### Brain Teasers

# Number Tricks

Ms. Arroyo asked the class to see if they could find the sum of the first 50 odd numbers. As everyone settled down to their addition, Terry ran to her and said, "The sum is 2,500." Ms. Arroyo thought, "Lucky guess," and gave him the task of finding the sum of the first 75 odd numbers. Within 20 seconds, Terry was back with the correct answer of 5,625.

How does Terry find the sum so quickly?

How does Terry find the sum so quickly?

### Hint

Consecutive odd numbers are 1, 3, 5, 7....Start with the first several odd numbers and look for a pattern.

### Answer

The following pattern holds: The sum is equal to n x n, when n is the number of consecutive odd numbers, starting with 1. For example, the sum of the first 3 odd numbers is equal to 3 x 3, or 9; the sum of the first 4 odd numbers is equal to 4 x 4, or 16; the sum of the first 5 odd numbers is equal to 5 x 5, or 25; and so on.Hide Hint Show Hint Hide Answer Show Answer

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## Comments

I don't get it, 50*50=2500

I don't understand it.

I don't understand it.

i think you worded it wrong...it's not 'the first 50 odd numbers', coz that'll mean [1,3,5...98,99]...should be 'the sum of all odd numbers up to 50' if you wish the answer to be 625, else it'd be 2500 instead...

Your explanation makes your example wrong....

Tut.

Tut.

ur wrong n da teaser. 50x50=2500

The "sum of all the odd numbers up to 75" is 1444.

morning rain8, i think you corrected some correct part, too...

morning_rain08, the puzzle is wrong. Use either of the following:

1.

The sum of the first 50 odd numbers (those between 1 and 100) is 50x50=2500.

The sum of the first 75 odd numbers (those between 1 and 150) is 75x75=5625.

2.

The sum of all odd numbers up to 50 (those between 1 and 50) is 25x25=625.

The sum of all odd numbers up to 75 (those between 1 and 75) is 38x38=1444.

1.

The sum of the first 50 odd numbers (those between 1 and 100) is 50x50=2500.

The sum of the first 75 odd numbers (those between 1 and 150) is 75x75=5625.

2.

The sum of all odd numbers up to 50 (those between 1 and 50) is 25x25=625.

The sum of all odd numbers up to 75 (those between 1 and 75) is 38x38=1444.

WOW, I THINK

"Explainers"-

Please replace "between 1 and 100" with "between 0 and 100" or "betweem 1 and 99, inclusive." Keep it consistant.

Please replace "between 1 and 100" with "between 0 and 100" or "betweem 1 and 99, inclusive." Keep it consistant.

why did i find this Easy?

75x75=5625

any number ending in 5 can easily be squared.

just take the number adjacent to the last 5:

75 => 7

145 => 14

35 => 3

then you multiply it with the number just one above it:

7x8= 56

then add 25 to the end

56 25 => 5625!

(exclamation point not factorial)

any number ending in 5 can easily be squared.

just take the number adjacent to the last 5:

75 => 7

145 => 14

35 => 3

then you multiply it with the number just one above it:

7x8= 56

then add 25 to the end

56 25 => 5625!

(exclamation point not factorial)

The reason this works is because

x^2 = (x-1)^2 + x + (x-1)

= x^2 - 2x + 1 + x + (x-1)

= x^2

this means that the difference between x^2 and (x+1)^2 is 2x + 1 (e.g. 10^2 = 100, 11^2 = 121, 121 - 100 = 21 = 10 + 11.)

1^2 - 0^2 = 1

2^2 - 1^2 = 3

3^2 - 2^2 = 5

4^2 - 3^2 = 7

. . .

So summing the first n odd numbers gives n^2.

x^2 = (x-1)^2 + x + (x-1)

= x^2 - 2x + 1 + x + (x-1)

= x^2

this means that the difference between x^2 and (x+1)^2 is 2x + 1 (e.g. 10^2 = 100, 11^2 = 121, 121 - 100 = 21 = 10 + 11.)

1^2 - 0^2 = 1

2^2 - 1^2 = 3

3^2 - 2^2 = 5

4^2 - 3^2 = 7

. . .

