Easy

Mini-Geek · 2006-10-19, 14:19

How do you calculate (and what is) 1+2+3+4...97+98+99+100 in under 10 seconds?

retina · 2006-10-19, 14:39

Divide into 2 sets, 1-50 & 51-100
Reverse the order of the second set
Imagine the second set written below the first set
Add each vertical pair (the pattern emerges), all pairs add to 101
Clearly there are 50 such pairs
Answer is 101*50

This is the arithmetic progression

Mini-Geek · 2006-10-19, 16:22

Quote:

Originally Posted by retina

Divide into 2 sets, 1-50 & 51-100
Reverse the order of the second set
Imagine the second set written below the first set
Add each vertical pair (the pattern emerges), all pairs add to 101
Clearly there are 50 such pairs
Answer is 101*50

The way I found was the same, except you added 1+99, 2+98...49+51 to get 49 100s (4900) then added the 50 and the 100 that were left over (because neither of them could be added to something different to make 100), to get 5050.

mfgoode · 2006-10-19, 17:14

Sorry to correct you but Retina is right, tho' you are also right tho' your method is not orthodox.
This problem was solved in a jiffy by Gauss, when he was 6 yrs. old , one of the three greatest mathematicians, who ever lived,[ Archimedes, Newton and Gauss]
I'm glad you mentioned it as it has refreshed my aging memory.

Mally

2006-10-19, 14:19	#1
Mini-Geek Account Deleted "Tim Sorbera" Aug 2006 San Antonio, TX USA 10253₈ Posts	Easy How do you calculate (and what is) 1+2+3+4...97+98+99+100 in under 10 seconds?

2006-10-19, 14:39	#2
retina Undefined "The unspeakable one" Jun 2006 My evil lair 1011101101010₂ Posts	Divide into 2 sets, 1-50 & 51-100 Reverse the order of the second set Imagine the second set written below the first set Add each vertical pair (the pattern emerges), all pairs add to 101 Clearly there are 50 such pairs Answer is 10150 This is the arithmetic progression Last fiddled with by retina on 2006-10-19 at 14:45 Reason: spoiler tags added*

2006-10-19, 17:14	#4
mfgoode Bronze Medalist Jan 2004 Mumbai,India 2²×3³×19 Posts	Arithmetic progression. Sorry to correct you but Retina is right, tho' you are also right tho' your method is not orthodox. This problem was solved in a jiffy by Gauss, when he was 6 yrs. old , one of the three greatest mathematicians, who ever lived,[ Archimedes, Newton and Gauss] I'm glad you mentioned it as it has refreshed my aging memory. Mally Last fiddled with by mfgoode on 2006-10-19 at 17:16 Reason: Add on

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