Consider the 800800800-digit integer
234523452345.....2345.234523452345.....2345.234523452345.....2345.
The first mmm digits and the last nnn digits of the above integer are crossed out so that the sum of the remaining digits is 234523452345. FInd the value of m+nm+nm+n.


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Answer:

130130130

Step by Step Explanation:
  1. Observe that the given number has 234523452345 repeated 200200200 times.
    2+3+4+5=142+3+4+5=142+3+4+5=14
    The sum of digits of the given number =14×200=2800=14×200=2800=14×200=2800
  2. After crossing out the first mmm digits and the last nnn digits, the sum is 234523452345.
    ⟹⟹⟹ the sum of first mmm and last nnn digits is 2800−2345=4552800−2345=4552800−2345=455
  3. Observe that 455=32×14+7455=32×14+7455=32×14+7. Thus we have to cross out 323232 blocks of 444 digits 234523452345 either from the front or the back, a 222 from the front that remains and a 555 from the back that remains. Thus, m+n=32×4+2=130m+n=32×4+2=130m+n=32×4+2=130
  4. Hence, the value of m+nm+nm+n is 130130130.

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