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Prime Time - Fun with Numbers

Grade 6CBSE

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

🔑Concepts

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A Factor of a number is an exact divisor of that number. For example, the factors of 1212 are 1,2,3,4,6,1, 2, 3, 4, 6, and 1212.

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A Multiple of a number is a number obtained by multiplying it by any natural number. For example, multiples of 55 are 5,10,15,20,…5, 10, 15, 20, \dots.

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Prime Numbers are numbers greater than 11 that have exactly two factors: 11 and the number itself. 22 is the only even prime number.

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Composite Numbers are numbers having more than two factors. The number 11 is neither prime nor composite.

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Perfect Number: A number for which the sum of all its factors is equal to twice the number. Example: 66 is a perfect number because its factors are 1,2,3,61, 2, 3, 6 and 1+2+3+6=12=2×61 + 2 + 3 + 6 = 12 = 2 \times 6.

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Divisibility Rule for 33: A number is divisible by 33 if the sum of its digits is a multiple of 33.

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Divisibility Rule for 1111: Find the difference between the sum of digits at odd places and the sum of digits at even places (starting from the right). If the difference is 00 or divisible by 1111, the number is divisible by 1111.

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HCF (Highest Common Factor): The greatest of the common factors of two or more given numbers.

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LCM (Lowest Common Multiple): The smallest of the common multiples of two or more given numbers.

📐Formulae

Product of two numbers=HCF×LCM\text{Product of two numbers} = \text{HCF} \times \text{LCM}

HCF(a,b)=a×bLCM(a,b)\text{HCF}(a, b) = \frac{a \times b}{\text{LCM}(a, b)}

LCM(a,b)=a×bHCF(a,b)\text{LCM}(a, b) = \frac{a \times b}{\text{HCF}(a, b)}

💡Examples

Problem 1:

Find the HCF of 2424 and 3636 using prime factorization.

Solution:

Prime factorization of 24=2×2×2×3=23×324 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3. Prime factorization of 36=2×2×3×3=22×3236 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2. Common factors are 2×2×3=122 \times 2 \times 3 = 12. So, HCF(24,36)=12\text{HCF}(24, 36) = 12.

Explanation:

HCF is the product of the lowest powers of common prime factors.

Problem 2:

Check if 13311331 is divisible by 1111.

Solution:

Sum of digits at odd places (from right): 1+3=41 + 3 = 4. Sum of digits at even places (from right): 3+1=43 + 1 = 4. Difference: 4−4=04 - 4 = 0. Since the difference is 00, 13311331 is divisible by 1111.

Explanation:

According to the divisibility rule of 1111, a difference of 00 or a multiple of 1111 indicates divisibility.

Problem 3:

Find the difference between the largest 55-digit number and the smallest 55-digit number formed using 1,0,2,3,41, 0, 2, 3, 4 without repetition.

Solution:

Largest number: 4321043210. Smallest number: 1023410234. Calculation: 43210−1023432976\begin{array}{r} 43210 \\ - 10234 \\ \hline 32976 \end{array}

Explanation:

We arrange the digits in descending order for the largest number and ascending order (keeping 00 at the second place) for the smallest number, then perform vertical subtraction.

Problem 4:

If the HCF of two numbers is 66 and their LCM is 3636, and one number is 1212, find the other number.

Solution:

Using the formula HCF×LCM=a×bHCF \times LCM = a \times b: 6×36=12×b6 \times 36 = 12 \times b 216=12×b216 = 12 \times b b=21612=18b = \frac{216}{12} = 18. The other number is 1818.

Explanation:

The product of the HCF and LCM of two numbers always equals the product of the numbers themselves.