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Number Play - Numbers Tell us Things

Grade 7CBSE

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

🔑Concepts

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Generalized Form: Any 2-digit number with tens digit aa and units digit bb can be written as 10a+b10a + b.

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Generalized Form: Any 3-digit number with digits a,b,ca, b, c can be written as 100a+10b+c100a + 10b + c.

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Reversing 2-digit Numbers: The sum of a 2-digit number and its reverse is always a multiple of 1111, i.e., (10a+b)+(10b+a)=11(a+b)(10a + b) + (10b + a) = 11(a + b).

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Reversing 2-digit Numbers: The difference between a 2-digit number and its reverse is always a multiple of 99, i.e., (10a+b)−(10b+a)=9(a−b)(10a + b) - (10b + a) = 9(a - b).

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

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Cryptarithmetic: These are number puzzles where letters take the place of digits. Each letter must represent a unique digit (0−90-9), and the leading digit of a number cannot be 00.

📐Formulae

N=10a+bN = 10a + b

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

(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)

💡Examples

Problem 1:

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

Solution:

A=7A = 7 and B=6B = 6

Explanation:

Looking at the units column: A+5A + 5 results in a units digit of 22. This means A+5=12A + 5 = 12, which gives A=7A = 7. Carrying over 11 to the tens column: 1+3+2=61 + 3 + 2 = 6. Therefore, B=6B = 6.

Problem 2:

If the 3-digit number 24x24x is divisible by 99, what is the value of xx?

Solution:

x=3x = 3

Explanation:

For a number to be divisible by 99, the sum of its digits must be a multiple of 99. Sum of digits =2+4+x=6+x= 2 + 4 + x = 6 + x. For 6+x6 + x to be divisible by 99, the smallest possible value for xx is 33 (since 6+3=96 + 3 = 9).

Problem 3:

Show that the sum of the digits of a 2-digit number abab and its reverse baba is divisible by 1111. Use the number 4747.

Solution:

47+74=12147 + 74 = 121, and 121=11×11121 = 11 \times 11

Explanation:

Representing 4747 as 10(4)+710(4) + 7 and 7474 as 10(7)+410(7) + 4. Sum =(40+7)+(70+4)=110+11=121= (40 + 7) + (70 + 4) = 110 + 11 = 121. Since 121121 is 11×1111 \times 11, it is divisible by 1111.