CBSE Class 10 Maths HOTs Real Numbers Set B

Very Short Answer Type Questions

Question. Explain why 13233343563715 is a composite number?
Answer: The given number ends in 5. Hence it is a multiple of 5. Therefore it is a composite number.

Question. \( a \) and \( b \) are two positive integers such that the least prime factor of \( a \) is 3 and the least prime factor of \( b \) is 5. Then calculate the least prime factor of \( (a + b) \).
Answer: \( a \) and \( b \) are two positive integers such that the least prime factor of \( a \) is 3 and the least prime factor of \( b \) is 5. Then least prime factor of \( (a + b) \) is 2.

Question. What is the HCF of the smallest composite number and the smallest prime number?
Answer: The smallest prime number is 2 and the smallest composite number is \( 4 = 2^2 \).
Hence, required \( HCF (2^2, 2) = 2 \).

Question. Calculate the HCF of \( 3^3 \times 5 \) and \( 3^2 \times 5^2 \).
Answer: We have \( 3^3 \times 5 = 3^2 \times 5 \times 3 \)
\( 3^2 \times 5^2 = 3^2 \times 5 \times 5 \)
\( HCF (3^3 \times 5, 3^2 \times 5^2) = 3^2 \times 5 \)
\( = 9 \times 5 = 45 \)

Question. If \( HCF (a, b) = 12 \) and \( a \times b = 1,800 \), then find \( LCM (a, b) \).
Answer: We know that
\( HCF(a, b) \times LCM (a, b) = a \times b \)
Substituting the values we have
\( 12 \times LCM (a, b) = 1800 \)
or, \( LCM(a, b) = \frac{1,800}{12} = 150 \)

Short Answer Type Questions - I

Question. Find HCF of the numbers given below: \( k, 2k, 3k, 4k \) and \( 5k \), where \( k \) is a Positive integer.
Answer: Here we can see easily that \( k \) is common factor between all and this is highest factor Thus HCF of \( k, 2k, 3k, 4k \) and \( 5k \), is \( k \).

Question. Find the HCF and LCM of 90 and 144 by the method of prime factorization.
Answer: We have \( 90 = 9 \times 10 \)
\( = 2 \times 3^2 \times 5 \)
and \( 144 = 16 \times 9 \)
\( = 2^4 \times 3^2 \)
\( HCF = 2 \times 3^2 = 18 \)
\( LCM = 2^4 \times 3^2 \times 5 = 720 \)

Question. Using Euclid’s algorithm, find the HCF of 240 and 288.
Answer: We have \( 240 = 228 \times 1 + 12 \)
and \( 288 = 12 \times 19 + 0 \)
Hence, HCF of 240 and 228 = 12

Question. Given that \( HCF (306, 1314) = 18 \). Find \( LCM (306, 1314) \)
Answer: We have \( HCF (306, 1314) = 18 \)
\( LCM (306, 1314) = ? \)
Let \( a = 306 \) and \( b = 1314 \), then we have
\( LCM(a, b) \times HCF (a, b) = a \times b \)
Substituting values we have
or, \( LCM(a, b) \times 18 = 306 \times 1314 \)
or, \( LCM (a, b) = \frac{306 \times 1314}{18} \)
\( LCM(306, 1314) = 22,338 \)

Question. Explain why \( (7 \times 13 \times 11) + 11 \) and \( (7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1) + 3 \) are composite numbers.
Answer: \( (7 \times 13 \times 11) + 11 = 11 \times (7 \times 13 + 1) \)
\( = 11 \times (91 + 1) \)
\( = 11 \times 92 \)
and
\( (7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1) + 3 = 3(7 \times 6 \times 5 \times 4 \times 2 \times 1 + 1) \)
\( = 3 \times (1681) = 3 \times 41 \times 41 \)
Since given numbers have more than two prime factors, both number are composite.

