CBSE Class 10 Mathematics Number Systems Assignment Set 02

Number System MCQ Questions with Answers Class 10 

NUMBER SYSTEMS

Question 1. If the H C F of 657 and 963 is expressible in the form of \(657x + 963(-15)\), find x.
Answer: Using Euclid's Division Lemma \(a = bq + r\), where \(0 \leq r < b\):
\(963 = 657 \times 1 + 306\)
\(657 = 306 \times 2 + 45\)
\(306 = 45 \times 6 + 36\)
\(45 = 36 \times 1 + 9\)
\(36 = 9 \times 4 + 0\)
Thus, \(\text{HCF}(657, 963) = 9\).
Given equation is \(9 = 657x + 963(-15)\):
\(657x = 9 - 963(-15)\)
\(657x = 9 + 14445\)
\(657x = 14454\)
\(x = \frac{14454}{657} = 22\)
In simple words: We first find the HCF of the two numbers using division steps, and then substitute that HCF into the linear equation to solve for \(x\).

Exam Tip: Always double-check your arithmetic when substituting back into linear combinations to avoid sign errors.

Question 2. Express the GCD of 48 and 18 as a linear combination.
Answer: Using Euclid's Division Lemma for 48 and 18:
\(48 = 18 \times 2 + 12\)
\(18 = 12 \times 1 + 6\)
\(12 = 6 \times 2 + 0\)
Thus, \(\text{HCF}(18, 48) = 6\).
Expressing 6 backwards:
\(6 = 18 - 12 \times 1\)
Substitute \(12 = 48 - 18 \times 2\):
\(6 = 18 - (48 - 18 \times 2) \times 1\)
\(6 = 18 - 48 + 18 \times 2\)
\(6 = 18 \times 3 + 48 \times (-1)\)
This gives one linear combination: \(18(3) + 48(-1) = 6\).
To show it is not unique, we can add and subtract \(18 \times 48\):
\(6 = 18 \times 3 + 48 \times (-1) + 18 \times 48 - 18 \times 48\)
\(6 = 18(3 + 48) + 48(-1 - 18)\)
\(6 = 18(51) + 48(-19)\)
Thus, \(x = 3, y = -1\) or \(x = 51, y = -19\), showing that the representation is not unique.
In simple words: Working backwards from the division steps allows us to write the HCF as a sum of multiples of the original numbers, and this can be done in multiple ways.

Exam Tip: Mention clearly that linear combinations for GCD are non-unique because multiples of the LCM can be added and subtracted.

Question 3. Prove that one of every three consecutive integers is divisible by 3.
Answer: Let \(n, n+1, n+2\) be three consecutive positive integers.
By Euclid's Division Lemma with divisor 3, any integer \(n\) can be represented in one of the following three forms:
Case I: When \(n = 3q\), then \(n\) is divisible by 3, while \(n+1 = 3q+1\) and \(n+2 = 3q+2\) are not divisible by 3.
Case II: When \(n = 3q+1\), then \(n+2 = 3q+1+2 = 3q+3 = 3(q+1)\), which is divisible by 3, while \(n\) and \(n+1\) are not divisible by 3.
Case III: When \(n = 3q+2\), then \(n+1 = 3q+2+1 = 3q+3 = 3(q+1)\), which is divisible by 3, while \(n\) and \(n+2\) are not divisible by 3.
Hence, exactly one out of every three consecutive integers is always divisible by 3.
In simple words: By checking all possible remainders when dividing by 3, we see that one number in every group of three will always be a multiple of 3.

Exam Tip: State division cases clearly using Euclid's lemma divisor equal to 3.

Question 4. Find the largest possible positive integer that will divide 398, 436, and 542 leaving remainder 7, 11, 15 respectively.
Answer: Subtract the respective remainders from each given number:
\(398 - 7 = 391\)
\(436 - 11 = 425\)
\(542 - 15 = 527\)
The required number is the HCF of 391, 425, and 527.
First, find the HCF of 391 and 425 using Euclid's algorithm:
\(425 = 391 \times 1 + 34\)
\(391 = 34 \times 11 + 17\)
\(34 = 17 \times 2 + 0\)
So, \(\text{HCF}(391, 425) = 17\).
Next, find the HCF of 17 and 527:
\(527 = 17 \times 31 + 0\)
So, \(\text{HCF}(17, 527) = 17\).
Therefore, the greatest number is 17.
In simple words: Subtract the remainders first so the numbers become completely divisible, then find their HCF.

