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A Story of Numbers - The Idea of a Base

Grade 8CBSE

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

πŸ”‘Concepts

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Our standard number system is the Decimal system (Base-10), which uses digits from 00 to 99.

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A two-digit number with digits aa and bb can be written in the generalized form as 10a+b10a + b.

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A three-digit number with digits aa, bb, and cc is expressed as 100a+10b+c100a + 10b + c.

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Reversing the digits of a two-digit number abab gives baba, which is 10b+a10b + a.

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The sum of a two-digit number and its reverse, (10a+b)+(10b+a)(10a + b) + (10b + a), is always a multiple of 1111 because it equals 11(a+b)11(a + b).

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The difference between a two-digit number and its reverse, (10a+b)βˆ’(10b+a)(10a + b) - (10b + a), is always a multiple of 99 because it equals 9(aβˆ’b)9(a - b).

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Cryptarithms are puzzles where letters replace digits in an arithmetic operation, and each letter must represent a unique digit.

πŸ“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)βˆ’(10b+a)=9(aβˆ’b)(10a + b) - (10b + a) = 9(a - b)

(100a+10b+c)βˆ’(100c+10b+a)=99(aβˆ’c)(100a + 10b + c) - (100c + 10b + a) = 99(a - c)

πŸ’‘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:

In the units column, A+5A + 5 results in a number ending in 22. The only digit for AA that satisfies A+5=12A + 5 = 12 is A=7A = 7. We carry over 11 to the tens column. In the tens column, 1(carry)+3+2=B1 (carry) + 3 + 2 = B, which gives B=6B = 6.

Problem 2:

Check if the difference between 7474 and its reverse is divisible by 99.

Solution:

74βˆ’47=2774 - 47 = 27, which is 9Γ—39 \times 3.

Explanation:

Using the generalized form: (10a+b)βˆ’(10b+a)=9(aβˆ’b)(10a + b) - (10b + a) = 9(a - b). Here a=7a=7 and b=4b=4. The difference is 9(7βˆ’4)=9Γ—3=279(7 - 4) = 9 \times 3 = 27. Since 2727 is a multiple of 99, it is divisible by 99.

Problem 3:

Write the number 582582 in expanded form using powers of 1010.

Solution:

5Γ—100+8Γ—10+2Γ—15 \times 100 + 8 \times 10 + 2 \times 1

Explanation:

In the base-10 system, the positions represent units (10010^0), tens (10110^1), and hundreds (10210^2). Thus, 582=(5Γ—102)+(8Γ—101)+(2Γ—100)582 = (5 \times 10^2) + (8 \times 10^1) + (2 \times 10^0).