Clean the blackboard!!
The integers 1, 2, …, 40 are written on a blackboard. The following operation is then repeated 39 times: In each repetition, any two numbers, say a and b, currently on the blackboard are erased and a new number a + b – 1 is written. What will be the number left on the board at the end?
Taken from CAT (Common admission test) paper.
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
Total sum of the numbers written on the blackboard =(40 times 41/ 2)=820 (sum of first N natural numbers is (n+1) times n/2) When two numbers ‘a’ and ‘b’ are erased and replaced by a new number a + b – 1, the total sum of the numbers written on the blackboard is reduced by 1. Since this operation is repeated 39 times, therefore, the total sum of the numbers will be reduced by 1 × 39 = 39. Therefore, after 39 operations there will be only 1 number that will be left on the blackboard and that will be 820 – 39 = 781
Hope this works :).