Samacheer Kalvi Class 10 Maths Solutions Chapter 2 Numbers and Sequences Exercise 2.2

Question 1. For what values of natural number n, \( 4^n \) can end with the digit 6?
Answer: To find the values of n for which \( 4^n \) ends in the digit 6, we can look at the pattern of powers of 4:
\( 4^1 = 4 \)
\( 4^2 = 16 \)
\( 4^3 = 64 \)
\( 4^4 = 256 \)
We observe that \( 4^n \) ends with the digit 6 when n is an even natural number. For example, \( 4^2 \) and \( 4^4 \) both end in 6. This pattern continues because an even number multiplied by 4 will always result in a number ending in 6 if the previous power ended in 6.
In simple words: When you multiply 4 by itself, the last digit goes back and forth between 4 and 6. It ends in 6 every time the power (n) is an even number.

🎯 Exam Tip: To determine the last digit of a number raised to a power, always look for the pattern of the last digits of its first few powers.

Question 2. If m, n are natural numbers, for what values of m, does \( 2^n \times 5^m \) end in 5?
Answer: Let's analyze the properties of \( 2^n \) and \( 5^m \). For any natural number n, \( 2^n \) will always be an even number. For any natural number m, \( 5^m \) will always be a number ending in 5 (e.g., 5, 25, 125, etc.), which means it is an odd number. When an even number is multiplied by an odd number that ends in 5, the resulting product will always end in 0. For example, \( 2 \times 5 = 10 \), \( 4 \times 5 = 20 \), \( 2 \times 25 = 50 \). Therefore, the product \( 2^n \times 5^m \) will always end in 0. This means that \( 2^n \times 5^m \) can never end in the digit 5 for any natural value of m.
In simple words: A number like \( 2^n \) is always even. A number like \( 5^m \) always ends in 5. When you multiply an even number by a number ending in 5, the answer will always end in 0, not 5. So, \( 2^n \times 5^m \) can never end in 5.

🎯 Exam Tip: Remember that any number multiplied by an even number will always be an even number, and if one of the factors is a multiple of 5, the product will end in either 0 or 5. If it's also even, it must end in 0.

Question 3. Find the H.C.F. of 252525 and 363636.
Answer: To find the Highest Common Factor (H.C.F.) of 252525 and 363636, we use Euclid's Division Algorithm. This method involves repeatedly dividing the larger number by the smaller number and replacing the larger number with the smaller number and the smaller number with the remainder until the remainder is zero.
\( 363636 = 252525 \times 1 + 111111 \)
Since the remainder \( 111111 \neq 0 \), we continue:
\( 252525 = 111111 \times 2 + 30303 \)
Since the remainder \( 30303 \neq 0 \), we continue:
\( 111111 = 30303 \times 3 + 20202 \)
Since the remainder \( 20202 \neq 0 \), we continue:
\( 30303 = 20202 \times 1 + 10101 \)
Since the remainder \( 10101 \neq 0 \), we continue:
\( 20202 = 10101 \times 2 + 0 \)
Here, the remainder is 0. The divisor at this step is 10101, which is the H.C.F.
Therefore, the H.C.F. of 252525 and 363636 is 10101.
In simple words: We find the H.C.F. by dividing the bigger number by the smaller number, and then dividing the smaller number by the leftover. We keep doing this until there is no leftover. The last number we used for dividing is the H.C.F. In this case, it's 10101.

🎯 Exam Tip: Clearly show each step of the Euclidean algorithm, noting the remainder at each stage until zero is reached. The final divisor is the H.C.F.

Question 4. If \( 13824 = 2^a \times 3^b \) then find a and b?
Answer: To find the values of a and b, we need to find the prime factorization of 13824. We can do this by repeatedly dividing 13824 by its prime factors.
Divide by 2: \( 13824 \div 2 = 6912 \)
Divide by 2: \( 6912 \div 2 = 3456 \)
Divide by 2: \( 3456 \div 2 = 1728 \)
Divide by 2: \( 1728 \div 2 = 864 \)
Divide by 2: \( 864 \div 2 = 432 \)
Divide by 2: \( 432 \div 2 = 216 \)
Divide by 2: \( 216 \div 2 = 108 \)
Divide by 2: \( 108 \div 2 = 54 \)
Divide by 2: \( 54 \div 2 = 27 \)
Now divide by 3: \( 27 \div 3 = 9 \)
Divide by 3: \( 9 \div 3 = 3 \)
Divide by 3: \( 3 \div 3 = 1 \)
So, the prime factorization of 13824 is \( 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \).
This can be written as \( 13824 = 2^9 \times 3^3 \).
Comparing this with the given form \( 13824 = 2^a \times 3^b \), we can see that:
\( a = 9 \)
\( b = 3 \)
In simple words: We break down the number 13824 into its prime factors. We find that it is 2 multiplied by itself 9 times, and 3 multiplied by itself 3 times. So, 'a' is 9 and 'b' is 3.

🎯 Exam Tip: Always perform prime factorization carefully, dividing by the smallest prime numbers first, to ensure no factors are missed.

