Consider the 600600600-digit integer
234234234.....234.234234234.....234.234234234.....234.
The first mmm and the last nnn digits of the above integer are crossed out so that the sum of the remaining digits is 234234234. Find the value of m+nm+nm+n.
Answer:
522522522
- Observe that the given number has 234234234 repeated 200200200 times.
The sum of the repeating digits =2+3+4=9=2+3+4=9=2+3+4=9
⟹⟹⟹The sum of digits of the given number =9×200=1800=9×200=1800=9×200=1800 - After crossing out the first mmm digits and the last nnn digits, the sum of the remaining digits is 234234234.
⟹⟹⟹ the sum of first mmm and last nnn digits is 1800−234=15661800−234=15661800−234=1566 - Observe that 1566=174×91566=174×91566=174×9. Thus, we have to cross out 174174174 blocks of 333 digits 2,3, and 42,3, and 42,3, and 4 either from the front or the back. Thus, m+n=174×3=522m+n=174×3=522m+n=174×3=522.
- Hence, the value of m+nm+nm+n is 522522522.