H.C.F. of $3240$, $3600$ and a third number is $36$ and their L.C.M. is $2^4 \times 3^5 \times 5^2 \times 7^2$. The third number is

### Concept & Formula When analyzing the H.C.F. and L.C.M. of three or more numbers via prime factorization: * **H.C.F.** is formed by taking the **lowest power** of each prime factor common to all numbers. * **L.C.M.** is formed by taking the **highest power** of every prime factor present across any of the numbers. By comparing the prime factorizations of the known numbers against the given H.C.F. and L.C.M., we can deduce the exact prime factor makeup of the unknown third number. ### Step-by-Step Solution * **Given:** * Number 1 ($N_1$) = $3240$ * Number 2 ($N_2$) = $3600$ * Unknown Number 3 ($N_3$) = ? * $H.C.F. = 36$ * $L.C.M. = 2^4 \times 3^5 \times 5^2 \times 7^2$ * **Step 1:** Prime factorize $N_1$, $N_2$, and the H.C.F. * $3240 = 324 \times 10 = 18^2 \times (2 \times 5) = (2 \times 3^2)^2 \times (2 \times 5) = 2^3 \times 3^4 \times 5^1$ * $3600 = 36 \times 100 = (2^2 \times 3^2) \times (2^2 \times 5^2) = 2^4 \times 3^2 \times 5^2$ * $H.C.F. (36) = 2^2 \times 3^2$ * **Step 2:** Let $N_3 = 2^a \times 3^b \times 5^c \times 7^d$. We will determine $a, b, c, d$ by applying H.C.F. and L.C.M. rules. * **Step 3 (Analyze base 2):** * H.C.F. has $2^2$. This means the *minimum* power of 2 among $N_1(2^3)$, $N_2(2^4)$, and $N_3(2^a)$ must be $2$. Thus, $a = 2$. (This eliminates options c and d). * **Step 4 (Analyze base 3):** * L.C.M. has $3^5$. This means the *maximum* power of 3 among $N_1(3^4)$, $N_2(3^2)$, and $N_3(3^b)$ must be $5$. Neither $N_1$ nor $N_2$ has $3^5$, so $N_3$ **must** provide it. Thus, $b = 5$. * **Step 5 (Analyze base 5):** * H.C.F. has no 5 (or $5^0$). $N_1$ has $5^1$ and $N_2$ has $5^2$. To force the minimum power of 5 to be 0, $N_3$ must not contain 5 as a factor. Thus, $c = 0$. (This eliminates option b). * **Step 6 (Analyze base 7):** * L.C.M. has $7^2$. Neither $N_1$ nor $N_2$ has any 7s. Thus, $N_3$ **must** provide the $7^2$. Thus, $d = 2$. * **Conclusion:** Bringing it all together, $N_3 = 2^2 \times 3^5 \times 7^2$. ### Exam Strategy & Shortcut **Surgical Elimination:** You do not need to calculate everything to find the answer. Look at the L.C.M.: it contains $7^2$ and $3^5$. By quick mental prime factorization, $3240$ maxes out at $3^4$, and $3600$ maxes out at $3^2$. Neither provides $3^5$, and neither provides $7^2$. Therefore, the third number **must** contain exactly $3^5 \times 7^2$. Only options (a) and (d) have this. Next, look at the H.C.F.: $36$ means $2^2$. Since $3240$ has $2^3$ and $3600$ has $2^4$, the third number **must** be the one providing the $2^2$ limitation. Therefore, it must be $2^2 \times 3^5 \times 7^2$. ### Common Pitfall A common error is confusing the minimum and maximum power rules, leading students to select an option like (d) $2^3 \times 3^5 \times 7^2$. If the third number had $2^3$, then the lowest power of 2 across all three numbers would be $2^3 = 8$. But the H.C.F. is $36$, which only contains $2^2 = 4$. Thus, the third number strictly anchors the H.C.F. to $2^2$. ### Final Answer **Therefore, the correct answer is $2^2 \times 3^5 \times 7^2$.**