Managing Time for Success
Using productivity and organizational tools, performance and learning can be maximized while leaving time for other important activities including down time. This talk includes Robin's sharing her own system of time management including setting goals, listening to powerful speakers such as Tony Robbins and Jim Rohn, implementing tools like Franklin Covey planner and Darren Hardy's Weekly Rhythm Register. This talk can be for students as well as people in the corporate world, parents and anyone who would like to take their life to the next level!
College of Mount Saint Vincent, Weds. 1/30/13 4PM Elizabeth Seton Library |
Sunday, December 30, 2012
Managing Time for Success: Math Confidence Workshop
Friday, December 28, 2012
December 2012 Brain Teaser Solution
Q: Paula and Georgia play a game with dice that have colors instead of numbers. Some faces are pink and some are green. Paula wins when the two top faces are the same color. Georgia wins when the colors are different. Their chances are even.
The first die has 5 pink faces and 1 green face. On the second die, how many faces are pink and how many are green?
A: 3 pink faces and 3 green faces
At first it may seem as if the second die should have the reverse of the first (1 pink and 5 green) but that would make Georgia win too often as the chance of different colors would be too high.
This can be done with a guess and check.
5 pink and 1 green matched with 2 pink and 4 green would give
P P
P P
P G
P G
P G
G G
PP, PP, PP, PP, PP, GP, PP, PP, PP, PP, PP, GP, PP, PP, PG, PG, PG, GG, PP, PP, PG, PG, PG, GG,PP, PP, PG, PG, PG, GG, GP, GP, GG, GG, GG, GG
still too many that match giving Paula an advantage
This can be done with a guess and check.
5 pink and 1 green matched with 3 pink and 3 green would give
P P
P P
P P
P G
P G
G G
(keep the second column steady while changing the first die)
PP, PP, PP, PP, PP, GP, PP, PP, PP, PP, PP, GP, PP, PP, PP, PP, PP, GP,
PG,PG,PG, PG, PG, GG, PG,PG,PG, PG, PG, GG, PG,PG,PG, PG, PG, GG
This makes 15 PP and 3 GG making 18 that are the same (and 18 different)
If you want to keep the first column steady and change the second:
PP, PP, PP, PG, PG, PG, PP, PP, PP, PG, PG, PG, PP, PP, PP, PG, PG, PG, PP, PP, PP, PG, PG, PG,, PP, PP, PP, PG, PG, PG, GP, GP, GP, GG, GG, GG
PG,PG,PG, PG, PG, GG, PG,PG,PG, PG, PG, GG, PG,PG,PG, PG, PG, GG
This makes 15 PP and 3 GG making 18 that are the same (and 18 different)
Solution 2:
Bill Adams said that he solved it using probability fractions:
Pink 5/6 x Pink 3/6 = 15/36 Pink Pink
Pink 5/6 x Green 3/6
Green1/6 x Pink 3/6
Green 1/6 x Green3/6 = 3/36 Green Green
Add those together and 15/36 + 3/36 = 18/36 = 1/2!
The first die has 5 pink faces and 1 green face. On the second die, how many faces are pink and how many are green?
A: 3 pink faces and 3 green faces
At first it may seem as if the second die should have the reverse of the first (1 pink and 5 green) but that would make Georgia win too often as the chance of different colors would be too high.
This can be done with a guess and check.
5 pink and 1 green matched with 2 pink and 4 green would give
P P
P P
P G
P G
P G
G G
PP, PP, PP, PP, PP, GP, PP, PP, PP, PP, PP, GP, PP, PP, PG, PG, PG, GG, PP, PP, PG, PG, PG, GG,PP, PP, PG, PG, PG, GG, GP, GP, GG, GG, GG, GG
still too many that match giving Paula an advantage
This can be done with a guess and check.
