# Solution 81: Fascinating Interplay between 100 and 001

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• Published on Feb 28, 2019
• Using careful experimentation and exploration, we craft the desired formula that guarantees an equal number of 100 and 001 in a binary string.
Congratulations to attyfarbuckle, alonamaloh, Rishav Gupta, Allaizn, Nicola C, Jaleb, and Sonal Kumari for successfully solving this math challenge question! attyfarbuckle was the first person to solve the question.
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For more Weekly Math Challenges:

• LetsSolveMathProblems  3 months ago +2

* At 7:45, I accidentally wrote 5112 as the final answer, even though it is 5122, as we found at 7:36. I apologize for the error. Thank you, GreenMeansGo, for notifying me!

• Jatmoz 3 months ago +3

It's fascinating how you make the solution sound like I could've actually solved it with enough patience. Cheers!

• Ravage PS 3 months ago +2

this was a fun problem!!! do you come up with these yourself or donyou find them on the internet?

• LetsSolveMathProblems  3 months ago +1

I have authored most of the Weekly Math Challenges so far, including this one. I'm glad you find them interesting! =)

• Atharva #breakthrough 3 months ago

Oh ok. I thought, as in CS, by substring you meant you could delete chars but have to keep them in sequence. With this interpretation of subsring can you solve the problem again?

• jkid1134 3 months ago +2

This is such a GOOD problem

• Aaron He 3 months ago +2

Why "find the last three digits" if the answer is four digits long?

• LetsSolveMathProblems  3 months ago +2

Originally, 2019 was the upper bound of the summation (instead of 12), so it was appropriate to ask for the final three digits of the answer. I did not change the statement of the problem after lowering the upper bound. However, looking back, perhaps it would have made more sense to ask for the entire four-digit answer. =)

• NoName 3 months ago

USA tests have 3 fields for numbers for each answer

• Certainly NOT the best pianist, but still 3 months ago +1

Maybe for testing, if the question is read properly and attentively?

• Serengeti Ghasa 3 months ago

Thanks a lot for providing
Such type challenge
I really could not understand the question so instead of last three digit of total nos I sum all the the nos and answer was 692.
Up to a4 we are 22 nos
From a5 to a12 we had 765×6 + 255×2=5100
So total nos are 5122
What a silly mistake done by me
With love and regards

• Sudheer Thunga 3 months ago +2

Hey how's it possible that you are sure that in the middle we will have 1's end

• Sudheer Thunga 3 months ago

@JackTheSpades Thank you so much!! :)

• JackTheSpades 3 months ago +4

He doesn't and it's irrelevant. If either left or right have 2 or more zeros (but not both) then they have an unequal amount of 100 and 001s.
So if the first two digits are 0, it doesn't matter if more zeros or ones follow.
for example: n=8:
00_0001_01 = 001x1, 100x0
00_1000_01 = 001x2, 100x1
Only if both left and right side have 11, 10 or 01 thus not allowing for more than 1 zero and 001/100 can't be formed outside of the 1's range.
Hope that helped.

• GreenMeansGO 3 months ago +4

You wrote 112 instead of 122

• LetsSolveMathProblems  3 months ago

You are absolutely right. The answer of 5122 at 7:36 is correct, but I mistakenly copied it as 5112 at 7:45. I just posted the correction as a pinned comment. Thank you for notifying me!

• vokuheila 3 months ago +16

All I see is

lol lolololol loooooll looolololooolll

• Max Haibara 3 months ago +1

Wow, that observation is very clever...