What is the natural number n for which (3^9 + 3^{12} + 3^{15} + 3^n) is a perfect cube of an integer?

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Q: 40 (CAPF/2019)
What is the natural number n for which (3^9 + 3^{12} + 3^{15} + 3^n) is a perfect cube of an integer?

question_subject: 

Maths

question_exam: 

CAPF

stats: 

0,2,10,5,4,1,2

To find the natural number n for which the given expression is a perfect cube, we need to analyze each option.

Option 1: 10

If we substitute n=10, the expression becomes (3^9 + 3^12 + 3^15 + 3^10). Since the exponents are different, it is unlikely that this expression will result in a perfect cube. Therefore, option 1 is likely incorrect.

Option 2: 11

If we substitute n=11, the expression becomes (3^9 + 3^12 + 3^15 + 3^11). Again, the exponents are different, so it is unlikely that this expression will be a perfect cube. Hence, option 2 may also be incorrect.

Option 3: 13

If we substitute n=13, the expression becomes (3^9 + 3^12 + 3^15 + 3^13). The exponents are still different, so this expression is not a perfect cube. Thus, option 3 is likely incorrect.

Option 4: 14

If we substitute n=14, the expression becomes (3^9 + 3^12 + 3^15 + 3^14).