5 pink and 1 green matched with 3 pink and 3 green would give
P P
P P
P P
P G
P G
G G
(keep the second column steady while changing the first die)
PP, PP, PP, PP, PP, GP, PP, PP, PP, PP, PP, GP, PP, PP, PP, PP, PP, GP,
PG,PG,PG, PG, PG, GG, PG,PG,PG, PG, PG, GG, PG,PG,PG, PG, PG, GG
This makes 15 PP and 3 GG making 18 that are the same (and 18 different)
If you want to keep the first column steady and change the second:
PP, PP, PP, PG, PG, PG, PP, PP, PP, PG, PG, PG, PP, PP, PP, PG, PG, PG, PP, PP, PP, PG, PG, PG,, PP, PP, PP, PG, PG, PG, GP, GP, GP, GG, GG, GG
PG,PG,PG, PG, PG, GG, PG,PG,PG, PG, PG, GG, PG,PG,PG, PG, PG, GG
This makes 15 PP and 3 GG making 18 that are the same (and 18 different)
Solution 2:
Bill Adams said that he solved it using probability fractions:
Pink 5/6 x Pink 3/6 = 15/36 Pink Pink
Pink 5/6 x Green 3/6
Green1/6 x Pink 3/6
Green 1/6 x Green3/6 = 3/36 Green Green
Add those together and 15/36 + 3/36 = 18/36 = 1/2!
Friday, December 07, 2012
The Math Confidence Challenge
Take The Math Confidence Challenge
Study for and take the SAT and/or ACT
Math Coaching by Math Confidence
Robin has worked with children, young adults and adults since 2000 building Math knowledge, confidence and study skills for academic, professional and personal success. Her refreshing and comforting approach engages the learner and enables them to accomplish their academic and personal goals.
Some students have needed a study partner while others have needed a turnaround.
Many learners have improved their knowledge, skills, and scores preparing with Robin for standardized exams (state tests, SHSAT, Regents, SAT, SAT II) and/or classroom tests while boosting their attitude and comfort with Math. Students and their families benefit from the Math Confidence advantage that helps students learn, retain and understand the material in a relaxed and secure manner.
College level programs of study, outside of math-specific fields still require challenging math courses. Achievement in these courses, such as calculus, teaches students perseverance and problem-solving skills essential in many professional careers (i.e. medicine). These skills provide professional candidates with a competitive edge in their chosen fields and give a sense of personal accomplishment.
Prof. Schwartz designs and delivers workshops both as a presenter and facilitator and attends many seminars and conferences for lifelong learning. Her memorable nicknames for Math concepts such as "the lonely girl at the dance" for the middle term in (x+3)^2 and "the Math police" who drive up when fractions are 'illegally' canceled make learners smile (and remember)!
FAQ about Math and Math Confidence
Q: Why take Math?
A: Even nontechnical positions like advertising are appreciating quantitative skills according to the New York Times Ad Companies Face a Widening Talent Gap. The actual Math one learns in school may or may not be used on the job, but the process of learning and doing Math teaches problem solving and critical thinking as well as persistence. These skills are useful in work and in life.
Q. How can people build their Math Confidence?
A: 1. Practice!! Solve Math problems (Math opportunities) including multiple choice as compare/contrast with "good wrong answers" can increase knowledge, skills, attitude and scores!
2. Treat Math exams like athletic events -- prepare by studying but also by eating breakfast, getting a good night's rest and packing up the night before.
3. Embrace Math mistakes! Learning from errors can be challenging emotionally but will improve critical thinking, build confidence and expand educational and career options.
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Q: What are the top factors for success in Math?
A: Here are some top tips for academic success and confidence:
Be prepared and proactive
A review sheet is a gift
Metacognition (raise self-awareness about your thinking)
Attention to Detail (reduce fuzzy errors)
Self-Care (sleep, food, stress management)
Effort expended (the feeling of a job well done)
Q. Are Math facts important?
Some of us from the 1900s remember standing up and reciting facts in class to learn times tables. Very few schools still do this but automaticity of Math facts is advantageous as it builds confidence and speed and makes Math more enjoyable. Although many types of calculators are available on cell phones, there are also Math challenges like
Android App Math Workout and other great tools available likeGreat Fact Practice at Dad's Worksheets , Developing Automaticity with Math Facts from Mrs Russ and FASTTMATH for Automaticity of Math Facts.
