Playing with Numbers Class 8 NCERT - Notes, Rules & Puzzles

Understanding Numbers in Generalised Form

In our daily lives, we write numbers using place value. For instance, when we write 52, we don't mean 5 + 2; we actually mean 5 tens and 2 ones, which is 50 + 2 = (10 * 5) + 2. This is called the generalised form of a number. For any two-digit number with tens digit 'a' and ones digit 'b', the number is represented as 10a + b. Similarly, a three-digit number with hundreds digit 'a', tens digit 'b', and ones digit 'c' is written as 100a + 10b + c. Writing numbers in this algebraic format helps us analyze their mathematical properties, prove divisibility rules, and solve fun number puzzles. For example, reversing the digits of a two-digit number 10a + b gives us 10b + a. If we add these two numbers, we get (10a + b) + (10b + a) = 11a + 11b = 11(a + b), which is always a multiple of 11! This shows why number games always work out perfectly.

Key Terms You Must Know

Generalised Form

Representing a number as the sum of the products of its digits and their respective place values (e.g., 10a + b for a 2-digit number).

Cryptarithm

A mathematical puzzle where letters or symbols are used to represent digits in an arithmetic operation, and we must find the unique digit for each letter.

Divisibility Rule

A shortcut method to determine whether a given number is divisible by a fixed divisor without performing long division.

Applying Divisibility Rules Logically

1. Divisibility by 10 and 5 — Check the units digit. If the units digit is 0, the number is divisible by 10. If the units digit is either 0 or 5, the number is divisible by 5. This is because any number in generalised form 10a + b is divisible by 5 or 10 depending entirely on the units place 'b'.
2. Divisibility by 2 — Look closely at the units digit. If the units digit is an even number (0, 2, 4, 6, 8), the entire number is divisible by 2. Odd units digits are never divisible by 2.
3. Divisibility by 9 and 3 — Sum up all the individual digits of the number. If the sum of the digits is divisible by 9, then the original number is divisible by 9. Similarly, if the sum of the digits is divisible by 3, the number is divisible by 3.

Exam Tips: Cracking Cryptarithms

When solving alphametic or cryptarithm puzzles (like 3A + 25 = B2) in your Class 8 exams, keep these golden rules in mind:

1. Digits are distinct: Each letter represents only one single digit from 0 to 9. Two different letters cannot represent the same digit.
2. Leading digit rule: The first digit of a number (like A in the number AB) can never be 0.
3. Do not confuse representations: In algebra, 'ab' means 'a multiplied by b'. But in playing with numbers, if A and B are digits, the representation AB represents a two-digit number, which is 10A + B. Always expand using place values before setting up your equations!

Practice Questions with Solutions

• Q: Find the value of A and B in the following addition puzzle:
  3 A
  + 2 5
  -----
  B 2
  A: Step 1: Look at the units column. We have A + 5 which must give a number ending in 2. Since A is a single digit, A + 5 must equal 12.
  Step 2: Solve for A: A = 12 - 5 = 7.
  Step 3: Since A + 5 = 12, there is a carryover of 1 to the tens column.
  Step 4: Now, add the tens column: 1 (carryover) + 3 + 2 = B.
  Step 5: Calculate B: B = 6.
  Final answer: A = 7 and B = 6.

• Q: If a three-digit number 24x is divisible by 9, what is the value of the digit x?
  A: Step 1: Use the divisibility rule for 9: the sum of the digits must be a multiple of 9.
  Step 2: Add the digits of the number: Sum = 2 + 4 + x = 6 + x.
  Step 3: Since x is a single digit (0 to 9), the possible sum (6 + x) must fall between 6 and 15.
  Step 4: The only multiple of 9 in this range is 9 itself.
  Step 5: Set up the equation: 6 + x = 9, which gives x = 3.
  Final answer: x = 3.

• Q: Find the values of A and B in the multiplication puzzle:
  A B
  x 6
  -----
  B B B
  A: Step 1: Represent the puzzle algebraically. The two-digit number AB is 10A + B. The product is 6 * (10A + B) = BBB, which is 111B.
  Step 2: Set up the equation: 60A + 6B = 111B.
  Step 3: Simplify by moving terms: 60A = 105B.
  Step 4: Divide both sides by 15: 4A = 7B.
  Step 5: Since A and B are non-zero single digits and have no common factors, B must be 4 and A must be 7 to satisfy the ratio.
  Final answer: A = 7 and B = 4.

• Q: Check if the number 21436587 is divisible by 3. Show your complete working.
  A: Step 1: Use the divisibility rule for 3: calculate the sum of all digits in the number.
  Step 2: Sum = 2 + 1 + 4 + 3 + 6 + 5 + 8 + 7.
  Step 3: Calculate the total: 2+1=3, 3+4=7, 7+3=10, 10+6=16, 16+5=21, 21+8=29, 29+7=36. The sum is 36.
  Step 4: Test if 36 is divisible by 3. Since 36 divided by 3 is exactly 12 with no remainder, 36 is divisible by 3.
  Step 5: Since the sum of the digits is divisible by 3, the original number is also divisible by 3.
  Final answer: Yes, the number 21436587 is divisible by 3.

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