5. Taking $2$ common out from the expression:
$m^2 + n^2 = 2[2p(p + 1) + 2q(q + 1) + 1]$
6. Let $k = 2p(p + 1) + 2q(q + 1) + 1$, which is an integer. Thus, $m^2 + n^2 = 2k$. This clearly shows that $m^2 + n^2$ is a multiple of $2$, meaning it is **even**.
7. To check for divisibility by $4$, let us look closely at the bracket: $[2p(p + 1) + 2q(q + 1) + 1]$.
* We know that the product of two consecutive integers $p(p + 1)$ is always even, so $2p(p + 1)$ is a multiple of $4$. Similarly, $2q(q + 1)$ is a multiple of $4$.
* Therefore, the term inside the bracket is (Multiple of $4$ + Multiple of $4$ + $1$), which is of the form $4(\text{integer}) + 1$, meaning it is an **odd integer**.
8. Substituting this back: $m^2 + n^2 = 2 \times (\text{odd integer})$. When an even number is multiplied by an odd integer, the result is of the form $4k + 2$, which leaves a remainder of $2$ when divided by $4$ and is **not divisible by $4$**.
Result: Hence, it is proved that $m^2 + n^2$ is even, but not divisible by $4$.
Q3. Let $p$ be a prime number. Prove that $\sqrt{p} + \sqrt{p+2}$ is always an irrational number.
Answer:
1. Let us assume to the contrary that $\sqrt{p} + \sqrt{p+2}$ is a rational number, say $r$, where $r \in \mathbb{Q}$ and $r \neq 0$.
2. Therefore:
$\sqrt{p} + \sqrt{p+2} = r$
3. Taking the reciprocal of both sides:
$\frac{1}{\sqrt{p} + \sqrt{p+2}} = \frac{1}{r}$
5. We are already given from our initial assumption that:
$\sqrt{p+2} + \sqrt{p} = r$ --- (Equation 2)
6. Adding Equation 1 and Equation 2:
$2\sqrt{p+2} = r + \frac{2}{r}$
$\sqrt{p+2} = \frac{1}{2}\left(r + \frac{2}{r}\right)$
7. Since $r$ is a rational number, the expression on the right-hand side ($\frac{1}{2}(r + \frac{2}{r})$) must also be a rational number. This implies that $\sqrt{p+2}$ is rational. However, since $p+2$ is not a perfect square for any prime number $p$, $\sqrt{p+2}$ is strictly an irrational number. This is a contradiction.
8. Similarly, subtracting Equation 1 from Equation 2 gives $\sqrt{p}$ as a rational number, which also contradicts the fact that $\sqrt{p}$ is irrational.
Result: This contradiction arises due to our incorrect assumption that $\sqrt{p} + \sqrt{p+2}$ is rational. Hence, $\sqrt{p} + \sqrt{p+2}$ is always an irrational number.
Q4. If $p$ is a prime number, prove that $\sqrt{p}$ is irrational.
Answer:
1. **Assumption:** Let us assume to the contrary that $\sqrt{p}$ is a rational number.
2. **Definition of Rational:** Therefore, there exist two integers, $a$ and $b$ (where $b \neq 0$), such that:
$\sqrt{p} = \frac{a}{b}$
Let $a$ and $b$ be coprime (i.e., their highest common factor, HCF, is $1$).
3. **Squaring both sides:**
$p = \frac{a^2}{b^2}$
$a^2 = p b^2 \quad \text{--- (Equation 1)}$
4. **Divisibility property:** From Equation 1, $p$ divides $a^2$. Since $p$ is a prime number, by the fundamental theorem of divisibility (if a prime divides the square of a number, it divides the number itself), $p$ also divides $a$.
Therefore, we can write $a = p c$ for some integer $c$.
5. **Substituting $a$ into Equation 1:**
$(p c)^2 = p b^2$
$p^2 c^2 = p b^2$
$b^2 = p c^2 \quad \text{--- (Equation 2)}$
6. **Conclusion on divisibility:** Equation 2 shows that $p$ divides $b^2$, which means $p$ also divides $b$.
7. **Contradiction:** From steps 4 and 6, both $a$ and $b$ have at least $p$ as a common factor. This contradicts our initial assumption that $a$ and $b$ are coprime (their HCF is $1$).
Result: This contradiction arises because of our incorrect assumption that $\sqrt{p}$ is rational. Hence, $\sqrt{p}$ is an irrational number.
