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Number Theory Level 3

A pack contain $n$ cards numbered from 1 to $n$ . Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is $1224$ .If the smaller of the numbers on the removed cards is $k$ ,then find the value of $k$

The answer is 25.

2 solutions

Abhishek Garg
Jan 28, 2015

This is jee advanced 2013 integer type question.

Prakhar Bindal
Jan 21, 2015

Method 1 (Examination Approach) Firstly We Should Note That Sum of first n natural numbers is n(n+1)/2. sum of numbers on the cards removed = 2k+1 sum on remaining cards n(n+1)/2 - (2k+1) = 1224
let us start taking values of n for which n(n+1)/2 is greater than or equal to 1224 Suppose n = 51 then n(n+1)/2 = 1326 from here we get non integral value of k suppose n = 52 then again k is non integral suppose = n = 53 then k is greater than n . and for greater values of n k will always be greater if n = 49 then sum is 1225 k = 0 which is not possible so lower values are not possible so we arrive at a conclusion that n lies between 49 and 51. so n=50 is our answer put values to get k =25. Method 2 (By Using Mod Notation and some exceptional theorems of number theory Method 3 (Exact Method) n(n+1)/2 - (2k+1) = 1224
After a bit of manipulation we get (n+50)(n-49) = 4k but k is greater than 1 and less than n also as k is integer one of n+50 or n-49 must be a multiple of 4 so k has to be greater than 49 and if n = 53 then k exceeds n again we need to put n = 50,51,52 to get that n = 52 and 51 are not allowed

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The answer is 25.

2 solutions

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