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)

Try it out, and leave your answers or steps you took in the comments.

*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!

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