Idaho Society of Professional Engineers
Friday Update - 11/11/05
UPCOMING EVENTS:
• November 14, 2005 –
ISPE Northern Chapter
Monthly Meeting - Ironhorse Restaurant, 407 E. Sherman Ave, Coeur d'Alene,
ID, 667-7314 – 5:30 PM
• November 17, 2005 (Please note
date change) - ISPE
Southwest Chapter Noon Meeting - 12:00 Noon - Washington Group International
Training Room - Boise Airport Master Plan Update – John Anderson
• January 20 - 23, 2006 - NSPE
Winter Meeting - Washington DC
• February 4, 2006 – ISPE
Northern Chapter (Lewiston) MATHCOUNTS Competition
• February 4, 2006 – ISPE
Southeast Chapter MATHCOUNTS Competition
• February 11, 2006 – ISPE
Southwest Chapter MATHCOUNTS Competition
• March 11, 2006 – State
MATHCOUNTS Competition – Boise State University - Boise
• March 16 - 17, 2006 - ISPE
Annual Meeting - Boise, ID
• July 6 - 11, 2006 - NSPE Summer
Meeting - Boston, MA
CALL FOR ISPE AWARD NOMINATIONS
Each year ISPE selects outstanding Idahoans in recognition of their engineering
accomplishments and contributions to the engineering profession. Awards will be
presented during the 2006 Annual Meeting in Boise. Nominations must be submitted
no later than February 1, 2006. Award criteria and nomination forms can
be obtained from the ISPE web
site, or by contacting the ISPE office at 208-426-0636.
The awards for which we are looking for nominees include:
Idaho Engineering
Hall of Fame: Given by ISPE to recognize Idahoans that have made
engineering contributions beyond Idaho i.e. nationally or world wide.
Idaho Excellence
in Engineering Award: To recognize an Idahoan who is distinguishing
themselves in engineering.
Idaho
Excellence in Engineering Educator Award: This award recognizes an
Engineering Educator who has had a significant impact on the engineering
profession in Idaho.
Young Engineer
of the Year Award: To recognize an engineer that is making a
contribution to their profession. Must be no more than 35 years old.
Self nominations are welcomed and encouraged.
Participate in the 2006 NSPE Professional Development Conference and earn 7.5
PDHs.
Click here for more information.
MATHCOUNTS PROBLEM OF THE WEEK
Can you solve this MATHCOUNTS problem? The answer will appear in next week's
edition of the Friday Update!
Falling in the Fall
When thinking of things that "fall in the fall," perhaps the first things
that come to your mind are leaves? For many people around the country, the
leaves are falling from their trees and hours of raking are necessary. Joe
decided that he would try to make some extra money by raking the leaves in his
neighbors’ yards. Joe figured he would need approximately seven bags per yard in
which to collect the leaves. The bags are sold 10 to a box, and Joe is planning
on raking 14 yards. How many boxes of bags must he buy?
--------------------------------------------------------------------------------
Another thing that seems to "fall in the fall" is the temperature. Heather
noticed that the high temperature on October 21st was 66 degrees, the high
temperature on October 26th was 63 degrees and the high temperature on October
31st was 60 degrees. It appears that the high temperatures on every fifth day
are forming an arithmetic sequence. If this sequence were to continue, on what
date would the high temperature be 33 degrees?
--------------------------------------------------------------------------------
A third thing that tends to "fall in the fall" is people’s amount of physical
activity. It is believed that people generally gain weight in the colder months
in large part due to this lack of physical activity. Lou decided to be
pro-active about this and promised himself he would exercise at least four days
of every calendar week (or every Sunday through Saturday time period). If Lou
never wants to miss exercising on consecutive days within a calendar week, and
he exercises exactly four days out of the seven, how many different workout
schedules are possible for a calendar week? (One possible workout schedule would
be to exercise on Monday, Wednesday, Friday and Saturday.)
Answer to last week’s MATHCOUNTS problem:
We can set up the expression 7 – 6 + 9 – a + 2 and simplify it to 12 – a. We now
know that in order for 76,9a2 to be a multiple of 11, the expression 12 – a must
be a multiple of 11. The multiples of 11 are 0, 11, 22, 33, etc. Since a must be
a digit, the only possibility is 12 – a = 11, and therefore a = 1. We can check
with our calculator to see that 76,912 is in fact divisible by 11.
--------------------------------------------------------------------------------
Though there are 11 possible sums, it is much "harder" to roll a sum of 12 than
a sum of 6. The 11 possible sums are not all equally likely. We could list out
the different outcomes of a roll of two dice. (It might be easier to think of
this as having a green die and a red die, so that we can clearly see that a roll
of green-1 & red-3 is different than green-3 & red-1.) If we list out all of the
outcomes, we see that there are 36 different possible roll outcomes resulting in
sums from 2 to 12. Of these 36 outcomes, only two of them result in a sum of 11.
They are green-5 & red-6 and green-6 & red-5. Therefore the probability of
rolling a sum of 11 with two dice is 2/36 = 1/18.
--------------------------------------------------------------------------------
This is tricky! The greatest four-digit palindrome is 9999. We need the
four-digit palindrome to be a multiple of 6, which means that the units digit
(and therefore the thousands digit) must be even. To make the number as large as
possible, we need to start with 8 _ _ 8. To be a multiple of 6, the number must
be even (a multiple of 2), which we already have, and the number must be a
multiple of 3. A multiple of three has a digit-sum that is divisible by 3.
Obviously, we’d like to have 8998, since that’s the largest number we can now
make. The digit-sum is 8 + 9 + 9 + 8 = 34, and 34 is not divisible by 3, so 8998
is not our answer. We can test 8888, but that also does not work since the
digit-sum is 32. Notice that our digit-sum is decreasing by 2 when we decrease
each of the missing digits by 1. When we get to 8778, we see that the digit-sum
is 30 (which is divisible by 3), so 8778 is the largest four-digit palindrome
divisible by 6.
If you want to see last week's problem again, click
http://www.mathcounts.org/webarticles/anmviewer.asp?a=748&z=104
Idaho Society of Professional Engineers
PO Box 170239
Boise, ID 83717-0239
208-426-0636
Fax: 208-426-0639
E-Mail: ispe@rmci.net
Web Site: www.Idahospe.org