Your May 2013 ShillerMath Tidbit by Larry
Shiller 
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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.

Funny bone

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

Brain booster

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 numbers, 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
Larry Shiller

Publisher

May special: Save $25 on any kit and get free shipping anywhere. Use promo code
KITSAVE25. Expires May 31, 2013. Only at [1]shillermath.com.

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copy (PDF is OK).

[2]Answer the puzzler!

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© Larry Shiller. All Rights Reserved.

References

1. http://www.shillermath.com/
2. http://www.shillermath.com/contact-us-2/
3. http://www.facebook.com/ShillerMath

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