Consider the 800800-digit integer
234523452345.....2345.234523452345.....2345.
The first mm digits and the last nn digits of the above integer are crossed out so that the sum of the remaining digits is 23452345. FInd the value of m+nm+n.
Answer:
130130
- Observe that the given number has 23452345 repeated 200200 times.
2+3+4+5=142+3+4+5=14
The sum of digits of the given number =14×200=2800=14×200=2800 - After crossing out the first mm digits and the last nn digits, the sum is 23452345.
⟹⟹ the sum of first mm and last nn digits is 2800−2345=4552800−2345=455 - Observe that 455=32×14+7455=32×14+7. Thus we have to cross out 3232 blocks of 44 digits 23452345 either from the front or the back, a 22 from the front that remains and a 55 from the back that remains. Thus, m+n=32×4+2=130m+n=32×4+2=130
- Hence, the value of m+nm+n is 130130.