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

Step by Step Explanation:
  1. 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
  2. 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 1800234=15661800234=15661800234=1566
  3. 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.
  4. Hence, the value of m+nm+nm+n is 522522522.

You can reuse this answer
Creative Commons License
whatsapp logo
Chat with us