Q5. Prove that $\frac{1}{\sqrt{5}}$ is irrational.
Answer:
1. **Assumption:** Let us assume to the contrary that $\frac{1}{\sqrt{5}}$ is a rational number.
2. **Definition of Rational:** Therefore, there exist two coprime integers $a$ and $b$ (where $b \neq 0$ and $a \neq 0$) such that:
$\frac{1}{\sqrt{5}} = \frac{a}{b}$
3. **Rearranging the equation:**
Taking the reciprocal of both sides:
$\sqrt{5} = \frac{b}{a}$
4. **Analyzing the components:**
• Since $a$ and $b$ are integers, their quotient $\frac{b}{a}$ is a rational number.
• This implies that $\sqrt{5}$ must also be a rational number.
5. **Contradiction:** This contradicts the known mathematical fact that $\sqrt{5}$ is an irrational number.
Result: This contradiction arises due to our incorrect assumption that $\frac{1}{\sqrt{5}}$ is rational. Hence, $\frac{1}{\sqrt{5}}$ is an irrational number.
Q6. Prove that for any positive integer $x$, $x^3 - x$ is divisible by $6$. (Hint: Factorize into consecutive integers and apply prime breakdown).
Answer:
1. Let the given expression be $f(x) = x^3 - x$.
3. Rearranging the factors in order of consecutive integers:
$f(x) = (x - 1) \cdot x \cdot (x + 1)$
4. This shows that $x^3 - x$ is the product of three consecutive positive integers: $(x - 1)$, $x$, and $(x + 1)$.
5. Now, we analyze the divisibility by $6$ (since $6 = 2 \times 3$, where $2$ and $3$ are coprime):
* **Divisibility by $2$:** Among any two consecutive integers, exactly one must be even. Since $(x-1)$ and $x$ (or $x$ and $x+1$) are consecutive, their product is always divisible by $2$.
* **Divisibility by $3$:** Among any three consecutive integers, exactly one must be a multiple of $3$. Therefore, the product of three consecutive integers is always divisible by $3$.
6. Since the expression is divisible by both $2$ and $3$, and $2$ and $3$ are coprime numbers, the expression must be divisible by their product ($2 \times 3 = 6$).
Result: Hence, it is proved that $x^3 - x$ is always divisible by $6$ for any positive integer $x$.
Q7. The HCF of two numbers $a$ and $b$ is $12$ and their product is $6340$. If the ratio of their prime powers allows up to two distinct common factors, determine all possible values of $(a, b)$.
Answer:
1. We are given that the HCF of two numbers $a$ and $b$ is $12$. Therefore, $12$ must be a factor of both $a$ and $b$.
2. Let the two numbers be written as multiples of their HCF:
$a = 12x$
$b = 12y$
where $x$ and $y$ are coprime integers ($\text{HCF}(x, y) = 1$).
3. Using the property that the product of two numbers is equal to the product of their HCF and LCM, or simply using their product directly:
$a \times b = 6340$
$(12x) \times (12y) = 6340$
$144xy = 6340$
5. Since $x$ and $y$ must be integers (as $a$ and $b$ are standard numbers where $12$ divides both), their product $xy$ must also be a rational number resulting from integer factors. However, $\frac{1585}{36}$ is not an integer, meaning no integer values of $x$ and $y$ can satisfy $144xy = 6340$.
6. Let us double check the division: $6340$ is not divisible by $144$ ($6340 = 144 \times 44 + 4$). Thus, no such integers $a$ and $b$ exist under standard conditions.
Result: There are **no possible values** for $(a, b)$ since the given product $6340$ is not divisible by the square of their HCF ($12^2 = 144$).
Q8. Prove that $\frac{3}{2\sqrt{5}}$ is irrational.
Answer:
1. **Assumption:** Let us assume to the contrary that $\frac{3}{2\sqrt{5}}$ is a rational number.
2. **Definition of Rational:** Therefore, there exist two coprime integers $a$ and $b$ (where $b \neq 0$ and $a \neq 0$) such that:
$\frac{3}{2\sqrt{5}} = \frac{a}{b}$
3. **Rearranging the equation:**
$\sqrt{5} = \frac{3b}{2a}$
4. **Analyzing the components:**
• Since $a$ and $b$ are integers, the expression $\frac{3b}{2a}$ is a rational number.
• This implies that $\sqrt{5}$ must also be a rational number.
5. **Contradiction:** This contradicts the known mathematical fact that $\sqrt{5}$ is an irrational number.
Result: This contradiction arises due to our incorrect assumption that $\frac{3}{2\sqrt{5}}$ is rational. Hence, $\frac{3}{2\sqrt{5}}$ is an irrational number.
Q9. Prove that $\sqrt{p} + \sqrt{q}$ is irrational, where $p$ and $q$ are distinct prime numbers.
Answer:
1. **Assumption:** Let us assume to the contrary that $\sqrt{p} + \sqrt{q}$ is a rational number, say $r$.
2. **Equation Setup:**
$\sqrt{p} + \sqrt{q} = r$
3. **Isolating a radical and squaring:**
$\sqrt{q} = r - \sqrt{p}$
Squaring both sides:
$q = (r - \sqrt{p})^2$
$q = r^2 - 2r\sqrt{p} + p$
4. **Rearranging to isolate the radical term:**
$2r\sqrt{p} = r^2 + p - q$
$\sqrt{p} = \frac{r^2 + p - q}{2r}$
5. **Analyzing the components:**
• Since $r$ is rational, and $p$ and $q$ are integers, the expression $\frac{r^2 + p - q}{2r}$ is a rational number.
• This implies that $\sqrt{p}$ must also be a rational number.
6. **Contradiction:** This contradicts the known mathematical fact that $\sqrt{p}$ (where $p$ is a prime number) is an irrational number.
Result: This contradiction arises due to our incorrect assumption that $\sqrt{p} + \sqrt{q}$ is rational. Hence, $\sqrt{p} + \sqrt{q}$ is an irrational number.
Q10. Show that for any positive integer $n$, $\sqrt{n}$ is either a rational or an irrational number.
Answer:
1. **Definition of Real Numbers:** Every positive integer $n$ can be categorized based on whether it is a perfect square or not a perfect square.
2. **Case 1: When $n$ is a perfect square:**
• If $n = k^2$ for some positive integer $k$, then:
$\sqrt{n} = \sqrt{k^2} = k$
• Since $k$ is an integer, it can be written as $\frac{k}{1}$, which means $\sqrt{n}$ is a **rational number**.
3. **Case 2: When $n$ is not a perfect square:**
• If $n$ is not a perfect square, its square root cannot be expressed as a simple integer or a finite/recurring decimal ratio of integers.
• By number theory properties (specifically involving prime factorizations where prime factors do not appear in even powers), $\sqrt{n}$ cannot be expressed in the form $\frac{a}{b}$ where $a$ and $b$ are integers ($b \neq 0$).
• Therefore, $\sqrt{n}$ is an **irrational number**.
Result: By the trichotomy of real numbers and the properties of integers, for any positive integer $n$, $\sqrt{n}$ is always strictly either a rational number or an irrational number.
Q11. Two tankers contain 850 liters and 680 liters of petrol respectively. Find the maximum capacity of a container which can measure the petrol of either tanker in exact number of times.
Answer:
1. **Understanding the Problem:** To find the maximum capacity of a container that can measure the petrol of both tankers an exact number of times, we need to find the Highest Common Factor (HCF) of the two quantities: $850$ liters and $680$ liters.
2. **Prime Factorization of each number:**
• Find the prime factors of $850$:
$850 = 2 \times 5 \times 5 \times 17 = 2 \times 5^2 \times 17$
• Find the prime factors of $680$:
$680 = 2 \times 2 \times 2 \times 5 \times 17 = 2^3 \times 5 \times 17$
3. **Determining the HCF:**
• The common prime factors with the lowest powers are $2$, $5$, and $17$.
• Lowest power of $2$ is $2^1$
• Lowest power of $5$ is $5^1$
• Lowest power of $17$ is $17^1$
Result: The maximum capacity of the container that can measure the petrol of either tanker an exact number of times is $170$ liters.
Q12. Let $\beta$ and $\delta$ be positive integers. The HCF of $\beta$ and $630$ is $210$, and the HCF of $\delta$ and $110$ is $55$. Find the maximum possible value of the HCF of $(\beta, 630, \delta, 110)$ using step-by-step Euclid's division algorithm logic.
Answer:
1. We are given that $\text{HCF}(\beta, 630) = 210$.
* Prime factorization of $630$: $630 = 2 \times 3^2 \times 5 \times 7 = 210 \times 3$.
* Since the HCF is $210$, $\beta$ must be a multiple of $210$ ($210 \times k_1$), but $\beta$ cannot be a multiple of larger common factors that share with $630$ beyond $210$.
