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Number Play - Digits in Disguise

Grade 8CBSE

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

🔑Concepts

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A two-digit number with tens digit aa and units digit bb is written in generalized form as 10a+b10a + b.

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A three-digit number with digits a,b,a, b, and cc (where aa is the hundreds digit) is expressed as 100a+10b+c100a + 10b + c.

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Cryptarithms are mathematical puzzles where digits are replaced by letters. Each letter represents a unique digit from 00 to 99.

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In any cryptarithm, the first digit of a number (the leftmost digit) cannot be 00.

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When reversing the digits of a two-digit number abab, the sum of the original and the reversed number (10a+b)+(10b+a)(10a + b) + (10b + a) is always divisible by 1111.

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The difference between a two-digit number and its reverse (10a+b)−(10b+a)(10a + b) - (10b + a) is always divisible by 99.

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The difference between a three-digit number abcabc and its reverse cbacba is (100a+10b+c)−(100c+10b+a)=99(a−c)(100a + 10b + c) - (100c + 10b + a) = 99(a - c), which is divisible by both 99 and 1111.

📐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} 3 A \\ + 2 5 \\ \hline B 2 \end{array}

Solution:

From the units column, we have A+5A + 5 resulting in a number with units digit 22. This means A+5=12A + 5 = 12, which gives A=7A = 7. Now, there is a carry over of 11 to the tens column. In the tens column: 1+3+2=B1 + 3 + 2 = B. Therefore, B=6B = 6. Final values: A=7,B=6A = 7, B = 6.

Explanation:

We analyze the units column first to find AA and then carry over the value to solve for BB in the tens column.

Problem 2:

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

Solution:

In the units column, A×AA \times A ends in AA. Possible values for AA are 0,1,5,0, 1, 5, or 66 (since 02=0,12=1,52=25,62=360^2=0, 1^2=1, 5^2=25, 6^2=36). If A=5A = 5, then 15×5=7515 \times 5 = 75, which does not match 9595. If A=6A = 6, then 16×6=9616 \times 6 = 96, which matches the form 9A9A exactly. Therefore, A=6A = 6.

Explanation:

We test single-digit numbers whose squares end in the same digit and verify which one satisfies the tens place result.

Problem 3:

A three-digit number abcabc is subtracted from its reverse cbacba. If a=7a=7 and c=2c=2, find the result.

Solution:

The generalized form is 99(a−c)99(a - c). Substituting the given values: 99(7−2)=99×5=49599(7 - 2) = 99 \times 5 = 495 Verification using vertical subtraction: 700+10b+2−(200+10b+7)495\begin{array}{r} 700 + 10b + 2 \\ - (200 + 10b + 7) \\ \hline 495 \end{array}

Explanation:

Using the property that (100a+10b+c)−(100c+10b+a)=99(a−c)(100a + 10b + c) - (100c + 10b + a) = 99(a - c) allows us to find the difference regardless of the value of bb.