Question. Explain whether \( 3 \times 12 \times 101 + 4 \) is a prime number or a composite number.
Answer: A prime number (or a prime) is a natural number greater than 1 that cannot be formed by multiplying two smaller natural numbers. A natural number greater than 1 that is not prime is called a composite number. For example, 5 is prime because the only ways of writing it as a product, \( 1 \times 5 \) or \( 5 \times 1 \), involve 5 itself. However, 6 is composite because it is the product of two numbers \( (2 \times 3) \) that are both smaller than 6. Every composite number can be written as the product of two or more (not necessarily distinct) primes.
\( 3 \times 12 \times 101 + 4 = 4(3 \times 3 \times 101 + 1) \)
\( = 4(909 + 1) \)
\( = 4(910) \)
\( = 2 \times 2 \times (10 \times 7 \times 13) \)
\( = 2 \times 2 \times 2 \times 5 \times 7 \times 13 \)
\( = \) a composite number

Question. Find the smallest natural number by which 1200 should be multiplied so that the square root of the product is a rational number.
Answer: We have \( 1200 = 12 \times 100 \)
\( = 4 \times 3 \times 4 \times 25 \)
\( = 4^2 \times 3 \times 5^2 \)
Here if we multiply by 3, then its square root will be a rational number because all power will be 2. Thus the required smallest natural number is 3.

Question. Show that any positive even integer can be written in the form \( 6q, 6q + 2 \) or \( 6q + 4 \), where \( q \) is an integer.
Answer: Let \( a \) be any positive integer, then by Euclid’s division algorithm \( a \) can be written as
\( a = bq + r \)
Take \( b = 6 \), then \( 0 \le r < 6 \) because \( 0 \le r < b \),
Thus \( a = 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4, 6q + 5 \)
Here \( 6q, 6q + 2 \) and \( 6q + 4 \) are divisible by 2 and so \( 6q, 6q + 2 \) and \( 6q + 4 \) are even positive integers.
Hence \( a \) is always an even integer if
\( a = 6q, 6q + 2, 6q + 4 \)

Question. Show that any positive odd integer is of the form \( 4q + 1 \) or \( 4q + 3 \), where \( q \) is some integer.
Answer: Let \( a \) be any positive integer, then by Euclid’s division algorithm \( a \) can be written as
\( a = bq + r \)
Take \( b = 4 \), then \( 0 \le r < 4 \) because \( 0 \le r < b \),
Thus \( a = 4q, 4q + 1, 4q + 2, 4q + 3 \)
Here we can see easily that \( a = 4q, 4q + 2 \) are even, as they are divisible by 2. Also \( 4q + 1, 4q + 3 \) are odd, as they are not divisible by 2.
Thus any positive integer which has the form of \( (4q + 1) \) or \( (4q + 3) \) is odd.

Question. Can two numbers have 15 as their HCF and 175 as their LCM? Give reasons.
Answer: LCM of two numbers should be exactly divisible by their HCF. Since, 15 does not divide 175, two numbers cannot have their HCF as 15 and LCM as 175.

Question. Check whether \( 4^n \) can end with the digit 0 for any natural number \( n \).
Answer: If the number \( 4^n \), for any \( n \), were to end with the digit zero, then it would be divisible by 5 and 2.
That is, the prime factorization of \( 4^n \) would contain the prime 5 and 2. This is not possible because the only prime in the factorization of \( 4^n = (2^2)^n = 2^{2n} \) is 2. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorization of \( 4^n \). So, there is no natural number \( n \) for which \( 4^n \) ends with the digit zero. Hence \( 4^n \) cannot end with the digit zero.

Question. Show that \( 7^n \) cannot end with the digit zero, for any natural number \( n \).
Answer: If the number \( 7^n \), for any \( n \), were to end with the digit zero, then it would be divisible by 5 and 2.
That is, the prime factorization of \( 7^n \) would contain the prime 5 and 2. This is not possible because the only prime in the factorization of \( 7^n = (1 \times 7)^n \) is 7. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorization of \( 7^n \). So, there is no natural number \( n \) for which \( 7^n \) ends with the digit zero. Hence \( 7^n \) cannot end with the digit zero.

Question. Check whether \( (15)^n \) can end with digit 0 for any \( n \in N \).
Answer: If the number \( (15)^n \), for any \( n \), were to end with the digit zero, then it would be divisible by 5 and 2.
That is, the prime factorization of \( (15)^n \) would contain the prime 5 and 2. This is not possible because the only prime in the factorization of \( (15)^n = (3 \times 5)^n \) are 3 and 5. The uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorization of \( (15)^n \). Since there is no prime factor 2, \( (15)^n \) cannot end with the digit zero.