Q. Isn't Math just formulas?
1. What is the equation of a line? 2. The slope formula?
3. Circumference of a circle? 4. Area of a circle?
If Math were just formulas then open book exams would be easy!
Many professors of Math and other technical subjects along with the Professional Engineering licensing exam allow open book exams.
A: No, Math is much more than formulas. Applying the formulas is the challenge -- many formulas are tricky (like exponent rules) so examples are best to improve recall. Progress can be made by thinking about and understanding the formulas. For example, volume is three dimensional, therefore, the units will be units cubed such as ft cubed or cm^3. For area , the answer will contain square units and perimeter (or circumference) will be just units.
Q: What are "fuzzy errors"?
A: How much is 2^0? (2 to the zero power) Answer
People often get the incorrect answer due to a lack of focus rather than a lack of understanding. Some common topics that illustrate this idea are: Area vs Perimeter, Exponent Rules, the word "NOT"
and the ubiquitous minus sign mistake.
One excellent way to reduce fuzzy errors is to study multiple choice items that have tempting good wrong answers."Is that your final answer?" is a way to help raise awareness and critical thinking -- it would be great if "Who Wants to be a Millionaire?" had an Algebra version!
Monday, November 19, 2012
November 2012 Brain Teaser Solution
Q: What mathematical symbol can be placed between 5 and 9, to get a number greater than 5 and smaller than 9?
A: A decimal point
A: A decimal point
Wednesday, October 24, 2012
October 2012 Brain Teaser Solution
Q:What value of X makes the seven digit number 6X92114 divisible by 11? (caveat: no calculator or long division)
A: 6
To determine if a number divisible by 11, add up the odd digits and compare that to the sum of the even digits. For example, 242 is divisible by 11 because the 1st digit (2) and 3rd digit (2) add up to 4 and is equivalent to the (only) even digit.
Zip Code 10471: the odd digits (digits 1, 3, 5) are 1, 4, 1 and add up to 6 while the even digits (digits 2 and 4) add up to 7. Therefore 10471 is not divisible by 11.
Zip Code 10065: the wealthiest zip code in the United States IS divisible by 11 as the odd digits (digits 1, 3, 5) are 1, 0, 5 and add up to 6 while the even digits (digits 2 and 4) add up to 6. Therefore 10065 is divisible by 11.
Therefore:
In the number 6X92114. the odd digits are: 6 9 1 4 and add up to 20
The even digits are x 2 1 and add up to x + 3
If we set x + 3 = 20 then x would be 17 which cannot fit in one digit
So we have to think about what happens when the numbers add up to more than 11
For example 990 is divisible by 11 as is 979 and 968 -- notice that in 979 the sum of the odd digits is 11 more than the even digit (9 + 9 is 11 more than 7). This is also true in 968 as the sum of the odd digits is 17 which is 11 more than the even digit of 6.
For 6X92114, if we set x + 3 = 20 then x would be 17 which cannot fit in one digit but just like 968, we can use a 6 instead of a 17 (notice that 6 is 11 away from 17!)
A: 6
To determine if a number divisible by 11, add up the odd digits and compare that to the sum of the even digits. For example, 242 is divisible by 11 because the 1st digit (2) and 3rd digit (2) add up to 4 and is equivalent to the (only) even digit.
Zip Code 10471: the odd digits (digits 1, 3, 5) are 1, 4, 1 and add up to 6 while the even digits (digits 2 and 4) add up to 7. Therefore 10471 is not divisible by 11.
Zip Code 10065: the wealthiest zip code in the United States IS divisible by 11 as the odd digits (digits 1, 3, 5) are 1, 0, 5 and add up to 6 while the even digits (digits 2 and 4) add up to 6. Therefore 10065 is divisible by 11.
