This week I tried a couple of ideas I got from MathNotations on my middle school group.

First I presented this problem:

### How much greater is the sum of 51+52+53+…+100 than the sum of 1+2+3+…+50?

Because I had predicted that they would all solve this by summing each series and subtracting, I explicitly asked them to find two ways, showing their work each time.

As predicted, they all tried summing each series using Gauss’ method (pairing up first and last, second and next-to-last, etc. gives 25 pairs, each with a sum of either 51 or 151, depending on which sequence you’re working on, so the sums are 51 x 25 and 151 x 25), with limited success (arithmetic errors, and multiplying the “pair totals” by 50 for the number of entries rather 25 for the number of pairs were the major stumbling blocks).

Only 2 (of 8 ) students came up with a second way. They were all pretty distressed that I wanted them to find another way. I hinted that it was possible to find the difference without adding up either series, which was totally mystifying to most.

One student came up with the method of comparing corresponding terms (each term in the larger sequence is 50 more than the corresponding term in the smaller sequence, so the total difference is 50 x 50), and another noticed that when using the Gauss method, the “pair sum” (151) for the larger sequence was 100 more than the pair sum (51) for the smaller sequence, so the total difference would be 100 x 25.

However, when I tried to show them that they could figure out (151 x 25) – (51 x 25) without multiplying out those terms, they were mystified. Note to the regular math teacher: review the distributive law!

Then we moved on to the fascinating facts activity. We went over the examples for 17 in class, then I had them work in pairs on 97. I assigned 153 for homework:

Some of the fun properties they found for 97:

• It is prime
• it is odd
• it can be represented as the sum of consecutive positive integers: 48 + 49
• The sum of the digits is a square (4^2), a 4th power (2^4) and unsummable!
• its palimage (79) is also prime
• take 97, subtract its palimage (79) to get 18, then sum the digits to get 9, which is the same as the sum of the digits in the product of its digits (63)
• it can be represented as a sum of squares: 9^2 + 4^2
• it can be represented as a sum of 4th powers: 3^4 + 2^4

Some of the properties they found for 153:

• it is odd
• it is composite
• all its digits are odd, as are the sum and product of its digits
• the product of the digits (15) can be written as 5^2 – 3^2 – 1^2 (notice use of the square of each digit)
• the product of the digits can be written as a difference of squares: 4^2 – 1^2
• the pairwise differences between digits are all even
• it can be written as the sum of 17 consecutive integers: 1 + 2 + 3 + … + 17
• it can be written as the sum of 3 consecutive integers: 50 + 51 + 52
• it can be written as the sum of 3 squares: 4^2 + 4^2 + 11^2 or 10^2 + 7^2 + 2^2
1. April 5, 2008 9:15 am

Another famous property of 153 is that 153 = 1^3 + 5^3 + 3^3.

2. April 5, 2008 11:39 am

Nice one, Joshua.

3. April 6, 2008 8:06 am

Thank you Mathmom for trying these with your students. i always depend on you to bring these investigations to life with anecdotal evidence. I also posted a long reply to your comment on my blog regarding why I chose 97 and 153. Enjoy!
Dave Marain