Question. The length, breadth and height of a room are 8 m 50 cm, 6 m 25 cm and 4 m 75 cm respectively. Find the length of the longest rod that can measure the dimensions of the room exactly.
Answer: Here we have to determine the HCF of all length which can measure all dimension.
Length, \( l = 8m\ 50cm = 850\ cm \)
\( = 50 \times 17 = 2 \times 5^2 \times 17 \)
Breadth \( b = 6m\ 25cm = 625\ cm \)
\( = 25 \times 25 = 5^2 \times 5^2 = 5^4 \)
Height \( h = 4m\ 75cm = 475\ cm \)
\( = 25 \times 19 = 5^2 \times 19 \)
\( HCF(l, b, h) = HCF (850, 625, 475) \)
\( = 5^2 = 25 \)

Question. If two positive integers \( p \) and \( q \) are written as \( p = a^2b^3 \) and \( q = a^3b \), where \( a \) and \( b \) are prime numbers than verify \( LCM(p,q) \times HCF(p,q) = p \cdot q \)
Answer: We have \( p = a^2b^3 = a \times a \times b \times b \times b \)
and \( q = a^3b = a \times a \times a \times b \)
Now \( LCM(p, q) = a \times a \times a \times b \times b \times b = a^3b^3 \)
and \( HCF(p, q) = a \times a \times b = a^2b \)
\( LCM(p, q) \times HCF(p, q) = a^3b^3 \times a^2b \)
\( = a^5b^4 \)
\( = a^2b^3 \times a^3b \)
\( = pq \)

Short Answer Type Questions - II

Question. Find the HCF of 180, 252 and 324 by Euclid’s Division algorithm.
Answer: We have \( 324 = 252 \times 1 + 72 \)
\( 252 = 72 \times 3 + 36 \)
\( 72 = 36 \times 2 + 0 \)
Thus \( HCF(324, 252) = 36 \)
Now \( 180 = 36 \times 5 + 0 \)
Thus \( HCF(36, 180) = 36 \)
Thus HCF of 180, 252, and 324 is 36.
Hence required number = \( 999999 - 63 = 999936 \)

Question. Use Euclid division lemma to show that the square of any positive integer cannot be of the form \( 5m + 2 \) or \( 5m + 3 \) for some integer \( m \).
Answer: Let \( a \) be any positive integer, then by Euclid’s division algorithm \( a \) can be written as
\( a = bq + r \), \( 0 \le r < b \) and \( q \in w \)
Take \( b = 5 \), then \( 0 \le r < 5 \) because \( 0 \le r < b \)
Thus \( a = 5q, 5q + 1, 5q + 2, 5q + 3 \) and \( 5q + 4 \),
Now \( a^2 = (5q)^2 = 25q^2 = 5(5q^2) = 5m \)
\( a^2 = (5q + 1)^2 = 25q^2 + 10q + 1 = 5(5q^2 + 2q) + 1 = 5m + 1 \)
\( a^2 = (5q + 2)^2 = 25q^2 + 20q + 4 = 5(5q^2 + 4q) + 4 = 5m + 4 \)
Similarly \( a^2 = (5q + 3)^2 = 5m + 4 \)
and \( a^2 = (5q + 4)^2 = 5m + 1 \)
Thus square of any positive integer cannot be of the form \( 5m + 2 \) or \( 5m + 3 \).

Question. Show that numbers \( 8^n \) can never end with digit 0 of any natural number \( n \).
Answer: If the number \( 8^n \), for any \( n \), were to end with the digit zero, then it would be divisible by 5 and 2. That is, the prime factorization of \( 8^n \) would contain the prime 5 and 2. This is not possible because the only prime in the factorization of \( (8)^n = (2^3)^n = 2^{3n} \) is 2. The uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorization of \( (8)^n \). Since there is no prime factor 5, \( (8)^n \) cannot end with the digit zero.

