1. Man Vs. Machine: A Prime Example of Number Sense
AMTNYS
October 27-29, 2011
Presented By:
Adam Sprague and Stephanie Wisniewski
SUNY Fredonia

2. Round 1: Repeat or Terminate
𝟏 𝟒𝟎
Terminate
𝟏 𝟐 𝟒𝟖 𝟓 𝟓𝟔 𝟗
Repeat
𝟏 𝟕𝟓
Repeat

3. The Number Sense Involved
We recognize a terminating decimal as a fraction with 𝟏𝟎 𝒑 in the denominator. Since the prime factorization of 𝟏𝟎 is 𝟐∙𝟓, the denominator can also be written as 𝟐 𝒑 𝟓 𝒑 where 𝒑 is any integer. For example: 𝟏 𝟑𝟐 = 𝟏 𝟐 𝟓 = 𝟓 𝟓 𝟓 𝟓 ∙ 𝟏 𝟐 𝟓 = 𝟓 𝟓 𝟓 𝟓 ∙ 𝟐 𝟓 = 𝟑𝟏𝟐𝟓 𝟏𝟎 𝟓 =𝟎.𝟎𝟑𝟏𝟐𝟓 Since the denominator 𝟑𝟐 could be deduced to a power of ten, 𝟏𝟎 𝟓 , we were able to determine this was a terminating decimal.

4. The Number Sense Involved
A fraction repeats when the denominator cannot be rewritten as a power of ten. For example: 𝟏 𝟕𝟓 = 𝒏 𝟏𝟎 𝟐 𝟐 𝒑 𝟓 𝒑 = 𝟏𝟎 𝒑 =𝟕𝟓𝒏=𝟑∙ 𝟓 𝟐 ∙𝒏 Due to the uniqueness of prime factorization it will repeat since 𝟑 is not a prime factor of 𝟏𝟎.

5. Round 2: Is This Number a Perfect Square𝟖𝟏
Yes, Perfect Square
𝟖𝟏𝟎
No, Not a Perfect Square
𝟖𝟏𝟎𝟎
Yes, Perfect Square
𝟖𝟏𝟎𝟎𝟎
No, Not a Perfect Square

6. The Number Sense Involved
Every perfect square can be broken down to its prime factors each raised to an even power. We have a perfect square when all the powers of the prime factors are even, such as in 𝟖𝟏, 𝟖𝟏𝟎𝟎, 𝟖𝟏𝟎𝟎𝟎𝟎 and 𝟐 𝟒𝟒 𝟓 𝟏𝟑𝟖 𝟑 𝟐𝟒𝟒 𝟕 𝟏𝟐 𝟗 𝟐 . This is not the case for 𝟖𝟏𝟎 or 𝟖𝟏𝟎𝟎𝟎 because their prime factors are not all of even powers.

7. The Number Sense Involved
𝟖𝟏= 𝟗 𝟐 = 𝟑 𝟒 𝟖𝟏𝟎= 𝟗 𝟐 ∙𝟏𝟎= 𝟑 𝟒 ∙𝟐∙𝟓 𝟖𝟏𝟎𝟎= 𝟗 𝟐 ∙ 𝟏𝟎 𝟐 = 𝟑 𝟒 ∙ 𝟐 𝟐 ∙ 𝟓 𝟐 𝟖𝟏𝟎𝟎𝟎= 𝟗 𝟐 ∙ 𝟏𝟎 𝟑 = 𝟑 𝟒 ∙ 𝟐 𝟑 ∙ 𝟓 𝟑 𝟐 𝟒𝟒 𝟓 𝟏𝟑𝟖 𝟑 𝟐𝟒𝟒 𝟕 𝟏𝟐 = ( 𝟐 𝟐𝟐 𝟓 𝟔𝟗 𝟑 𝟏𝟐𝟐 𝟕 𝟔 ) 𝟐

8. Round 3: Sums of Consecutive Integers and Perfect Squares
In 2002, the 12th-grade American Mathematics Competition (AMC 12) asked the following problem: The sum of 18 consecutive positive integers is a perfect square. What is the smallest possible value for this sum? [

9. Guess & Check Method 18 Consecutive Integers
𝟏+𝟐+𝟑+𝟒+𝟓+𝟔+𝟕+𝟖+𝟗+𝟏𝟎+𝟏𝟏+𝟏𝟐+𝟏𝟑+𝟏𝟒+𝟏𝟓+𝟏𝟔+𝟏𝟕+𝟏𝟖 =𝟏𝟕𝟏 𝟐+𝟑+𝟒+𝟓+𝟔+𝟕+𝟖+𝟗+𝟏𝟎+𝟏𝟏+𝟏𝟐+𝟏𝟑+𝟏𝟒+𝟏𝟓+𝟏𝟔+𝟏𝟕+𝟏𝟖+𝟏𝟗 =𝟏𝟖𝟗 𝟑+𝟒+𝟓+𝟔+𝟕+𝟖+𝟗+𝟏𝟎+𝟏𝟏+𝟏𝟐+𝟏𝟑+𝟏𝟒+𝟏𝟓+𝟏𝟔+𝟏𝟕+𝟏𝟖+𝟏𝟗+𝟐𝟎 =𝟐𝟎𝟕 𝟒+𝟓+𝟔+𝟕+𝟖+𝟗+𝟏𝟎+𝟏𝟏+𝟏𝟐+𝟏𝟑+𝟏𝟒+𝟏𝟓+𝟏𝟔+𝟏𝟕+𝟏𝟖+𝟏𝟗+𝟐𝟎+𝟐𝟏 =𝟐𝟐𝟓= 𝟏𝟓 𝟐

