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Number Play - Digits in Disguise

Grade 7CBSE

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

🔑Concepts

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Generalised Form of Numbers: A two-digit number abab is represented as 10a+b10a + b, and a three-digit number abcabc is represented as 100a+10b+c100a + 10b + c.

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Letters for Digits: In these puzzles, letters replace digits in arithmetic operations. Each letter must represent only one digit, and the first digit of a number cannot be 00.

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Rules for Addition: In vertical addition, if the sum of a column is greater than 99, the tens digit is carried over to the next column on the left.

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Rules for Multiplication: In problems like A×A=...AA \times A = ...A, we look for digits whose square ends in the same digit. The possibilities are 0,1,5,60, 1, 5, 6 since 02=00^2=0, 12=11^2=1, 52=255^2=25, and 62=366^2=36.

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Divisibility Logic: Cryptarithmetic puzzles often use divisibility rules (like 2, 3, 5, 9, and 10) to narrow down the possible values for letters.

📐Formulae

ab=10a+bab = 10a + b

abc=100a+10b+cabc = 100a + 10b + c

abcd=1000a+100b+10c+dabcd = 1000a + 100b + 10c + d

💡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:

In the units column, we have A+5A + 5 which results in a units digit of 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, we have 1(carry)+3+2=B1 (\text{carry}) + 3 + 2 = B, which gives B=6B = 6. Therefore, A=7A = 7 and B=6B = 6.

Explanation:

We solve the units column first to find the carry-over, then apply it to the tens column to find the missing digit.

Problem 2:

Find the value of AA in the following multiplication: 1A×A9A\begin{array}{r} 1A \\ \times A \\ \hline 9A \end{array}

Solution:

We need to find a digit AA such that A×AA \times A ends in AA. The possible digits are 0,1,5,60, 1, 5, 6. If A=1A = 1, then 11×1=11≠9111 \times 1 = 11 \neq 91. If A=5A = 5, then 15×5=75≠9515 \times 5 = 75 \neq 95. If A=6A = 6, then 16×6=9616 \times 6 = 96. This matches the format 9A9A. Thus, A=6A = 6.

Explanation:

By checking the unit digit property of squares, we narrow down the search and verify the result by performing the full multiplication.

Problem 3:

Find AA and BB in the addition: AA+ABA\begin{array}{r} A \\ A \\ +A \\ \hline BA \end{array}

Solution:

The sum is A+A+A=3AA + A + A = 3A. The problem states 3A=BA3A = BA, which in generalised form is 3A=10B+A3A = 10B + A. Subtracting AA from both sides gives 2A=10B2A = 10B, or A=5BA = 5B. Since AA and BB are non-zero digits, if B=1B = 1, then A=5A = 5. If B=2B = 2, then A=10A = 10 (not a single digit). Therefore, A=5A = 5 and B=1B = 1.

Explanation:

Express the vertical addition as an algebraic equation using the place value logic to solve for the digits.