Find the smallest integer n such that (for some number of 1's)
9 9 9 9 9 9 × n = 1 1 1 … 1 1 1 .
It can be expressed as
b ( 1 0 6 − 1 ) 1 0 5 4 − 1 .
Find 1 0 + b .
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Yeah! That's one correct solution.
So simple and is level 5! Overrated!
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The problem when initially phrased was extremely hard to understand and so many people gave up on it / answered wrongly. I've now rephrased the question for clarity, and it is much easier to read.
I rated it level 3 , but it got level 5!
FYI: It was not necessary that a = 1 0 . There are actually multiple ways to express that number in the given form. We could have used a as any power of 10, and then use b to divide out the difference (while still being an integer).
As such, I have updated the problem to indicate that a = 1 0 .
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yeah! I never thought about that too! I now realized that we have to be much careful , before posting a problem. :)
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Right :)
This is a good example where clear phrasing of a problem would allow others to actually understand what you are intending to say, and hence be able to work on / solve the problem. The question, as initially phrased, was extremely confusing, and I had difficulty figuring out what you meant. I had to look at your solution, and then backtrack out what was in your mind.
Great job with posting problems, I look forward to seeing more good problems, as you learn from your experience.
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@Calvin Lin – Thank you sir!, I will try my best .
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We know that: 1 1 1 1 . . . 1 1 = i = 0 ∑ n 1 0 i = 9 1 0 n + 1 − 1
We multiply and divide by 1 0 6 − 1 = 9 9 9 9 9 9 to get: 1 1 1 1 1 . . . 1 1 = 1 0 6 − 1 9 9 9 9 9 9 ∗ 9 1 0 n + 1 − 1 Or: 1 1 1 1 . . . 1 1 = 9 9 9 9 9 9 ∗ 9 ∗ ( 1 0 6 − 1 ) 1 0 n + 1 − 1 Now, we observe that no matter what is the number of ones, the base will always remain 1 0 so a = 1 0 and as we see b = 9 . Hence the answer is: a + b = 1 9