111 is an amazing number!
1 1 1 2 + 1 1 1 3 + 1 1 1 4 + … + 1 1 1 1 6 = ?
Give your answer in decimal representation.
The answer is 1645.
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2 solutions
Moderator note:
Yes, this is the standard approach. For the sake of clarity, can you show your steps to obtain the value of 1 6 4 5 ?
Bonus question : Evaluate this expression below as well. And generalize your result for m = 1 ∑ n m m m m + 1 .
1 1 1 2 + 2 2 2 3 + 3 3 3 4 + 4 4 4 5 + … + 8 8 8 9 + 9 9 9 1 0
Note that ∑ k 2 = n ( n + 1 ) ( 2 n + 1 ) / 6 , ∑ k = n ( n + 1 ) / 2 and of course ∑ 1 = n . We get
6 1 6 ( 1 6 + 1 ) ( 2 × 1 6 + 1 ) + 2 1 6 ( 1 6 + 1 ) + 1 6 − 3 = 1 6 4 5
having to subtracting 3 since we're missing 1 1 1 1 .
Alternatively, you could do something similar but taking advantage of the fact that ( k + 1 ) 2 − k = 1 1 1 k .
1 1 1 2 in decimal system = 7
1 1 1 3 in decimal system = 1 3
1 1 1 4 in decimal system = 2 1
So we see that the numbers are in a series where the difference increases by 2 :
7 + 6 → 1 3 + 8 → 2 1 + 1 0 → 3 1 + 1 2 → 4 3 . . . . . . . .
Summing these upto the 1 6 t h term, we get 1 6 4 5
Moderator note:
You have only shown that it appears to be true for the first few terms of the expression. How do you know that the common difference between each consecutive terms increases by 2?
Despite that, can you think of an simpler approach to this problem?
For any radix r , as ( 1 1 1 ) r = r 2 + r + 1 , hence the given summation is r = 2 ∑ 1 6 ( r 2 + r + 1 ) = 1 6 4 5 Please see Jake Lai's first three lines for steps to get 1645.