Another very easy problem, Apoor, a four year old child, had to write \(1\) to \( n\) numbers as a punishment. If he had made a mistake while writing any number, say \(r\), then he had to write that number again \((r-1)\) times, which he always wrote correctly. If he had made mistakes while writing all except one number and the total numbers he wrote is \(111\), then find the number that he wrote correctly.

I have a feeling that @KingGeorge will solve this one under 2 minutes.

I'm gonna say that he wrote 10 correctly.

Haha I knew it! ;)

Very nice :)

Since he wrote very number except one incorrectly, we can say he wrote \[1+2+3+4+...+n-(r-1)=111\]numbers total. Where \(r\) is the number he wrote correctly. If we look at the triangular numbers at http://oeis.org/A000217 we see that \(T_{15}=120\) is the smallest triangular number greater than 111. The next is \(T_{16}=136\). If it were \(T_{16}\) he would have had to have written 25 down correctly, but he can't have done that since \(25>16\). Thus, he wrote the numbers up to 15, and wrote \(120-(r-1)=111\) numbers total. Solving this, we get \(r=10\).

Typo. If it were \(T_{16}\) he would have had to written 26 down correctly. Not 25.

I think it's worth noting that this problem has a solution for every possible number of "total numbers" he wrote.

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