Question. Find the HCF, by Euclid’s division algorithm of the numbers 92690, 7378 and 7161.
Answer: By using Euclid’s Division Lemma, we have
\( 92690 = 7378 \times 12 + 4154 \)
\( 7378 = 4154 \times 1 + 3224 \)
\( 4154 = 3224 \times 1 + 930 \)
\( 3224 = 930 \times 3 + 434 \)
\( 930 = 434 \times 2 + 62 \)
\( 434 = 62 \times 7 + 0 \)
\( HCF (92690, 7378) = 62 \)
Now, using Euclid’s Division Lemma on 7161 and 62, we have
\( 7161 = 62 \times 115 + 31 \)
\( 62 = 31 \times 2 + 0 \)
Thus \( HCF (7161, 62) = 31 \)
Hence, HCF of 92690, 7378 and 7161 is 31.

Question. 144 cartons of Coke cans and 90 cartons of Pepsi cans are to be stacked in a canteen. If each stack is of the same height and if it equal contain cartons of the same drink, what would be the greatest number of cartons each stack would have?
Answer: The required answer will be HCF of 144 and 90.
\( 144 = 2^4 \times 3^2 \)
\( 90 = 2 \times 3^2 \times 5 \)
\( HCF(144, 90) = 2 \times 3^2 = 18 \)
Thus each stack would have 18 cartons.

Question. Three bells toll at intervals of 9, 12, 15 minutes respectively. If they start tolling together, after what time will they next toll together?
Answer: The required answer is the LCM of 9, 12, and 15 minutes.
Finding prime factor of given number we have,
\( 9 = 3 \times 3 = 3^2 \)
\( 12 = 2 \times 2 \times 3 = 2^2 \times 3 \)
\( 15 = 3 \times 5 \)
\( LCM(9, 12, 15) = 2^2 \times 3^2 \times 5 \)
\( = 180 \text{ minutes} \)
The bells will toll next together after 180 minutes.

Question. Find HCF and LCM of 16 and 36 by prime factorization and check your answer.
Answer: Finding prime factor of given number we have,
\( 16 = 2 \times 2 \times 2 \times 2 = 2^4 \)
\( 36 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2 \)
\( HCF(16, 36) = 2^2 = 4 \)
\( LCM (16, 36) = 2^4 \times 3^2 \)
\( = 16 \times 9 = 144 \)
To check HCF and LCM by using formula
\( HCF(a, b) \times LCM(a, b) = a \times b \)
or, \( 4 \times 144 = 16 \times 36 \)
\( 576 = 576 \)
Thus LHS = RHS

Question. Find the HCF and LCM of 510 and 92 and verify that HCF × LCM = Product of two given numbers.
Answer: Finding prime factor of given number we have,
\( 92 = 2^2 \times 23 \)
\( 510 = 30 \times 17 = 2 \times 3 \times 5 \times 17 \)
\( HCF (510, 92) = 2 \)
\( LCM (510, 92) = 2^2 \times 23 \times 3 \times 5 \times 17 \)
\( = 23460 \)
\( HCF (510, 92) \times LCM (510, 92) \)
\( = 2 \times 23460 = 46920 \)
Product of two numbers \( = 510 \times 92 = 46920 \)
Hence, \( HCF \times LCM = \text{Product of two numbers} \)

Question. The HCF of 65 and 117 is expressible in the form \( 65m - 117 \). Find the value of \( m \). Also find the LCM of 65 and 117 using prime factorization method.
Answer: Finding prime factor of given number we have,
\( 117 = 13 \times 3^2 \)
\( 65 = 13 \times 5 \)
\( HCF(117, 65) = 13 \)
\( LCM(117, 65) = 13 \times 5 \times 3^2 = 585 \)
\( HCF = 65m - 117 \)
\( 13 = 65m - 117 \)
\( 65m = 117 + 13 = 130 \)
\( m = \frac{130}{65} = 2 \)

Question. Show that any positive odd integer is of the form \( 6q + 1, 6q + 3 \) or \( 6q + 5 \), where \( q \) is some integer.
Answer: Let \( a \) be any positive integer, then by Euclid’s division algorithm \( a \) can be written as
\( a = bq + r \)
Take \( b = 6 \), then \( 0 \le r < 6 \) because \( 0 \le r < b \),
Thus \( a = 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4, 6q + 5 \)
Here \( 6q, 6q + 2 \) and \( 6q + 4 \) are divisible by 2 and so \( 6q, 6q + 2 \) and \( 6q + 4 \) are even positive integers.
But \( 6q + 1, 6q + 3, 6q + 5 \) are odd, as they are not divisible by 2.
Thus any positive odd integer is of the form \( 6q + 1, 6q + 3 \) or \( 6q + 5 \).