Question 5. If \( p_1^{x_1} \times p_2^{x_2} \times p_3^{x_3} \times p_4^{x_4} = 113400 \) where \( p_1, p_2, p_3, p_4 \) are primes in ascending order and \( x_1, x_2, x_3, x_4 \) are integers, find the value of \( p_1, p_2, p_3, p_4 \) and \( x_1, x_2, x_3, x_4 \).
Answer: We need to find the prime factorization of 113400.
First, divide 113400 by prime numbers in ascending order:
\( 113400 \div 2 = 56700 \)
\( 56700 \div 2 = 28350 \)
\( 28350 \div 2 = 14175 \) (So, \( 2^3 \) is a factor)
Next, try dividing by 3:
\( 14175 \div 3 = 4725 \)
\( 4725 \div 3 = 1575 \)
\( 1575 \div 3 = 525 \)
\( 525 \div 3 = 175 \) (So, \( 3^4 \) is a factor)
Next, try dividing by 5:
\( 175 \div 5 = 35 \)
\( 35 \div 5 = 7 \) (So, \( 5^2 \) is a factor)
Finally, divide by 7:
\( 7 \div 7 = 1 \) (So, \( 7^1 \) is a factor)
Thus, the prime factorization of 113400 is \( 2^3 \times 3^4 \times 5^2 \times 7^1 \).
Comparing this with \( p_1^{x_1} \times p_2^{x_2} \times p_3^{x_3} \times p_4^{x_4} \), where \( p_1, p_2, p_3, p_4 \) are primes in ascending order, we get:
\( p_1 = 2 \), \( x_1 = 3 \)
\( p_2 = 3 \), \( x_2 = 4 \)
\( p_3 = 5 \), \( x_3 = 2 \)
\( p_4 = 7 \), \( x_4 = 1 \)
In simple words: We break down the number 113400 into its smallest building blocks (prime factors). We find that it is made of three 2s, four 3s, two 5s, and one 7. So, the prime numbers are 2, 3, 5, 7, and their counts (powers) are 3, 4, 2, 1.

🎯 Exam Tip: Always list prime factors in ascending order for standard representation, and ensure all exponents are correctly identified from the factorization.

Question 6. Find the L.C.M. and H.C.F. of 408 and 170 by applying the fundamental theorem of Arithmetic.
Answer: First, we find the prime factorization of each number using the fundamental theorem of Arithmetic.
For 408:
\( 408 \div 2 = 204 \)
\( 204 \div 2 = 102 \)
\( 102 \div 2 = 51 \)
\( 51 \div 3 = 17 \)
\( 17 \div 17 = 1 \)
So, \( 408 = 2 \times 2 \times 2 \times 3 \times 17 = 2^3 \times 3^1 \times 17^1 \).
For 170:
\( 170 \div 2 = 85 \)
\( 85 \div 5 = 17 \)
\( 17 \div 17 = 1 \)
So, \( 170 = 2 \times 5 \times 17 = 2^1 \times 5^1 \times 17^1 \).
To find the H.C.F. (Highest Common Factor), we take the product of the common prime factors raised to the lowest power they appear in either factorization.
Common prime factors are 2 and 17.
Lowest power of 2 is \( 2^1 \).
Lowest power of 17 is \( 17^1 \).
\( \text{H.C.F.} = 2^1 \times 17^1 = 2 \times 17 = 34 \).
To find the L.C.M. (Least Common Multiple), we take the product of all prime factors (common and uncommon) raised to the highest power they appear in either factorization.
Highest power of 2 is \( 2^3 \).
Highest power of 3 is \( 3^1 \).
Highest power of 5 is \( 5^1 \).
Highest power of 17 is \( 17^1 \).
\( \text{L.C.M.} = 2^3 \times 3^1 \times 5^1 \times 17^1 = 8 \times 3 \times 5 \times 17 = 24 \times 85 = 2040 \).
Therefore, H.C.F. = 34 and L.C.M. = 2040.
In simple words: First, we break down both 408 and 170 into their prime factors. For H.C.F., we pick common prime factors with their smallest power. For L.C.M., we pick all prime factors (common and not common) with their biggest power. We find that the H.C.F. is 34 and the L.C.M. is 2040.

🎯 Exam Tip: Always list all prime factors for each number first. Remember that H.C.F. uses the *lowest* powers of *common* primes, while L.C.M. uses the *highest* powers of *all* primes.

Question 7. Find the greatest number consisting of 6 digits which is exactly divisible by 24, 15, 36?
Answer: To find the greatest 6-digit number exactly divisible by 24, 15, and 36, we first need to find the Least Common Multiple (L.C.M.) of these three numbers.
Prime factorization of each number:
\( 24 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3^1 \)
\( 15 = 3 \times 5 = 3^1 \times 5^1 \)
\( 36 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2 \)
The L.C.M. is found by taking the highest power of all prime factors present:
Highest power of 2 is \( 2^3 \).
Highest power of 3 is \( 3^2 \).
Highest power of 5 is \( 5^1 \).
\( \text{L.C.M.} = 2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 = 72 \times 5 = 360 \).
Now, the greatest 6-digit number is 999999.
To find the greatest 6-digit number divisible by 360, we divide 999999 by 360 and subtract the remainder from 999999.
\( 999999 \div 360 \)
We perform the division:
\( 999999 = 360 \times 2777 + 279 \)
The remainder is 279.
So, the greatest 6-digit number exactly divisible by 360 is \( 999999 - 279 = 999720 \).
In simple words: First, we find the smallest number that 24, 15, and 36 can all divide into, which is 360. Then, we take the biggest 6-digit number (999999) and divide it by 360. We get a leftover part. We subtract this leftover part from 999999 to find the largest 6-digit number that 360 can divide perfectly. This number is 999720.