Therefore:
In the number 6X92114. the odd digits are: 6 9 1 4 and add up to 20
The even digits are x 2 1 and add up to x + 3
If we set x + 3 = 20 then x would be 17 which cannot fit in one digit
So we have to think about what happens when the numbers add up to more than 11
For example 990 is divisible by 11 as is 979 and 968 -- notice that in 979 the sum of the odd digits is 11 more than the even digit (9 + 9 is 11 more than 7). This is also true in 968 as the sum of the odd digits is 17 which is 11 more than the even digit of 6.
For 6X92114, if we set x + 3 = 20 then x would be 17 which cannot fit in one digit but just like 968, we can use a 6 instead of a 17 (notice that 6 is 11 away from 17!)
Saturday, September 29, 2012
Promoting Academic Integrity: Prof Somerset Syracuse U
In Statics and Dynamics, the one Mechanical Engineering class we had to take was MEE 225 first semester sophomore year in the Fall of 1982. Our professor, James Somerset, programmed his computer to generate unique exams for each student. There were 25 problems and you had to match them with answers in a 36 answer grid -- there were 11 extra answers. We all got our own data. For example, a _A__kg car is traveling at __B_m/s. We also got different As and Bs.
This promoted academic integrity as there was absolutely no way to cheat as we were all doing different problems!!
This promoted academic integrity as there was absolutely no way to cheat as we were all doing different problems!!
Saturday, September 22, 2012
September 2012 Brain Teaser Solution
Q: When a four digit number wxyz is multiplied by 4, the digits are reversed and it becomes zyxw. All the digits are different. What is the number?
A: 2178
If the number is still 4 digits when multiplied by 4, then the number has to be less than 2500 (as 2500 x 4 = 10000 and you would then have 5 digits).
That makes w = 1 or 2
But when multiplying by 4, the last digit will never be 1 so when we reverse to zyxw, the w must be 2 so that the last digit will then be 2. The multiples of 4 that have a last digit of 2 are 3 x 4 and 8 x 4.
Since we are multiplying a number in the 2000s by 4, the number is becomes will start with an 8.
w = 2 and z = 8
The second digit x must be 4 or less as it says above, the number must be less than 2500.
Because 2250 x 4 = 9000, x must be 0 1 or 2
Through some trial and error, you can find that the number is 2178 and its reverse is 8712.
If the number is still 4 digits when multiplied by 4, then the number has to be less than 2500 (as 2500 x 4 = 10000 and you would then have 5 digits).
That makes w = 1 or 2
But when multiplying by 4, the last digit will never be 1 so when we reverse to zyxw, the w must be 2 so that the last digit will then be 2. The multiples of 4 that have a last digit of 2 are 3 x 4 and 8 x 4.
Since we are multiplying a number in the 2000s by 4, the number is becomes will start with an 8.
w = 2 and z = 8
The second digit x must be 4 or less as it says above, the number must be less than 2500.
Because 2250 x 4 = 9000, x must be 0 1 or 2
Through some trial and error, you can find that the number is 2178 and its reverse is 8712.
Friday, September 07, 2012
Regents Scoring and new 3 - 8 Math sample items
http://cityroom.blogs.nytimes.com/2012/09/06/at-a-new-school-a-math-lesson-from-the-mayor/#postComment
Jeff S is correct: if students get 30 points out of 87 (34%), it gets scaled up to a 65.
Here is the conversion chart for the most recent Integrated Algebra Regents -- it is still thankfully up on line as the actual exam ;)
http://www.nysedregents.org/IntegratedAlgebra/612/ialg62012-cc.pdf
Unfortunately, the latest 3 - 8 exams have not been posted due to the changeover to the Common Core. The latest exams for grades 3 - 8 are 2010: Here is an example: http://www.nysedregents.org/Grade5/Mathematics/home.html
However, there are new sample items for Grades 3 - 8 and they look more challenging than the old exams -- check them out:
http://www.p12.nysed.gov/assessment/common-core-sample-questions/ (scroll to the bottom)
Sunday, August 19, 2012
August 2012 Brain Teaser Solution
Q: A group of four people has to cross a bridge at night. A flashlight is needed to light the path and they only have one flashlight. No more than two people can cross the bridge simultaneously, and it takes different time for the people in the group to cross the bridge: Abby crosses the bridge in 1 minute, Bob crosses the bridge in 2 minutes, Cathy crosses the bridge in 5 minutes, David crosses the bridge in 10 minutes. How can the group cross the bridge in 17 minutes?