Question. Show that exactly one of the number \( n, n + 2 \) or \( n + 4 \) is divisible by 3.
Answer: If \( n \) is divisible by 3, clearly \( n + 2 \) and \( n + 4 \) is not divisible by 3.
If \( n \) is not divisible by 3, then two case arise as given below.
Case 1: \( n = 3k + 1 \)
\( n + 2 = 3k + 1 + 2 = 3k + 3 = 3(k + 1) \)
and \( n + 4 = 3k + 1 + 4 = 3k + 5 = 3(k + 1) + 2 \)
We can clearly see that in this case \( n + 2 \) is divisible by 3 and \( n + 4 \) is not divisible by 3. Thus in this case only \( n + 2 \) is divisible by 3.
Case 2: \( n = 3k + 2 \)
\( n + 2 = 3k + 2 + 2 = 3k + 4 = 3(k + 1) + 1 \)
and \( n + 4 = 3k + 2 + 4 = 3k + 6 = 3(k + 2) \)
We can clearly see that in this case \( n + 4 \) is divisible by 3 and \( n + 2 \) is not divisible by 3. Thus in this case only \( n + 4 \) is divisible by 3.
Hence, exactly one of the numbers \( n, n + 2, n + 4 \) is divisible by 3.

Long Answer Type Questions

Question. Find HCF and LCM of 378, 180 and 420 by prime factorization method. Is HCF × LCM of these numbers equal to the product of the given three numbers?
Answer: Finding prime factor of given number we have,
\( 378 = 2 \times 3^3 \times 7 \)
\( 180 = 2^2 \times 3^2 \times 5 \)
\( 420 = 2^2 \times 3 \times 7 \times 5 \)
\( HCF(378, 180, 420) = 2 \times 3 = 6 \)
\( LCM(378, 180, 420) = 2^2 \times 3^3 \times 5 \times 7 \)
\( = 4 \times 27 \times 5 \times 7 = 3780 \)
\( HCF \times LCM = 6 \times 3780 = 22680 \)
Product of given numbers \( = 378 \times 180 \times 420 = 28576800 \)
No, \( HCF \times LCM \neq \text{Product of three numbers} \)

Question. State Fundamental theorem of Arithmetic. Find LCM of numbers 2520 and 10530 by prime factorization by 3.
Answer: The fundamental theorem of arithmetic (FTA), also called the unique factorization theorem or the unique prime-factorization theorem, states that every integer greater than 1 either is prime itself or is the product of a unique combination of prime numbers.
                    OR
Every composite number can be expressed as the product powers of primes and this factorization is unique.
Finding prime factor of given number we have,
\(2520 = 20 \times 126 = 20 \times 6 \times 21\)
\(= 2^3 \times 3^2 \times 5 \times 7\)
\(10530 = 30 \times 351 = 30 \times 9 \times 39\)
\(= 30 \times 9 \times 3 \times 13\)
\(= 2 \times 3^4 \times 5 \times 13\)
\(LCM(2520, 10530) = 2^3 \times 3^4 \times 5 \times 7 \times 13\)
\(= 294840\)

Question. Can the number \(6^n\), \(n\) being a natural number, end with the digit 5 ? Give reasons.
Answer: If the number \(6^n\) for any \(n\), were to end with the digit five, then it would be divisible by 5.
That is, the prime factorization of \(6^n\) would contain the prime 5. This is not possible because the only prime in the factorization of \(6^n = (2 \times 3)^n\) are 2 and 3. The uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorization of \(6^n\). Since there is no prime factor 5, \(6^n\) cannot end with the digit five.

