# Fun Math Fridays – 101 problems, Math aint one

My client and friend I-eXcel brought me a practice exam with no answers.  We went through it, and I enjoyed this problem so much, I decided to bless your soul with it!

Which statement is true about 101 consecutive integers ?  A. mean > median, B. mean = median, C. mean < median, D. cannot determine (based on real facts, lol)

My solution:

I. Start with the most common 101 consecutive integers (counting numbers): 1 2 3 … 48 49 50 51 52 … 97 98 99 100 101

Mean With an engineering mind, I spy that you always get a sum of 100 when you add: 1+99, 2+98, 3+97 … 48+52, 49+51.  So you have 49 x 100 = 4900.

Now all you have to add is the leftovers.  4900+50+100+101 = 5151

Divide that by you number of integers (101). Mean = 5151/101 = 51

Median Make it trivial, use integers 1 thru 11.  1 2 3 4 5 6 7 8 9 10 11.  Median is 6.  Similarly, 1 thru 101, Median = 51.

II. Another set of 101 consecutive integers: 0 1 2 3 … 48 49 50 51 52 … 97 98 99 100

Mean 0+100, 1+99, 2+98, 3+97 … 48+52, 49+51.  50 x 100 = 5000.

Leftovers. 5000+50 = 5050

Mean = 5050/101 = 50

Median 0 1 2 3 4 5 6 7 8 9 10. Median is 5. Similarly, 0 thru 100, Median = 50.

III. Last one, integers include negative numbers, too: -1 0 1 2 3 … 48 49 50 51 52 … 97 98 99

Mean 1+99, 2+98, 3+97 … 48+52, 49+51.  49 x 100 = 4900.

Leftovers. 4900+ -1 +0 = 4899

Mean = 4899/101 = 48.5

Median -1 0 1 2 3 4 5 6 7 8 9. Median is 4. Similarly, -1 thru 99, Median = 49.

AHHHHHHHHHHHH!! What happened?! Those old negatives make thing tricky!

The correct answer is D-cannot determine!

Victory!  You win!

#FunMathFridays #MathRoxMySox #101ProblemsMathAintOne  Enjoy your weekend!

## 1 thought on “Fun Math Fridays – 101 problems, Math aint one”

1. Pingback: 101 Problems | ADORN THE WORLD