Interesting Summation

Algebra Level 5

When M M is a digit, M + M M ‾ + M M M ‾ + ⋯ + M M M ⋯ M ‾ ⏟ n M ’s = M a [ 1 0 n + 1 − 10 − 9 n ] . M + \overline{MM} + \overline{MMM} + \cdots + \underbrace{\overline{MMM\cdots M}}_{n \ M\text{'s}} = \frac{M}{a}\left [ 10^{n+1}-10-9n \right ]. Find the value of a . a.

Note that M M ‾ \overline{MM} represents a two-digit number with both digits being M . M.


The answer is 81.

Tanuj Ravi by Tanuj Ravi , 22, India

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1 solution

Tanuj Ravi
Feb 3, 2015

Taking m common 

                   m(\quad 1+11+111+1111..........upto\quad n\quad terms\quad )

Multiplying and dividing by 9 

               \frac { m }{ 9 } \left[ 9+99+999+9999............ \right]

This can be written as 

              \frac { m }{ 9 } \left[ 10-1+100-1+1000-1............... \right]

            \frac { m }{ 9 } \left[ \sum _{ k=1 }^{ n }{ { 10 }^{ k } } \quad -\quad (1+1+1......to\quad n\quad terms) \right] 

          \frac { m }{ 9 } \left[ 10\frac { \left[ { 10 }^{ n }-1 \right]  }{ 10-1 } \quad -\quad n \right]

simplifying further we get 

           \frac { m }{ 81 } \left[ { 10 }^{ n+1 }-10\quad -\quad 9n \right]

so the value of a 81

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Alternatively, you can interpret the problem as computing ∑ k = 1 n ∑ j = 0 k ( m ⋅ 1 0 j ) \displaystyle\sum_{k=1}^n\sum_{j=0}^k\left(m\cdot 10^j\right) . This is quite trivial using GP summation formula and basic summation laws.

Prasun Biswas - 5 years, 8 months ago

Hey cant we use inverse mathematical induction here...

Mohit Gupta - 5 years, 3 months ago

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