UAE
Let ^@S^@ be the smallest positive multiple of ^@15^@ that comprises exactly ^@3k^@ digits with k ‘0’s, k ‘3’s and k ‘8’s. Find the remainder when ^@S^@ is divided by ^@8^@.
Answer:
^@0^@
- If a number is a multiple of ^@15^@, it is a multiple of ^@3^@ and ^@5^@ both.
We are given that ^@S^@ is the smallest positive multiple of ^@15^@ which comprises exactly ^@3k^@ digits. Also, ^@S^@ has ^@k^@ ‘^@0^@’^@s^@, ^@k^@‘^@3^@’^@s, ^@ and ^@k^@‘^@8^@’^@s.^@
Observe that ^@S^@ must end with ^@0^@ as it is a multiple of ^@5^@. - The sum of all the digits of ^@S = k \times 0 + k \times 3 + k \times 8 = 3k + 8k = 11k^@
Since ^@S^@ is a multiple of ^@3^@, the sum of all its digits must be a multiple of ^@3^@.
The smallest value of ^@k^@ such that ^@11k^@ is a multiple of ^@3^@ is ^@3^@. Therefore, there are ^@3^@‘^@0^@’^@s^@, ^@3^@‘^@3^@’^@s, ^@ and ^@3^@‘^@8^@’^@s^@ in ^@S^@.
^@\implies S = 300338880^@ The remainder when ^@S^@ is divided by ^@8^@ ^@=^@ Remainder of (Last ^@3^@ digits of ^@S \div 8^@)
^@=^@ Remainder of (^@880 \div 8^@)
^@= 0^@- Hence, the remainder when ^@S^@ is divided by ^@8^@ is ^@0^@.