2. We are given that $\text{HCF}(\delta, 110) = 55$.
* Prime factorization of $110$: $110 = 2 \times 5 \times 11 = 55 \times 2$.
* Since the HCF is $55$, $\delta$ must be a multiple of $55$ ($55 \times k_2$), but not sharing extra factors with $110$.
3. We need to find the maximum possible value of the HCF of all four numbers: $\text{HCF}(\beta, 630, \delta, 110)$.
4. Using the associative property of HCF:
$\text{HCF}(\beta, 630, \delta, 110) = \text{HCF}(\text{HCF}(\beta, 630), \text{HCF}(\delta, 110))$
5. Substituting the known HCF values:
$= \text{HCF}(210, 55)$
7. Since the remainder is zero at the last step, the HCF of $210$ and $55$ is **5**.
Result: Therefore, the maximum possible value of the HCF of $(\beta, 630, \delta, 110)$ is **5**.
Q13. Show that any number of the form $4^n$, where $n$ is a natural number, can never end with the digit zero.
Answer:
1. **Condition for ending with zero:** If a number of the form $4^n$ ends with the digit zero ($0$), it must be divisible by $10$. This means its prime factorization must contain both $2$ and $5$ as prime factors.
2. **Prime factorization of the base:**
• Let us consider the base $4$.
• Expressing $4$ in terms of its prime factors:
$4 = 2^2$
3. **Applying the exponent $n$:**
$4^n = (2^2)^n = 2^{2n}$
4. **Analyzing the prime factors:**
• By the Fundamental Theorem of Arithmetic, the prime factorization of $4^n$ is unique.
• The only prime factor in the prime factorization of $4^n$ is $2$.
• The prime factor $5$ is completely absent from its factorization.
5. **Conclusion:** Since $5$ is not present in the prime factorization of $4^n$ for any natural number $n$, it is impossible for $4^n$ to be divisible by $10$.
Result: Therefore, any number of the form $4^n$, where $n$ is a natural number, can never end with the digit zero.
Q14. If $x = \frac{3 + \sqrt{5}}{2}$ and $y = \frac{3 - \sqrt{5}}{2}$, show that $x^3 + y^3$ is a rational number, whereas $x^3 - y^3$ is an irrational number. Find their exact values.
Answer:
1. First, let us find the sum ($x + y$) and the product ($xy$) of the given values:
* $x + y = \frac{3 + \sqrt{5}}{2} + \frac{3 - \sqrt{5}}{2} = \frac{3 + \sqrt{5} + 3 - \sqrt{5}}{2} = \frac{6}{2} = 3$
* $xy = \left(\frac{3 + \sqrt{5}}{2}\right)\left(\frac{3 - \sqrt{5}}{2}\right) = \frac{3^2 - (\sqrt{5})^2}{4} = \frac{9 - 5}{4} = \frac{4}{4} = 1$
2. Now, we use algebraic identities to find $x^3 + y^3$:
$x^3 + y^3 = (x + y)^3 - 3xy(x + y)$
3. Substituting the values of $(x + y) = 3$ and $xy = 1$:
$x^3 + y^3 = (3)^3 - 3(1)(3)$
$x^3 + y^3 = 27 - 9 = 18$
4. Since $18$ is a rational number, **$x^3 + y^3$ is a rational number** with an exact value of **$18$**.
5. Next, we find $x^3 - y^3$ using the algebraic identity:
$x^3 - y^3 = (x - y)(x^2 + xy + y^2) = (x - y)[(x + y)^2 - xy]$
6. To find $(x - y)$, we can use the identity $(x - y)^2 = (x + y)^2 - 4xy$:
$(x - y)^2 = (3)^2 - 4(1) = 9 - 4 = 5$
$x - y = \sqrt{5}$
Q15. Find the largest number that divides 2053 and 967 and leaves a remainder of 5 and 7 respectively.
Answer:
1. **Understanding the Problem:** We need to find the largest number that divides $2053$ and $967$ leaving respective remainders of $5$ and $7$. This means if we subtract the respective remainders from the given numbers, the resulting numbers will be completely divisible by the required number.
2. **Adjusting the given numbers:**
• First number after subtracting its remainder:
$2053 - 5 = 2048$
• Second number after subtracting its remainder:
$967 - 7 = 960$
3. **Determining the HCF:** The largest number that divides both $2048$ and $960$ completely is their Highest Common Factor (HCF).
