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Consider the ^@800^@-digit integer
^@234523452345 ..... 2345.^@
The first ^@m^@ digits and the last ^@n^@ digits of the above integer are crossed out so that the sum of the remaining digits is ^@2345^@. FInd the value of ^@ m + n ^@.
Answer:
^@130^@
- Observe that the given number has ^@2345^@ repeated ^@200^@ times.
^@2 + 3 + 4 + 5 = 14^@
The sum of digits of the given number ^@ = 14 \times 200 = 2800^@ - After crossing out the first ^@m^@ digits and the last ^@n^@ digits, the sum is ^@2345^@.
^@\implies ^@ the sum of first ^@m^@ and last ^@n^@ digits is ^@2800 - 2345 = 455^@ - Observe that ^@455 = 32 \times 14 + 7^@. Thus we have to cross out ^@32^@ blocks of ^@4^@ digits ^@ 2345^@ either from the front or the back, a ^@2^@ from the front that remains and a ^@5^@ from the back that remains. Thus, ^@m + n = 32 \times 4 + 2 = 130 ^@
- Hence, the value of ^@m + n^@ is ^@130^@.