With many thanks to:
http://www.math.utah.edu/~cherk/puzzles.html
A: Abby and Bob go across: 2 minutes
Bob returns with flash: 2 minutes
Cathy and David go across: 10 minutes
----------
14 minutes
Now Abby, Cathy and David have crossed, so one of these needs to return
with the flashlight and afterwards go across with Bob in 3 minutes:
With many thanks to:
http://www.math.utah.edu/~cherk/puzzles.html
Tuesday, July 24, 2012
July 2012 Brain Teaser Solution
Q:What are the largest and smallest 5-digit numbers that satisfy the following conditions?
1. Each digit of the number is a prime digit.
2. Each successive pair of digits forms a 2-digit number that is NOT a prime number.
3. Each of the prime digits must appear at least once in the 5-digit number.
A: Largest: 35772
Smallest: 32257
Definition of Prime Numbers from Coolmath.com
Prime digits are 2, 3, 5 and 7. 1 is not considered a prime number or digit.
Taken as pairs, the only combinations that satisfy condition 2 are:
22, 25, 27, 32, 33, 35, 52, 55, 57, 72, 75 and 77.
Of that list, 33 is the only number that contains a 3 as the second digit and so for condition 3 to be satisfied, 3 can only appear at the start of the 5-digit numbers.
Many thanks to Braingle
1. Each digit of the number is a prime digit.
2. Each successive pair of digits forms a 2-digit number that is NOT a prime number.
3. Each of the prime digits must appear at least once in the 5-digit number.
A: Largest: 35772
Smallest: 32257
Definition of Prime Numbers from Coolmath.com
Prime digits are 2, 3, 5 and 7. 1 is not considered a prime number or digit.
Taken as pairs, the only combinations that satisfy condition 2 are:
22, 25, 27, 32, 33, 35, 52, 55, 57, 72, 75 and 77.
Of that list, 33 is the only number that contains a 3 as the second digit and so for condition 3 to be satisfied, 3 can only appear at the start of the 5-digit numbers.
Many thanks to Braingle
Wednesday, June 27, 2012
Math Confidence ACT Scores and Studying Methodology
The exam was just like the ACT books and the content and sample test online http://www.act.org/. I started to study about 4 months before June 9. I would do one section a week then two, then three. In the last month before the exam, I completed 1 section each day. During the last week, I completed 2 sections each day. I would rotate the sections to get a variety of practice and always timed myself to make sure I was being efficient. The Math topics were all familiar to me and the Reading Comp was just fine. The two sections that I had to study the most were English (Grammar) and Science.
I used the Real ACT books -- the one with 3 tests and then the newer one with 5 tests (the same 3 tests from the first book and 2 new tests). The other book I used was Barron's ACT 36. The content online included questions for each section and a full-length practice test which I completed the day before the exam.
I used the Real ACT books -- the one with 3 tests and then the newer one with 5 tests (the same 3 tests from the first book and 2 new tests). The other book I used was Barron's ACT 36. The content online included questions for each section and a full-length practice test which I completed the day before the exam.
All those months of studying paid off. The day of the exam, Science was the most challenging section as I was not confident about 2 of the 7 sections. I was not surprised to see that Science was my lowest score.
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Thursday, June 07, 2012
June 2012 Brain Teaser Solution
Q: My daughter has many sisters. She has as many sisters as she has brothers. Each of her brothers has twice as many sisters as brothers. How many sons and daughters do I have?
A: Four daughters and three sons.
A: Four daughters and three sons.
Saturday, May 26, 2012
May 2012 Brain Teaser Solution
Q:A clock is observed. The hour hand is exactly at the minute mark, and the minute hand is six minutes ahead of it. Later, the clock is observed again. This time, the hour hand is exactly on a different minute mark, and the minute hand is seven minutes ahead of it. How much time elapsed between the first and second observations?
