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�Your May 2013�
ShillerLearning Tidbit
�Bringing the Bling to Math
�Not all manipulatives are created equal
University of Notre Dame Associate Professor of Psychology Nicole McNeil studies how children learn (or don't learn) math. She just came up with some fascinating results that have repercussions for parents who want their preschoolers to learn - and love - math.
Manipulatives work - but only when they are unfamiliar to the child. When children know a manipulative and its purpose, they focus on that - and not on its alternative purpose in helping to learn math.
When a child doesn't know a manipulative, what the authors call "perceptual richness" takes place - a good thing because it brings more parts of the brain to bear on the mathematical concept and not on any pre-conceived notion of what the manipulative normally does.
In other words, when it comes to preschooler math manipulatives, don't use that thing if it don't have that bling.
They said it
It's a lot of bling to play with. You got to have the bling. Serena Williams
When girls are asking themselves "Who am I?" for the first time and they hear all this bad PR about math, they think, "Well, whoever I am, I'm not somebody who likes math."�Danica McKellar
Math: The only place where a person can buy 80 watermelons without anyone thinking they're a weirdo.�Anonymous
May 2013 Puzzler�[Grades 4-8]
Say you divide a circle into many equal parts by drawing line segments from the center to points on the circle. If no two line segments form a diameter, what can you say about the number of parts?
Provide the correct answer by May 25, 2013�to be this month's puzzler winner.
Answer to previous Puzzler�[Grades 10-12]
Why is it true that a number whose digits sum to a multiple of 3 will also be divisible by 3 (and vice versa)?
Solution:�
Here's a proof for 3-digit whole number, which can be generalized to whole numbers of any size:
The number abc may be represented as 100*a + 10*b +1*c. For example 485 is 100*4 + 10*8 + 1*5. But 100*a + 10*b + 1*c is the same as (99+1)*a + (9+1)*b + (0+1)*c, which is 99a + 9b + (a+b+c).
99a + 9b is always multiple of 3. So if a+b+c is a multiple of 3, abc is a multiple of 3 and otherwise not; and if abc is a multiple of 3, so is a+b+c and otherwise not.
I hope you enjoyed this short math break.
Sincerely,
Larry Shiller
Publisher
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