Okay, so I kind of remember how to solve these problems… However, I forgot the quick method.
For positive integers a and b let a ~ b be define by a~b = a + (a-1) + (a-2) + … +b. For example, 9~4 = 9+8+7+6+5+4=39
What is the value of (100~5) -(98-3)
@avneety Hint: don’t actually compute (100~5) and (98~3) (you can still solve this way but it’ll take a bit longer).
Instead, think about what numbers are subtracted and which ones remain. By approaching it correctly, this problem can be solved in under 15 seconds.
Alternatively, you can solve the problem quickly by simply rearranging the numbers around.
108?
@Babuji answer is actually 192.
I’snt it 100+99+4+5=208 @MITer94
"What is the value of (100~5) -(98-3) "
This isn’t just about an answer, it’s about an approach. This type of problem comes up a lot on the SAT
Line up the numbers: 100 + 99 + 98 + 97 + … + 6 + 5
minus 98 + 97 + … + 6 + 5 + 4 + 3 (sorry, I can’t get them to line up here)
Everything from 98 to 5 cancels, leaving you with 100 + 99 + -( 4 + 3)… remember, everything in the bottom row is subtracted.
I get 192
Any time you find yourself doing tons of calculations on an SAT problem, there’s a trick that you’re missing.
I.e.,
100 + 99 + 98 + 97 + ... + 6 + 5
___________98 + 97 + … + 6 + 5 + 4 + 3 =
100 + 99 - 4 - 3 = 192
If you don’t notice that you could apply this trick to the question, use the brute force -
sum of arithmetic sequence formula
http://www.regentsprep.org/regents/math/algtrig/atp2/ArithG3.gif
(100-4)×(100+5)/2 - (98-2)×(98+3)/2 = 96×(105-101)/2 = 96×2 = 192
The other way is to notice that (100~5) - (98~3) = (100-98) + (99-97) + … + (5-3) = 2*96 = 192.
@MITer94
Exactly - this is the most common approach to this type of questions, like
Find the difference between sums of all even numbers and all odd numbers in the range of 1 to 100 (inclusive):
(100-99) + (98-97) + … + (2-1) = 1×50 = 50.
Another way to express the @bjkmom 's short cutting solution:
(100~5) - (98~3) = (100~99) + (98~5) - ((98~5) + (4~3)) =
199 - 7 = 192. Same difference :D.