Thread: middle school logic question

Looking at the topics here I am sure this is an easy question for you!
This is a middle school logic question to test insight and shouldn’t require hectic calculations.

Is it possible that both 99 and 100 could be correct answers?
If so why?
If not both options what is the correct answer?
thanks Phoebe

2-1+3-2+4-3+5-4+6-5+…+101-100

There's at least two ways to do this problem. One is to note that every other term looks like n - (n-1) = 1. How many of these terms are there? Well, there are 101 - 2 + 1 = 100 terms. Hence, the sum is 100.

The other way of doing this, that I know of, is to see that you really have a telescoping series. That is, there's a -2 for the 2, a -3 for the 3, and so on. Not every number has a corresponding negative, however. To see which ones don't, let me write out just a few more terms explicitly:

2-1+3-2+4-3+5-4+6-5+…+99-98+100-99+101-100.

So we can see that the -1 in the beginning doesn't have a +1 to cancel it out. Also, the 101 doesn't have a canceling term. Hence, the sum is 101 - 1 = 100.

Originally Posted by phoebe

Looking at the topics here I am sure this is an easy question for you!
This is a middle school logic question to test insight and shouldn’t require hectic calculations.

Is it possible that both 99 and 100 could be correct answers?
If so why?
If not both options what is the correct answer?
thanks Phoebe

2-1+3-2+4-3+5-4+6-5+…+101-100

2-1+3-2+4-3+5-4+6-5+…+101-100= (2-1)+(3-2)+(4-3)+(5-4)+(6-5)+…+(101-100)

each bracket has value 1 and there are 100 such brackets.

CB

Is it possible that 1+1 = both 2 and 3 ?