Question. State Fundamental theorem of Arithmetic. Is it possible that HCF and LCM of two numbers be 24 and 540 respectively. Justify your answer.
Answer: Fundamental theorem of Arithmetic : Every integer greater than one either is prime itself or is the product of prime numbers and that this product is unique. Up to the order of the factors.
LCM of two numbers should be exactly divisible by their HCF. In other words LCM is always a multiple of HCF. Since, 24 does not divide 540 two numbers cannot have their HCF as 24 and LCM as 540.
\(HCF = 24\)
\(LCM = 540\)
\(\frac{LCM}{HCF} = \frac{540}{24} = 22.5\) not an integer

Question. Find the HCF of 256 and 36 using Euclid’s Division Algorithm. Also, find their LCM and verify that HCF \(\times\) LCM = Product of the two numbers.
Answer: By using Euclid’s Division Lemma, we have
\(256 = 36 \times 7 + 4\)
\(36 = 4 \times 9 + 0\)
Hence, the HCF of 256 and 36 is 4.
\(LCM : 256 = 2^8\)
\(36 = 2^2 \times 3^2\)
\(LCM (36, 256) = 2^8 \times 3^2 = 256 \times 9 = 2304\)
\(HCF \times LCM = \text{Product of the two numbers}\)
\(4 \times 2304 = 256 \times 36\)
\(9216 = 9216\)
Hence verified.

Question. A fruit vendor has 990 apples and 945 oranges. He packs them into baskets. Each basket contains only one of the two fruits but in equal number. Find the number of fruits to be put in each basket in order to have minimum number of baskets.
Answer: Required answer is the HCF of 990 and 945.
By using Euclid’s Division Lemma, we have
\(990 = 945 \times 1 + 45\)
\(945 = 45 \times 21 + 0\)
Thus HCF of 990 and 945 is 45. The fruit vendor should put 45 fruits in each basket to have minimum number of baskets.

Question. For any positive integer \(n\), prove that \(n^3 - n\) is divisible by 6.
Answer: We have \(n^3 - n = n(n^2 - 1) = (n - 1)n(n + 1)\)
Thus \(n^3 - n\) is product of three consecutive positive integers.
Since, any positive integers \(a\) is of the form \(3q\), \(3q + 1\), or \(3q + 2\) for some integer \(q\).
Let \(a\), \(a + 1\), \(a + 2\) be any three consecutive integers.
Case I : \(a = 3q\)
If \(a = 3q\) then,
\(a(a + 1)(a + 2) = 3q(3q + 1)(3q + 2)\)
Product of two consecutive integers \((3q + 1)\) and \((3q + 2)\) is an even integer, say \(2r\).
Thus \(a(a + 1)(a + 2) = 3q(2r) = 6qr\), which is divisible by 6.
Case II : \(a = 3q + 1\)
If \(a = 3q + 1\) then
\(a(a + 1)(a + 2) = (3q + 1)(3q + 2)(3q + 3)\)
\(= (3q + 1)(3q + 2)3(q + 1)\)
\(= 3(q + 1)(3q + 1)(3q + 2)\)
\(= 3(q + 1)(2r)\)
\(= 6r(q + 1)\) which is divisible by 6.
Case III : \(a = 3q + 2\)
If \(a = 3q + 2\) then
\(a(a + 1)(a + 2) = (3q + 2)(3q + 3)(3q + 4)\)
\(= 3(3q + 2)(q + 1)(3q + 4)\)
Here \((3q + 2)\) and \((3q + 4)\) are product of two consecutive even integers or product of consecutive integers divisible by 2.
\(= \text{multiple of 6 every } q\)
\(= 6r \text{ (say)}\)
which is divisible by 6. Hence, the product of three consecutive integers is divisible by 6 and \(n^3 - n\) is also divisible by 6.

Question. Prove that \(n^2 - n\) is divisible by 2 for every positive integer \(n\).
Answer: We have \(n^2 - n = n(n - 1)\)
Thus \(n^2 - n\) is product of two consecutive positive integers.
Any positive integer is of the form \(2q\) or \(2q + 1\), for some integer \(q\).
Case 1 : \(n = 2q\)
If \(n = 2q\) we have
\(n(n - 1) = 2q(2q - 1) = 2m\),
where \(m = q(2q - 1)\) which is divisible by 2.
Case 2 : \(n = 2q + 1\)
If \(n = 2q + 1\), we have
\(n(n - 1) = (2q + 1)(2q + 1 - 1) = 2q(2q + 1) = 2m\)
where \(m = q(2q + 1)\) which is divisible by 2.
Hence, \(n^2 - n\) is divisible by 2 for every positive integer \(n\).

