Question

If L.C.M. of (a, b )=52×34, L.C.M. of (b,c)=53×34

Answer 1

5^3 \times 3^4

Answer 2

The problem statement is incomplete. Assuming the question asks for the Least Common Multiple (L.C.M.) of (a, b, c) given the L.C.M. of (a, b) and (b, c).

Let the prime factorization of a, b, and c be: $a = 5^{x_a} \times 3^{y_a} \times K_a$

$b = 5^{x_b} \times 3^{y_b} \times K_b$

$c = 5^{x_c} \times 3^{y_c} \times K_c$

where $K_a, K_b, K_c$ are integers whose prime factors are neither 5 nor 3.

The L.C.M. of two numbers is found by taking the highest power of each prime factor present in either number.

L.C.M.(a, b) = $5^{\max(x_a, x_b)} \times 3^{\max(y_a, y_b)} \times \text{L.C.M.}(K_a, K_b)$

L.C.M.(b, c) = $5^{\max(x_b, x_c)} \times 3^{\max(y_b, y_c)} \times \text{L.C.M.}(K_b, K_c)$

We are given: L.C.M.(a, b) = $5^2 \times 3^4$

L.C.M.(b, c) = $5^3 \times 3^4$

Comparing the given L.C.M.s with the prime factorization form:

From L.C.M.(a, b) = $5^2 \times 3^4$:

1. $\max(x_a, x_b) = 2$
2. $\max(y_a, y_b) = 4$
3. $\text{L.C.M.}(K_a, K_b) = 1$. This implies that $K_a$ and $K_b$ have no prime factors other than 5 and 3 (which are already accounted for by the powers of 5 and 3) and their common factors must be 1. If they had any prime factor $p \ne 5, 3$, then $p$ would appear in L.C.M.(a,b). Since L.C.M.(a,b) only has factors 5 and 3, $K_a$ and $K_b$ must not have any other prime factors. Assuming $a, b, c$ only consist of prime factors 5 and 3, we take $K_a=K_b=K_c=1$.

From L.C.M.(b, c) = $5^3 \times 3^4$:

4. $\max(x_b, x_c) = 3$
5. $\max(y_b, y_c) = 4$
6. $\text{L.C.M.}(K_b, K_c) = 1$. With $K_b=1$, this implies $K_c=1$.

So, we assume $a = 5^{x_a} 3^{y_a}$, $b = 5^{x_b} 3^{y_b}$, $c = 5^{x_c} 3^{y_c}$.

We need to find L.C.M.(a, b, c) = $5^{\max(x_a, x_b, x_c)} \times 3^{\max(y_a, y_b, y_c)}$.

Let's determine the exponents for the prime factor 5:

From (1), $\max(x_a, x_b) = 2$. This means $x_a \le 2$ and $x_b \le 2$.

From (4), $\max(x_b, x_c) = 3$. This means $x_b \le 3$ and $x_c \le 3$.

Combining the constraints on $x_b$: $x_b \le 2$ and $x_b \le 3$. This implies $x_b \le 2$.

From $\max(x_b, x_c) = 3$, at least one of $x_b$ or $x_c$ must be equal to 3. Since $x_b \le 2$, it must be that $x_c = 3$.

So, we have $x_a \le 2$, $x_b \le 2$, and $x_c = 3$.

Now, we find the maximum of these exponents: $\max(x_a, x_b, x_c) = \max(\le 2, \le 2, 3) = 3$.

Let's determine the exponents for the prime factor 3:

From (2), $\max(y_a, y_b) = 4$. This means $y_a \le 4$ and $y_b \le 4$. Also, at least one of $y_a$ or $y_b$ must be 4.

From (5), $\max(y_b, y_c) = 4$. This means $y_b \le 4$ and $y_c \le 4$. Also, at least one of $y_b$ or $y_c$ must be 4.

We need to find $\max(y_a, y_b, y_c)$. We know $y_a \le 4$, $y_b \le 4$, and $y_c \le 4$.

Consider two cases for $y_b$:

Case 1: $y_b = 4$. Then $\max(y_a, y_b, y_c) = \max(y_a, 4, y_c)$. Since $y_a \le 4$ and $y_c \le 4$, the maximum is 4.

Case 2: $y_b < 4$. From $\max(y_a, y_b) = 4$, it must be $y_a = 4$. From $\max(y_b, y_c) = 4$, it must be $y_c = 4$. Then $\max(y_a, y_b, y_c) = \max(4, y_b, 4) = 4$ (since $y_b < 4$).

In both cases, $\max(y_a, y_b, y_c) = 4$.

Therefore, L.C.M.(a, b, c) = $5^{\max(x_a, x_b, x_c)} \times 3^{\max(y_a, y_b, y_c)} = 5^3 \times 3^4$.