IPMAT 2025 Exam Solved Papers

The given expression, $3^{10}+2$, when cubed, gives us $\left(3^{10}+2\right)^{3}=3^{30}+8+\left(6 \times 3^{20}\right)+\left(12 \times 3^{10}\right)$This expansion can be rewritten $\operatorname{as}\left(\underline{\left.3^{10}+2\right)^{3}}=\left(3^{30}+8\right)+\underline{\left(6 \times 3^{10}\right)\left(3^{10}+2\right)}\right.$Since the two underlined parts are both divisible by $\left(3^{10}+2\right)$, we can conclude that the remaining nonunderlined part, $\left(3^{30}+8\right)$ should also be divisible by $\left(3^{10}+2\right)$. Therefore option D, is correct. We can do the same for other options.We can try $\left(3^{10}+2\right)^{2}$ to verify option A, the expansion will be $\left(3^{20}+4+4 * 3^{10}\right)$. Thus, after dividing this by ( $3^{10}+2$ ) the remainder will clearly be $4 * 3^{10}$, since this is definitely not divisible by ( $3^{10}+2$ ), as $3^{10}$ is not divisible by $\left(3^{10}+2\right)$, which is an odd number, we can conclude option A is incorrect.Option B is also incorrect because if $3^{30}+8$ is divisible after the cubic expansion, a numbe 6 less than it cannot be.Finally, if we assume option C to be divisible, then the difference between option C and option D should also be divisible by $\left(3^{10}+2\right)$, this difference is nothing but $\left(3^{30}-3^{20}\right)=3^{20}\left(3^{10}-1\right)$, since none of the two parts in this difference are divisible, the difference itself is not divisible, and option C will not be divisible either.Alternate Explanation:Let $3^{10}=a$ and $2=b$Then the given expression is of forma $+b$So, option A is of form $a^{2}+b^{2}$option B is of form $a^{3}+b$option C is of form $a^{2}+b^{3}$option D is of forma $a^{3}+b^{3}$Among all these polynomials, we know only $a^{3}+b^{3}$ has a factor $a+b$So, option D is the correct answer.