4. **Prime Factorization of each number:**
• Find the prime factors of $2048$ (which is $2^{11}$):
$2048 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{11}$
• Find the prime factors of $960$:
$960 = 2 \times 480 = 2^6 \times 15 = 2^6 \times 3 \times 5 = 32 \times 3 \times 5$
5. **Calculating the HCF:**
• The only common prime factor between $2048$ ($2^{11}$) and $960$ ($2^6 \times 3 \times 5$) is $2$.
• The lowest power of $2$ common to both is $2^6$.
$\text{HCF}(2048, 960) = 2^6 = 64$
Result: The largest number that divides 2053 and 967 leaving remainders of 5 and 7 respectively is $64$.
Q16. Three sets of English, Mathematics, and Science books containing 336, 240, and 96 books respectively have to be stacked in such a way that all the books are stored subject-wise and the height of each stack is the same. Find the number of stacks.
Answer:
1. **Understanding the Problem:** To ensure that the height of each stack is the same and the maximum number of books are in each stack (meaning the minimum total number of stacks), we need to find the maximum number of books per stack. This is given by the Highest Common Factor (HCF) of the number of books in each set: $336$, $240$, and $96$.
2. **Prime Factorization of each number:**
• Prime factorization of $336$:
$336 = 2^4 \times 3 \times 7$
• Prime factorization of $240$:
$240 = 2^4 \times 3 \times 5$
• Prime factorization of $96$:
$96 = 2^5 \times 3$
3. **Determining the HCF:**
• The common prime factors with the lowest powers are $2$ and $3$.
• Lowest power of $2$ is $2^4$
• Lowest power of $3$ is $3^1$
$\text{HCF}(336, 240, 96) = 2^4 \times 3 = 16 \times 3 = 48$
• Therefore, the maximum number of books in each stack is $48$.
4. **Calculating the Number of Stacks:**
• Number of stacks for English books = $\frac{336}{48} = 7$
• Number of stacks for Mathematics books = $\frac{240}{48} = 5$
• Number of stacks for Science books = $\frac{96}{48} = 2$
• Total number of stacks = $7 + 5 + 2 = 14$
Result: The total number of stacks required to store all the books subject-wise with equal height is $14$.
Q17. Prove that the product of three consecutive positive integers is divisible by 6.
Answer:
1. **Understanding the Problem:** Let any three consecutive positive integers be represented as $n$, $n+1$, and $n+2$, where $n$ is a positive integer. We need to prove that their product, $n(n+1)(n+2)$, is divisible by $6$.
2. **Divisibility Condition:** For a number to be divisible by $6$, it must be divisible by both $2$ and $3$ (since $2$ and $3$ are coprime numbers).
3. **Checking Divisibility by 2:**
• Out of any two consecutive integers, one is always even and the other is odd.
• Therefore, among any three consecutive integers $n$, $n+1$, and $n+2$, at least one integer must be even.
• Since it is even, the product $n(n+1)(n+2)$ is always divisible by $2$.
4. **Checking Divisibility by 3:**
• Out of any three consecutive integers, one integer is always a multiple of $3$.
• Therefore, the product $n(n+1)(n+2)$ is always divisible by $3$.
5. **Conclusion:** Since the product $n(n+1)(n+2)$ is divisible by both $2$ and $3$, and $\text{HCF}(2, 3) = 1$, it must be divisible by their product ($2 \times 3 = 6$).
Result: Hence, the product of three consecutive positive integers is always divisible by 6.
Q18. Find the smallest number which when increased by 17 is exactly divisible by 520 and 468.
Answer:
1. **Understanding the Problem:** Let the required smallest number be $x$. According to the question, when the number is increased by $17$, it becomes $(x + 17)$, which must be exactly divisible by both $520$ and $468$.
2. **Finding the LCM:** The smallest number that is divisible by both $520$ and $468$ is their Lowest Common Multiple (LCM).
3. **Prime Factorization of each number:**
• Prime factorization of $520$:
$520 = 2 \times 2 \times 2 \times 5 \times 13 = 2^3 \times 5^1 \times 13^1$
• Prime factorization of $468$:
$468 = 2 \times 2 \times 3 \times 3 \times 13 = 2^2 \times 3^2 \times 13^1$
4. **Calculating the LCM:**
• Take the highest power of each prime factor involved ($2$, $3$, $5$, and $13$).
