I want to know about the number system playing with number chapter

Answer: The chapter 'Playing with Numbers' introduces the generalized form of numbers ( $10a + b$ and $100a + 10b + c$ ), explains digit-reversal games and divisibility tests through algebraic decomposition, and solves arithmetic puzzles (cryptarithms).

Step-by-step solution

Step 1: Express numbers in generalized form

In our base-10 number system, any two-digit number with tens digit $a$ and units digit $b$ is written as $10a + b$ . Similarly, a three-digit number with hundreds digit $a$ , tens digit $b$ , and units digit $c$ is expressed as $100a + 10b + c$ . This generalized form lets us explore algebraic patterns and divisibility rules.

Step 2: Analyze the number reversal property

When we reverse the digits of a two-digit number $10a + b$ , we get $10b + a$ . Adding the original number to its reverse gives $11a + 11b = 11(a + b)$ , which is always divisible by $11$ and by the sum of its digits $(a + b)$ . Subtracting the reverse gives $(10a + b) - (10b + a) = 9(a - b)$ , which is always divisible by $9$ .

Step 3: Derive divisibility rules using generalized forms

We can rewrite a three-digit number $100a + 10b + c$ by decomposing the coefficients into multiples of $9$ . Since $9(11a + b)$ is always divisible by $3$ and $9$ , the entire number is divisible by $9$ or $3$ if and only if the remaining term $(a + b + c)$ , which is the sum of the digits, is divisible by $9$ or $3$ .

Step 4: Solve letter puzzles and cryptarithms

In cryptarithms or letter puzzles, each letter stands for a unique single digit from $0$ to $9$ . In this addition puzzle, the units column gives $A + 5 = 12$ because the sum ends in $2$ , giving $A = 7$ with a carry of $1$ . Then the tens column gives $1 + 3 + 2 = B$ , which yields $B = 6$ .

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