10. What If We Asked…
Can the sum of 16 consecutive positive integers be a perfect square?
Let us try our guess and check approach;
𝟏+𝟐+𝟑+𝟒+𝟓+𝟔+𝟕+𝟖+𝟗+𝟏𝟎+𝟏𝟏+𝟏𝟐+𝟏𝟑+𝟏𝟒+𝟏𝟓+𝟏𝟔=𝟏𝟑𝟔
𝟐+𝟑+𝟒+𝟓+𝟔+𝟕+𝟖+𝟗+𝟏𝟎+𝟏𝟏+𝟏𝟐+𝟏𝟑+𝟏𝟒+𝟏𝟓+𝟏𝟔+𝟏𝟕=𝟏𝟓𝟐
𝟑+𝟒+𝟓+𝟔+𝟕+𝟖+𝟗+𝟏𝟎+𝟏𝟏+𝟏𝟐+𝟏𝟑+𝟏𝟒+𝟏𝟓+𝟏𝟔+𝟏𝟕+𝟏𝟖=𝟏𝟔𝟖
𝟒+𝟓+𝟔+𝟕+𝟖+𝟗+𝟏𝟎+𝟏𝟏+𝟏𝟐+𝟏𝟑+𝟏𝟒+𝟏𝟓+𝟏𝟔+𝟏𝟕+𝟏𝟖+𝟏𝟗=𝟏𝟖𝟒
𝟓+𝟔+𝟕+𝟖+𝟗+𝟏𝟎+𝟏𝟏+𝟏𝟐+𝟏𝟑+𝟏𝟒+𝟏𝟓+𝟏𝟔+𝟏𝟕+𝟏𝟖+𝟏𝟗+𝟐𝟎=𝟐𝟎𝟎
𝟔+𝟕+𝟖+𝟗+𝟏𝟎+𝟏𝟏+𝟏𝟐+𝟏𝟑+𝟏𝟒+𝟏𝟓+𝟏𝟔+𝟏𝟕+𝟏𝟖+𝟏𝟗+𝟐𝟎+𝟐𝟏=𝟐𝟏𝟔 𝟕+𝟖+𝟗+𝟏𝟎+𝟏𝟏+𝟏𝟐+𝟏𝟑+𝟏𝟒+𝟏𝟓+𝟏𝟔+𝟏𝟕+𝟏𝟖+𝟏𝟗+𝟐𝟎+𝟐𝟏+𝟐𝟐=𝟐𝟑𝟐
𝟖+𝟗+𝟏𝟎+𝟏𝟏+𝟏𝟐+𝟏𝟑+𝟏𝟒+𝟏𝟓+𝟏𝟔+𝟏𝟕+𝟏𝟖+𝟏𝟗+𝟐𝟎+𝟐𝟏+𝟐𝟐+𝟐𝟑=𝟐𝟒𝟖 ⋮

11. The Number Sense Involved
Let 𝒏, 𝒏+𝟏, 𝒏+𝟐, 𝒏+𝟑, …, 𝒏+𝟏𝟒, 𝒏+𝟏𝟓 represent the 𝟏𝟔 consecutive positive integers. The sum of 𝟏𝟔 consecutive positive integers is
=𝟖 𝟐𝒏+𝟏𝟓
= 𝟐 𝟑 (𝟐𝒏+𝟏𝟓)
Now, what can we determine from this product? 𝐧
+ 𝐧+𝟏
+ 𝐧+𝟐
+ 𝐧+𝟑
+ 𝐧+𝟒
+ 𝐧+𝟓
+ 𝐧+𝟔
+ 𝐧+𝟕
+(𝐧+𝟏𝟓)
+ 𝐧+𝟏𝟒
+ 𝐧+𝟏𝟑
+ 𝐧+𝟏𝟐
+ 𝐧+𝟏𝟏
+ 𝐧+𝟏𝟎
+ 𝐧+𝟗
+ 𝐧+𝟖
𝟐𝐧+𝟏𝟓

12. The Number Sense Involved
Since 𝟖= 𝟐 𝟑 and we know that (𝟐𝒏+𝟏𝟓) is an odd number, the prime factorization of our sum will always have a factor of 𝟐 𝟑 . Since we also know that a perfect square has all prime factors with even exponents we know that it is not possible for 𝟏𝟔 consecutive integers to have a sum which is a perfect square.

13. In Conclusion… Thank you. Adam Sprague Stephanie Wisniewski
SUNY Fredonia