Exam Tip: Always subtract remainders before calculating the HCF for word problems involving remainders.

Question 5. Find the least number that is divisible by all numbers between 1 and 10 (both inclusive).
Answer: The required number is the LCM of numbers from 1 to 10: \(1, 2, 3, 4, 5, 6, 7, 8, 9, 10\).
Prime factorization of each number:
\(1 = 1\)
\(2 = 2\)
\(3 = 3\)
\(4 = 2^2\)
\(5 = 5\)
\(6 = 2 \times 3\)
\(7 = 7\)
\(8 = 2^3\)
\(9 = 3^2\)
\(10 = 2 \times 5\)
Taking the highest power of each prime factor:
\(\text{LCM} = 2^3 \times 3^2 \times 5^1 \times 7^1\)
\(\text{LCM} = 8 \times 9 \times 5 \times 7 = 2520\)
In simple words: The least divisible number is the lowest common multiple, found by multiplying the highest powers of all prime factors from 1 to 10.

Exam Tip: List prime factorizations and pick the highest exponent for each prime.

Question 6. Show that 571 is a prime number.
Answer: Let \(x = 571\). Then \(\sqrt{571}\) lies between \(\sqrt{529}\) and \(\sqrt{576}\), which are \(23^2\) and \(24^2\) respectively.
All prime numbers less than 24 are: \(2, 3, 5, 7, 11, 13, 17, 19, 23\).
Checking divisibility of 571 by these primes:
- Not divisible by 2 (not even)
- Not divisible by 3 (sum of digits \(5 + 7 + 1 = 13\), not divisible by 3)
- Not divisible by 5 (does not end in 0 or 5)
- \(571 = 7 \times 81 + 4\) (not divisible by 7)
- \(571 = 11 \times 51 + 10\) (not divisible by 11)
- \(571 = 13 \x 43 + 12\) (not divisible by 13)
- \(571 = 17 \times 33 + 10\) (not divisible by 17)
- \(571 = 19 \times 30 + 1\) (not divisible by 19)
- \(571 = 23 \times 24 + 19\) (not divisible by 23)
Since 571 is not divisible by any prime number less than or equal to \(\sqrt{571}\), it is a prime number.
In simple words: To check if a number is prime, test division by all primes up to its square root. If none divide it evenly, it is prime.

Exam Tip: State clearly that checking up to the square root of the number is sufficient.

Question 7. If d is the HCF of 30 and 72, find the value of x and y satisfying \(d = 30x + 72y\).
Answer: Using Euclid's algorithm for 30 and 72:
\(72 = 30 \times 2 + 12\)
\(30 = 12 \times 2 + 6\)
\(12 = 6 \times 2 + 0\)
Thus, \(d = \text{HCF}(30, 72) = 6\).
Expressing 6 backwards:
\(6 = 30 - 12 \times 2\)
Substitute \(12 = 72 - 30 \times 2\):
\(6 = 30 - (72 - 30 \times 2) \times 2\)
\(6 = 30 - 2 \times 72 + 30 \times 4\)
\(6 = 30 \times 5 + 72 \times (-2)\)
Thus, one solution is \(x = 5, y = -2\).
General solutions can be generated, and another valid pair is \(x = 77, y = -32\). Hence, \(x\) and \(y\) are not unique.
In simple words: Work backwards from the Euclidean algorithm steps to express the HCF in terms of 30 and 72.

Exam Tip: Show all substitution steps clearly to earn full marks in Bézout's identity problems.

Question 8. Show that the product of 3 consecutive positive integers is divisible by 6.
Answer: Let the three consecutive integers be \(n, n+1, n+2\).
Out of any three consecutive integers:
- At least one is divisible by 2.
- Exactly one is divisible by 3 (as proven in Question 3).
Since 2 and 3 are coprime, their product \(2 \times 3 = 6\) must divide the product of the three consecutive integers \(n(n+1)(n+2)\).
Hence, the product of 3 consecutive positive integers is always divisible by 6.
In simple words: Since the group contains a multiple of 2 and a multiple of 3, their product 6 must divide the whole expression.

Exam Tip: Mention coprimality of 2 and 3 to conclude divisibility by 6.