Question. Find HCF of 81 and 237 and express it as a linear combination of 81 and 237 i.e. HCF (81, 237) = \(81x + 237y\) for some \(x\) and \(y\).
Answer: By using Euclid’s Division Lemma, we have
\(237 = 81 \times 2 + 75\) ...(1)
\(81 = 75 \times 1 + 6\) ...(2)
\(75 = 6 \times 12 + 3\) ...(3)
\(6 = 3 \times 2 + 0\) ...(4)
Hence, HCF (81, 237) = 3.
In order to write 3 in the form of \(81x + 237y\),
\(3 = 75 - 6 \times 12\)
\(= 75 - (81 - 75 \times 1) \times 12\) [Replace 6 from (2)]
\(= 75 - 81 \times 12 + 75 \times 12\)
\(= 75 + 75 \times 12 - 81 \times 12\)
\(= 75(1 + 12) - 81 \times 12\)
\(= 75 \times 13 - 81 \times 12\)
\(= (237 - 81 \times 2) \times 13 - 81 \times 12\) [Replace 75 from (1)]
\(= 237 \times 13 - 81 \times 2 \times 13 - 81 \times 12\)
\(= 237 \times 13 - 81(26 + 12)\)
\(= 237 \times 13 - 81 \times 38\)
\(= 81(-38) + 237(13)\)
\(= 81x + 237y\)
Hence \(x = -38\) and \(y = 13\). These values of \(x\) and \(y\) are not unique.

Question. Show that the square of any positive integer is of the forms \(4m\) or \(4m + 1\), where \(m\) is any integer.
Answer: Let \(a\) be any positive integer, then by Euclid’s division algorithm \(a\) can be written as
\(a = bq + r\)
Take \(b = 4\), then \(0 \le r < 4\) because \(0 \le r < b\),
Thus \(a = 4q, 4q + 1, 4q + 2, 4q + 3\)
Case 1 : \(a = 4q\)
\(a^2 = (4q)^2 = 16q^2 = 4(4q^2) = 4m\)
where \(m = 4q^2\)
Case 2 : \(a = 4q + 1\)
\(a^2 = (4q + 1)^2 = 16q^2 + 8q + 1 = 4(4q^2 + 2q) + 1 = 4m + 1\)
where \(m = 4q^2 + 2q\)
Case 3 : \(a = 4q + 2\)
\(a^2 = (4q + 2)^2 = 16q^2 + 16q + 4 = 4(4q^2 + 4q + 1) = 4m\)
where \(m = 4q^2 + 4q + 1\)
Case 4 : \(a^2 = (4q + 3)^2 = 16q^2 + 24q + 9 = 16q^2 + 24q + 8 + 1 = 4(4q^2 + 6q + 2) + 1 = 4m + 1\)
where \(m = 4q^2 + 6q + 2\)
From cases 1, 2, 3 and 4 we conclude that the square of any +ve integer is of the form \(4m\) or \(4m + 1\).

Question. Use Euclid’s Division Lemma to show that the cube of any positive integer is of the form \(9m\), \(9m + 1\), or \(9m + 8\), for some integer \(m\).
Answer: Let \(a\) be any positive integer, then by Euclid’s division algorithm \(a\) can be written as
\(a = bq + r\)
Take \(b = 3\), then \(0 \le r < 3\) because \(0 \le r < b\),
Thus \(a = 3q, 3q + 1\), and \(3q + 2\)
Case 1 : \(a = 3q\)
\(a^3 = (3q)^3 = 27q^3 = 9(3q^3) = 9m\) where \(m = 3q^3\)
Case 2 : \(a = 3q + 1\)
\(a^3 = (3q + 1)^3 = 27q^3 + 27q^2 + 9q + 1 = 9(3q^3 + 3q^2 + q) + 1\)
or \(a^3 = 9m + 1\) where \(m = 3q^3 + 3q^2 + q\)
Case 3 : \(a = 3q + 2\)
\(a^3 = (3q + 2)^3 = 27q^3 + 54q^2 + 36q + 8 = 9(3q^3 + 6q^2 + 4q) + 8\)
or \(a^3 = 9m + 8\) where \(m = 3q^3 + 6q^2 + 4q\)
From Case 1, 2 and 3, we conclude that the cube of any positive integer is of the form \(9m, 9m + 1\) or \(9m + 8\) for some integer \(m\).