Find the least value of (3K2 − 2) for which six digits number 7K8508 is divisible by 6.
10
A number is divisible by 6 when it is divisible by both 2 and 3. The last digit of 7K8508 is 8, so divisibility by 2 is satisfied for every K.
Digit sum = 7 + K + 8 + 5 + 0 + 8 = 28 + K. For divisibility by 3, (28 + K) ≡ 0 (mod 3), i.e. K ≡ 2 (mod 3). The smallest such single-digit K is K = 2.
Therefore 3K² − 2 = 3 × 4 − 2 = 12 − 2 = 10.
Hence, the answer is 10.
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2. S is always divisible by 9.
select the correct answer using the code given below:
