# Writer’s Block – 4 ## 6 thoughts on “Writer’s Block – 4”

1. Shalini Chaudhuri |

can u pls help me out with ds ques given below:-
{2^2+(22)^2+(222)^2+(2…49 digits)^2}/9.need to find out the remainder

• catcracker |

Shalini,

We can see a pattern here. 2, 22, 222, 2222 etc are equivalent to 2, 4, 6, 8 etc when divided by 9. So their squares will also be equivalent to 2^2, 4^2, 6^2 and so on when divided by 9. Overall we can say the required answer will be equivalent to that of (2^2 + 4^2 + 6^2 +….98^2) when divided by 9, which is the same as [2^2 * (1^2+2^2+3^2+…+49^2)] or [4 * (49*50*99)/6] when divided by 9. I guess it will come 6 (no access to paper right now so…)

regards
J

2. Shalini Chaudhuri |

thnk u 🙂

3. latticesam |

Dear J,
a. What is the intuition and/or logic behind the statement ” the third order differences are same, so it will be a cubic equation” ? How do I arrive at this? in other questions as well ?
b. In Question 3 it is stated that n>11 , then how can we eliminate the options by using n<11.
I have seen you do this trick in an earlier question also on the blog. What is the legitimacy of it, is it foolproof ? If not what do we have to look for so as to be sure that taking a value that violates the condition will not affect the answer?

Regards,
Kartik

• catcracker |

It is based on the same logic as “the third derivative of a cubic polynomial is a constant”. If you think about it, a cubic sequence is a discrete form of a cubic polynomial…so will exhibit similar behaviour. And yes, it is pretty much foolproof 🙂

regards
J

• latticesam |

Perfect! I got it regarding the cubic sequence. Thanks a lot =D
I request you to publish some practice questions as well on the concepts discussed. Like the one’s in the Writer’s block and Unwrite, there are only about 3-4 in each of them. If you could provide only questions, we could solve it as an assignment. I hope you consider it 🙂
Did you say that the cubic method is foolproof or about the trick in Q3?
Regards,
Kartik