Problem 4 : first , we find the sum of numbers from 1 to 9 .
By using the arithmetic progression summation formula we get the sum of first n natural numbers as (n)(n+1)/2.
By putting 9(as there are 9 terms) in the same , we get 9*10/2=45.
Now we find the nearest perfect sq. LESS THAN 45 ( coz no sum from 1 to 9 can be greater than which we have taken as no number here is omitted and since our protagonist has missed a number from one to nine his sum is bound to less than 45 ).
We see that the number satisfying these conditions is 36 . Now we subtract: 45-36 =9 . So He must've missed 9 .