In base 10, 5 divides 30.
In base _ _ , 5 divides 30 too.
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5 is divisible by numbers that ends with 0 or 5. Clearly from the options, 5 < 11, 12, 13, 14, 15. Therefore 5 remains 5 when converted to base ten from any of the options given. However, 30 > 11, 12, 13, 14, 15. Therefore converting to base ten would involve expansion by the powers of base. Hence base15 as it is the only option that gives a number that ends with 5, which is therefore divisible by 5.
We note that 5 b 3 0 b = 5 1 0 3 1 0 b 1 0 , where b denotes the number base. Since b > 5 , 5 b = 5 1 0 . 3 0 b is divisible by 5 b only if b is divisible by 5, therefore, it is divisible when b = 5 , 1 0 , 1 5 , …
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3 0 in base n is equal to 3 n .
To get an integer n , 3 n has to be equal to a number divisible by 3 (as well as by 5 as stated in the problem).
3 n needs to be larger than 3 0 , since all potential bases are larger than 1 0 .
The first number divisible by both 5 and 3 and larger than 3 0 is 4 5 .
3 n = 4 5 from which n = 1 5