So summing the first n odd numbers gives n^2.

the sum of the first n odd numbers is n^2

Ok, my comment looks really lame after reading the others but yeah, the answer is correct

My experience as an actuary made this quite easy. But I'm sure it was hard for most people.

I figured out this trick in college (I was a math major) so this was easy for me. Good teaser!

And for the sum of even numbers its ( N x N ) + N

I love math, but never knew this trick. Very cool !

Jun 25, 2009

1st 50 odd numbers are:

1, 3, ..., 97, 99

Add 1+99=100, 3+97=100, ..., 49+51=100 (25 times).

25*100=2500

1, 3, ..., 97, 99

Add 1+99=100, 3+97=100, ..., 49+51=100 (25 times).

25*100=2500

wow this teaser brings lots of comments...nothing to say, it was pretty hard for me . but once i saw the answer, it amazed me that math is so precise!

huh? totally don't get it!

The early comments are confusing, since it must have been worded incorrectly at first.

I, too, am a bit of a math nerd but never knew this before, so it's a neat trick for me.

I, too, am a bit of a math nerd but never knew this before, so it's a neat trick for me.

sbminerb is the first comment that makes sense to me. Thanks!

Interesting.

Also, if you wanted to do even numbers it would be n^2 + n

Brilliant, i got it with my first try. For the guys that are confised, it's the first 50 ODD numbers, not just the first 50 numbers. So in conclusion, the range is to a hundred.

I guess this must have been corrected - no problem there now, the way I see it.

I used a 'series' method of finding the sum. The numbers are 1+3+5+ ...+97+99. Adding 1+99, 3+97, 5+95 etc gives us 25 pairs of numbers whose sum is 100. 25x100=2500.

I used a 'series' method of finding the sum. The numbers are 1+3+5+ ...+97+99. Adding 1+99, 3+97, 5+95 etc gives us 25 pairs of numbers whose sum is 100. 25x100=2500.

Oct 08, 2010

I used the summation technique:

Let Sij[pattern] = "Summation from i to j of pattern"

First 50 odd numbers = S0,49[2n+1]

= 2S0,49[n] + S0,49[1]

= 2x[49(49+1)/2] + 50

= 49(50) + 50

= 50(49 + 1)

= 50(50), 2500, or n^2

**note that Sij[n] refers to the arithmetic sequence

**also note that Sij[CONSTANT NUMBER] = Constant*(j-i+1)

Let Sij[pattern] = "Summation from i to j of pattern"

First 50 odd numbers = S0,49[2n+1]

= 2S0,49[n] + S0,49[1]

= 2x[49(49+1)/2] + 50

= 49(50) + 50

= 50(49 + 1)

= 50(50), 2500, or n^2

**note that Sij[n] refers to the arithmetic sequence

**also note that Sij[CONSTANT NUMBER] = Constant*(j-i+1)

Nov 08, 2010

There is no sum of the first 50 odd numbers, nor is there a sum of the first 75 odd numbers, BECAUSE

odd numbers are also negative.

Rephrase part of the question with

"the sum of the first 50 odd POSITIVE numbers," etc.

odd numbers are also negative.

Rephrase part of the question with

"the sum of the first 50 odd POSITIVE numbers," etc.

Nov 19, 2010

For some reason I found the answer by taking (n-1)(n+1)+1.

(50-1)(50+1)+1

49*51+1

2499+1

2500

I have no idea how I came to this conclusion, but what the hey.. It works, doesn't it?

(50-1)(50+1)+1

49*51+1

2499+1

2500

I have no idea how I came to this conclusion, but what the hey.. It works, doesn't it?

vintagepotato: read my explanation above if you want to understand why your solution also works.

COOLNESS!!! I used to use that all the time, but then I forgot about it. Now I realize.... it's always a square! I also know a handy technique for squaring ANY number ending in 5 in a matter of seconds.

Good grief! Another math problem to solve. These are not even fun. At least not to me. Don't suggest I not even try it as I DID NOT!!!

boring sorry not a math person.

It can be solved in seconds by opening an Excel Spreadsheet and dragging down the first 50 & 75 odd numbers and hitting auto-sum LOL!

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