Do JEE use bashing?

Algebra Level 4

A pack contains n n cards numbered from 1 to n \text{1 to n} .Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224.Find the smaller of the number on removed cards.


The answer is 25.

Mohit Gupta by Mohit Gupta , 20, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Sum of the numbers = n ( n + 1 ) 2 \frac{n(n+1)}{2} , Let the numbers removed be p , p + 1 p,p+1

n ( n + 1 ) 2 − ( 2 p + 1 ) = 1224 \frac{n(n+1)}{2} - (2p+1) = 1224

or, n ( n + 1 ) 2 = 1225 + 2 p \frac{n(n+1)}{2} = 1225 + 2p

So n ( n + 1 ) 2 > 1225 \frac{n(n+1)}{2} > 1225

Solving we get n > 49 n>49 .

For n = 50 n=50 we have The total sum as 1275.

So p + ( p + 1 ) = 1275 − 1224 p+(p+1)=1275-1224

Therefore p = 25 .

2 Helpful 0 Interesting 0 Brilliant 0 Confused

I think the topic should be number theory. It would be fine

Aditya Narayan Sharma - 5 years, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...