A: The hour hand moves as the minute hand moves. For example, at 2:30, the hour hand is halfway between 2 and 3. The hour hand is exactly at the minute mark five times an hour on the 12, 24, 36, 48 and on the hour.
So 1:12 would make the hour hand pointing at 6 minutes and the minute hand pointing at 12 minutes.
24 minutes after on the minute hand would mean that the hour hand would have to be pointing at the 18 minute mark which would be more than halfway between 3 and 4.
36 minutes after the hour would mean that the hour hand would point exactly at the 6 and the hour hand only points exactly at the 6 when it is exactly at 6 o'clock.
48 minutes after the hour would mean that the hour hand would point at the 42 which would be a bit after 8 and not closer to nine.
So for the first observation the clock must be at 1:12.
The second observation cannot be 1:12 (since that it a 6minute difference) so we will now check the 24 minute after -- this would put the hour hand at the 17 which would work perfect with 3 as at 3:24, the hour hand has moved 2/5 between the 2 and 3.
Therefore the time elapsed between the first observation, 1:12, and the second observation, 3:24 is 2 hours and 12 minutes.
A: The hour hand moves as the minute hand moves. For example, at 2:30, the hour hand is halfway between 2 and 3. The hour hand is exactly at the minute mark five times an hour on the 12, 24, 36, 48 and on the hour.
So 1:12 would make the hour hand pointing at 6 minutes and the minute hand pointing at 12 minutes.
24 minutes after on the minute hand would mean that the hour hand would have to be pointing at the 18 minute mark which would be more than halfway between 3 and 4.
36 minutes after the hour would mean that the hour hand would point exactly at the 6 and the hour hand only points exactly at the 6 when it is exactly at 6 o'clock.
48 minutes after the hour would mean that the hour hand would point at the 42 which would be a bit after 8 and not closer to nine.
So for the first observation the clock must be at 1:12.
The second observation cannot be 1:12 (since that it a 6minute difference) so we will now check the 24 minute after -- this would put the hour hand at the 17 which would work perfect with 3 as at 3:24, the hour hand has moved 2/5 between the 2 and 3.
Therefore the time elapsed between the first observation, 1:12, and the second observation, 3:24 is 2 hours and 12 minutes.
Saturday, April 28, 2012
April 2012 Brain Teaser Solution
Q:You are in a pitch dark room selecting socks from a drawer that has only six socks, a mixture of black and white. If you choose two socks, the chances that you draw out a white pair is 2/3. What are the chances that you draw out a black pair?
A: 0 (it is impossible)
The good wrong answer is 1/3 because that is the remainder of the probability. This assumes that the pair of black socks has the same probability as a pair of white socks but there is also a chance that you pull out one black and one white. The combinations with 4 white (W1, W2, W3 and W4) and 2 black (B1 and B2) are:
A: 0 (it is impossible)
The good wrong answer is 1/3 because that is the remainder of the probability. This assumes that the pair of black socks has the same probability as a pair of white socks but there is also a chance that you pull out one black and one white. The combinations with 4 white (W1, W2, W3 and W4) and 2 black (B1 and B2) are:
W1W2, W1W3, W1W4, W2W3, W2W4, W3W4, W1B1, W2B1, W3B1, W4B1, W1B2, W2B2, W3B2, W4B2, and (last but not least) B1B2. Far less than two-thirds of the aforementioned are a white pair.
So let's try 5 white and 1 black: W1W2, W1W3, W1W4, W1W5, W2W3, W2W4, W2W5, W3W4, W3W5, W4W5, W1B1, W2B1, W3B1, W4B1, W5B1makes a total of 15 combination of which 10 are white pairs. This simplifies to 2/3 so there must be 5 white and 1 black. But with only one black sock, you can never get a black PAIR!
