b + c

💡Examples

Problem 1:

Check if the number 2143658721436587 is divisible by 99.

Solution:

Sum of digits =2+1+4+3+6+5+8+7=36= 2 + 1 + 4 + 3 + 6 + 5 + 8 + 7 = 36. Since 3636 is divisible by 99 (36÷9=436 \div 9 = 4), the original number is divisible by 99.

Explanation:

The divisibility rule for 99 states that if the sum of the digits is a multiple of 99, the number itself is divisible by 99.

Problem 2:

Find the values of AA and BB in the following addition: 3A+25B2\begin{array}{r} 3A \\ + 25 \\ \hline B2 \end{array}

Solution:

In the units column, A+5A + 5 ends in 22. This means A+5=12A + 5 = 12, so A=12−5=7A = 12 - 5 = 7. We carry over 11 to the tens column. In the tens column, 1(carry)+3+2=B1 (carry) + 3 + 2 = B. Therefore, B=6B = 6. So, A=7,B=6A = 7, B = 6.

Explanation:

We use basic addition rules. Since A+5A+5 results in a units digit of 22, AA must be 77 (as 7+5=127+5=12). The tens digit is then the sum of the carry and the digits in the tens place.

Problem 3:

If 24x24x is a multiple of 33, where xx is a digit, what are the possible values of xx?

Solution:

For 24x24x to be divisible by 33, the sum of its digits 2+4+x2 + 4 + x must be divisible by 33. So, 6+x6 + x must be 6,9,12, or 156, 9, 12, \text{ or } 15. If 6+x=6,x=06 + x = 6, x = 0. If 6+x=9,x=36 + x = 9, x = 3. If 6+x=12,x=66 + x = 12, x = 6. If 6+x=15,x=96 + x = 15, x = 9. Possible values of xx are 0,3,6,90, 3, 6, 9.

Explanation:

We apply the divisibility rule for 33 and solve for the single-digit variable xx such that the total sum remains a multiple of 33.