Question 9. Show that for an odd positive integer to be a perfect square, it should be of the form \(8k + 1\).
Answer: Let \(a\) be an odd positive integer, so \(a = 2m + 1\) for some integer \(m\).
Squaring both sides:
\(a^2 = (2m + 1)^2\)
\(a^2 = 4m^2 + 4m + 1\)
\(a^2 = 4m(m + 1) + 1\)
Since \(m(m+1)\) is the product of two consecutive integers, it is always even, so let \(m(m+1) = 2k\) for some integer \(k\).
Substituting this back:
\(a^2 = 4(2k) + 1\)
\(a^2 = 8k + 1\)
Hence, the square of any odd positive integer is always of the form \(8k + 1\).
In simple words: Squaring an odd number and factoring out 4 along with the consecutive even product shows it always leaves a remainder of 1 when divided by 8.

Exam Tip: Use the property that the product of two consecutive integers is always even.

Question 10. Find the greatest number of 6 digits exactly divisible by 24, 15 and 36.
Answer: First, find the LCM of 24, 15, and 36:
\(24 = 2^3 \times 3\)
\(15 = 3 \times 5\)
\(36 = 2^2 \times 3^2\)
\(\text{LCM} = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360\)
The greatest 6-digit number is 999999.
Divide 999999 by 360:
\(999999 = 360 \times 2777 + 279\)
Remainder \(R = 279\).
Subtract the remainder from 999999 to find the greatest 6-digit number divisible by 360:
Required number = \(999999 - 279 = 999720\).
In simple words: Find the LCM of the divisors, divide the largest 6-digit number by this LCM, and subtract the remainder to get the answer.

Exam Tip: Always subtract the remainder from the dividend to get the largest multiple below a given number.

Question 11. If a and b are positive integers, show that \(\sqrt{2}\) always lies between \(\frac{a}{b}\) and \(\frac{a+2b}{a+b}\).
Answer: To compare the fractions, let us compute the difference:
\(\frac{a}{b} - \frac{a+2b}{a+b} = \frac{a(a+b) - b(a+2b)}{b(a+b)}\)
\(= \frac{a^2 + ab - ab - 2b^2}{b(a+b)} = \frac{a^2 - 2b^2}{b(a+b)}\)
Depending on whether \(a^2 - 2b^2\) is positive or negative, we have two cases:
Case 1: If \(a > \sqrt{2}b\), then \(a^2 > 2b^2\), so \(\frac{a}{b} > \frac{a+2b}{a+b}\).
In this case, it can be proven that \(\frac{a+2b}{a+b} < \sqrt{2} < \frac{a}{b}\).
Case 2: If \(a < \sqrt{2}b\), then \(\frac{a}{b} < \frac{a+2b}{a+b}\), and we have \(\frac{a}{b} < \sqrt{2} < \frac{a+2b}{a+b}\).
Combining both results, \(\sqrt{2}\) always lies between \(\frac{a}{b}\) and \(\frac{a+2b}{a+b}\).
In simple words: By taking the difference between the two fractions, we show that \(\sqrt{2}\) is always sandwiched between them for any positive integers \(a\) and \(b\).

Exam Tip: Use cross-multiplication and difference of fractions to establish inequalities involving square roots.

Question 12. Prove that \((\sqrt{n-1} + \sqrt{n+1})\) is irrational, for every \(n \in N\).
Answer: Let us assume to the contrary that \(\sqrt{n-1} + \sqrt{n+1}\) is a rational number equal to \(r\), where \(r \in \mathbb{Q}\).
Squaring both sides:
\((\sqrt{n-1} + \sqrt{n+1})^2 = r^2\)
\((n-1) + (n+1) + 2\sqrt{(n-1)(n+1)} = r^2\)
\(2n + 2\sqrt{n^2 - 1} = r^2\)
\(2\sqrt{n^2 - 1} = r^2 - 2n\)
\(\sqrt{n^2 - 1} = \frac{r^2 - 2n}{2}\)
Since \(r\) is rational, the right-hand side is a rational number. However, for any natural number \(n \geq 2\), \(\sqrt{n^2 - 1}\) is strictly irrational because \(n^2 - 1\) is never a perfect square.
This contradicts our assumption. Hence, \((\sqrt{n-1} + \sqrt{n+1})\) is irrational for every \(n \in \mathbb{N}\).
In simple words: Assume the expression is rational, square both sides to isolate the square root, and show that a rational number would equal an irrational root, which is impossible.

Exam Tip: Use proof by contradiction by squaring the sum of radicals to isolate the irrational term.