🎯 Exam Tip: Remember to calculate the L.C.M. of all divisors first. Then, divide the largest number of the given digits by the L.C.M. and subtract the remainder from it.

Question 8. What is the smallest number that when divided by three numbers such as 35, 56 and 91 leaves remainder 7 in each case?
Answer: To find the smallest number that leaves a remainder of 7 when divided by 35, 56, and 91, we first need to find the Least Common Multiple (L.C.M.) of these three numbers.
Prime factorization of each number:
\( 35 = 5 \times 7 \)
\( 56 = 2 \times 2 \times 2 \times 7 = 2^3 \times 7 \)
\( 91 = 7 \times 13 \)
The L.C.M. is found by taking the highest power of all prime factors present:
Highest power of 2 is \( 2^3 \).
Highest power of 5 is \( 5^1 \).
Highest power of 7 is \( 7^1 \).
Highest power of 13 is \( 13^1 \).
\( \text{L.C.M.} = 2^3 \times 5 \times 7 \times 13 = 8 \times 5 \times 7 \times 13 = 40 \times 91 = 3640 \).
This L.C.M. (3640) is the smallest number that is exactly divisible by 35, 56, and 91.
If a number leaves a remainder of 7 in each case, it means the number is 7 more than the L.C.M.
Required number = L.C.M. + Remainder
Required number = \( 3640 + 7 = 3647 \).
In simple words: First, we find the smallest number that 35, 56, and 91 can all divide perfectly into. This number is 3640. Since we want a leftover of 7 each time, we just add 7 to this smallest perfect number. So, the answer is 3647.

🎯 Exam Tip: When a number leaves a common remainder for multiple divisors, first find the L.C.M. of the divisors, then add the common remainder to the L.C.M.

Question 9. Find the least number that is divisible by the first ten natural numbers?
Answer: To find the least number divisible by the first ten natural numbers (1, 2, 3, 4, 5, 6, 7, 8, 9, 10), we need to find their Least Common Multiple (L.C.M.).
First, list the prime factorization of each number:
\( 1 = 1 \)
\( 2 = 2^1 \)
\( 3 = 3^1 \)
\( 4 = 2^2 \)
\( 5 = 5^1 \)
\( 6 = 2^1 \times 3^1 \)
\( 7 = 7^1 \)
\( 8 = 2^3 \)
\( 9 = 3^2 \)
\( 10 = 2^1 \times 5^1 \)
To find the L.C.M., we take the highest power of each prime factor that appears in any of the factorizations:
Highest power of 2 is \( 2^3 \) (from 8).
Highest power of 3 is \( 3^2 \) (from 9).
Highest power of 5 is \( 5^1 \) (from 5 or 10).
Highest power of 7 is \( 7^1 \) (from 7).
Multiply these highest powers together:
\( \text{L.C.M.} = 2^3 \times 3^2 \times 5^1 \times 7^1 = 8 \times 9 \times 5 \times 7 \)
\( \text{L.C.M.} = 72 \times 35 \)
\( \text{L.C.M.} = 2520 \).
Therefore, the least number that is divisible by the first ten natural numbers is 2520.
In simple words: We need to find the smallest number that 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 can all divide into without any leftover. We do this by listing out the prime factors of each number and taking the highest power of each unique prime factor. When we multiply these highest powers together, we get 2520.

🎯 Exam Tip: To find the L.C.M. of a series of numbers, always use prime factorization, ensuring you select the highest power for each unique prime factor present across all numbers.

Modular Arithmetic

Two integers "a" and "b" are said to be congruent modulo n if their difference \( (b - a) \) is an integer multiple of n. This means that \( b - a = k \times n \) for some integer k. We can write this relationship as \( a \equiv b \pmod n \). In simpler terms, a and b have the same remainder when divided by n. Modular arithmetic is very useful in computer science for tasks like hashing and cryptography.

Euclid's Division Lemma and Modular Arithmetic

Euclid's Division Lemma states that for any two positive integers, say n and m, there exist unique integers q (quotient) and r (remainder) such that \( n = m \times q + r \), where \( 0 \le r < m \). This lemma is the basis for understanding modular arithmetic. In modular arithmetic, if \( n = mq + r \), then n is congruent to r modulo m, written as \( n \equiv r \pmod m \). This means that n and r have the same remainder when divided by m. For instance, if you divide 10 by 3, you get a quotient of 3 and a remainder of 1. So, \( 10 \equiv 1 \pmod 3 \).