S Korea
Consider the ^@600^@-digit integer
^@234234234 ..... 234.^@
The first ^@m^@ and the last ^@n^@ digits of the above integer are crossed out so that the sum of the remaining digits is ^@234^@. Find the value of ^@ m + n ^@.
Answer:
^@522^@
- Observe that the given number has ^@234^@ repeated ^@200^@ times.
The sum of the repeating digits ^@=2 + 3 + 4 = 9^@
^@ \implies ^@The sum of digits of the given number ^@ = 9 \times 200 = 1800^@ - After crossing out the first ^@m^@ digits and the last ^@n^@ digits, the sum of the remaining digits is ^@234^@.
^@\implies ^@ the sum of first ^@m^@ and last ^@n^@ digits is ^@1800 - 234 = 1566^@ - Observe that ^@1566 = 174 \times 9 ^@. Thus, we have to cross out ^@174^@ blocks of ^@3^@ digits ^@ 2, 3, \text{ and } 4^@ either from the front or the back. Thus, ^@m + n = 174 \times 3 = 522 ^@.
- Hence, the value of ^@m + n^@ is ^@522^@.