• Highest power of $2$ is $2^3$
• Highest power of $3$ is $3^2$
• Highest power of $5$ is $5^1$
• Highest power of $13$ is $13^1$
5. **Solving for the required number ($x$):**
$x + 17 = 4,680$
$x = 4,680 - 17$
$x = 4,663$
Result: The smallest number which when increased by 17 is exactly divisible by 520 and 468 is $4,663$.
Q19. Two positive integers $a$ and $b$ are expressible in the form $a = p \cdot q^2$ and $b = p^3 \cdot q$, where $p$ and $q$ are prime numbers. If $\text{LCM}(a, b) = p^3 q^3$ and $\text{HCF}(a, b) = p \cdot q$, verify whether $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$ holds true, and find the smallest natural number $n$ for which $(pq)^n$ ends with zero.
Answer:
1. **Given data:**
* $a = p \cdot q^2$
* $b = p^3 \cdot q$
* $\text{HCF}(a, b) = p \cdot q$
* $\text{LCM}(a, b) = p^3 q^3$
2. **Part 1: Verification of $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$**
* **Calculate Left-Hand Side (LHS):**
$\text{LHS} = \text{HCF}(a, b) \times \text{LCM}(a, b) = (p \cdot q) \times (p^3 q^3) = p^{1+3} q^{1+3} = p^4 q^4$
* **Calculate Right-Hand Side (RHS):**
$\text{RHS} = a \times b = (p \cdot q^2) \times (p^3 \cdot q) = p^{1+3} q^{2+1} = p^4 q^3$
* **Comparison:**
Since $\text{LHS} = p^4 q^4$ and $\text{RHS} = p^4 q^3$, we can see that $\text{LHS} \neq \text{RHS}$ because the powers of $q$ do not match ($q^4 \neq q^3$).
* *Note:* While the standard property $\text{HCF} \times \text{LCM} = a \times b$ holds for any two numbers where $\text{LCM} = p^3 q^2$, the explicitly given $\text{LCM}$ in this specific problem ($p^3 q^3$) makes the property fail. Thus, it **does not hold true** for the given values.
3. **Part 2: Finding the smallest natural number $n$ for which $(pq)^n$ ends with zero**
* For any number to end with the digit zero, its prime factorization must contain both $2$ and $5$ as prime factors.
* Therefore, the prime factors $p$ and $q$ must be $2$ and $5$ (in some order).
* The term becomes $(2 \times 5)^n = (10)^n = 10^n$.
* For $10^n$ to end with zero, the smallest natural number $n$ is **$1$** (since $10^1 = 10$, which ends in zero).
Result: The relationship $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$ **does not hold true** for the given values, and the smallest natural number $n$ is **$1$**.
Q20. A merchant has 120 liters of oil of one kind, 180 liters of another kind, and 240 liters of a third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity. What should be the greatest capacity of such a tin?
Answer:
1. **Understanding the Problem:** To find the greatest capacity of a tin that can measure all three quantities of oil an exact number of times, we need to find the Highest Common Factor (HCF) of the three quantities: $120$ liters, $180$ liters, and $240$ liters.
2. **Prime Factorization of each number:**
• Prime factorization of $120$:
$120 = 2 \times 2 \times 2 \times 3 \times 5 = 2^3 \times 3^1 \times 5^1$
• Prime factorization of $180$:
$180 = 2 \times 2 \times 3 \times 3 \times 5 = 2^2 \times 3^2 \times 5^1$
• Prime factorization of $240$:
$240 = 2 \times 2 \times 2 \times 2 \times 3 \times 5 = 2^4 \times 3^1 \times 5^1$
3. **Determining the HCF:**
• The common prime factors involved are $2$, $3$, and $5$.
• Lowest power of $2$ is $2^2$
• Lowest power of $3$ is $3^1$
• Lowest power of $5$ is $5^1$
Result: The greatest capacity of the tin that can fill the three kinds of oil in an exact number of times is $60$ liters.
Academic Repository Disclaimer & எச்சரிக்கை அறிவிப்பு :
English: These model solutions are structured to global academic standards for self-assessment and practice purposes. Students are advised to verify with official textbooks for board examinations.
தமிழ்: இந்த மாதிரி கணிதத் தீர்வுகள் மாணவர்களின் சுய கற்றல் மற்றும் பயிற்சித் திறனுக்காக உலகளாவிய கல்வித் தரத்தில் உருவாக்கப்பட்டுள்ளன. தேர்வுகளுக்குத் தயாராகும் போது உங்கள் அதிகாரப்பூர்வ பாடப்புத்தகத்துடன் சரிபார்க்கவும்.