So let's try 5 white and 1 black: W1W2, W1W3, W1W4, W1W5, W2W3, W2W4, W2W5, W3W4, W3W5, W4W5, W1B1, W2B1, W3B1, W4B1, W5B1makes a total of 15 combination of which 10 are white pairs. This simplifies to 2/3 so there must be 5 white and 1 black. But with only one black sock, you can never get a black PAIR!
Friday, March 23, 2012
March 2012 Brain Teaser Solution
Which two whole numbers, neither containing any zeros, when multiplied together equal exactly 1,000,000,000?
A: 512 and 1953125
The factorization of this number is 10^9
which can be broken down into its prime factors of (2 x 5)^9
so 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5
If we mix any 2 with any 5 we will get 10 and the number will end in a zero due to multiplication by 10
Therefore we must keep the 2s and the 5s separate from one another
2^9 is 512 and 5^9 is 1953125
These two factors (each with no zeros) will multiply out to 1 billion!
A: 512 and 1953125
The factorization of this number is 10^9
which can be broken down into its prime factors of (2 x 5)^9
so 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 5
If we mix any 2 with any 5 we will get 10 and the number will end in a zero due to multiplication by 10
Therefore we must keep the 2s and the 5s separate from one another
2^9 is 512 and 5^9 is 1953125
These two factors (each with no zeros) will multiply out to 1 billion!
Monday, February 20, 2012
February 2012 Brain Teaser Solution
Q: Find three-digit numbers that the number itself, its double and its triple contain each digit from 1 to 9 exactly once. For example, 192 works because 192, 384, 576 contain 1 to 9 each once. 273 also works because 273, 546, 819 contain 1 to 9 each once.
A: 219 and 327.
We need to keep the numbers low because if we are going to double and triple and maintain only 3 digits, then that means that the number must be less than 334 (334 tripled makes 1002 -- 4 digits).
We can rule out any numbers where the digits repeats as we have to use all 9 digits (1 through 9).
I used a guess and check to solve this problem. I kept the first digit 1, 2 or 3 and kept out any repeated digits. This led me to 219 and 327 but those are also the same digits as those in the given (192 and 273)!!!
It was then that I realized that if you swap 19 and 2, and swap 27 and 3, you'll get them.
What are the other two numbers that also have this property?
A: 219 and 327.
We need to keep the numbers low because if we are going to double and triple and maintain only 3 digits, then that means that the number must be less than 334 (334 tripled makes 1002 -- 4 digits).
We can rule out any numbers where the digits repeats as we have to use all 9 digits (1 through 9).
I used a guess and check to solve this problem. I kept the first digit 1, 2 or 3 and kept out any repeated digits. This led me to 219 and 327 but those are also the same digits as those in the given (192 and 273)!!!
It was then that I realized that if you swap 19 and 2, and swap 27 and 3, you'll get them.
Thursday, January 19, 2012
January 2012 Brain Teaser Solution
Andy and Sandy had to take a make-up class in math over the summer, a two-month, self-paced course with a test at the end of each of 12 chapters. The course requires a 70% grade to pass. In the first month, they both had difficulties with the concepts. Andy averaged 60% on his exams; Sandy averaged 50%. In the second month, Andy averaged 90% on his exams; Sandy averaged 80%. While Sandy got a passing grade of 75% in the class; Andy failed with 65%. How did Sandy pass while Andy flunked?
Their grades were weighted averages.
This can be solved using guess-and-check or through using algebra.
Let x = # of tests in the first month, therefore 12 - x would be the # of tests in the second month.
so Andy's tests can be modeled with algebra using: .6x + .9(12 - x) = .65 (12)
.6x + 10.8 -.9x = 7.8
-.3x + 10.8 = 7.8
-.3x = -3
x = 10 and 12 - x = 2
so Andy took 10 tests in the first month and 2 in the second month.
10(60) + 2(90) = 780 divided by 12 = 65
Sandy's tests can be modeled using .5x + .8(12 - x) = .75(12)
.5x + 9.6 - .8x = 9
9.6 - .3x = 9
-.3x = -.6
x = 2
so Sandy took 2 tests in the first month and 10 in the second month
2(50) + 10(80) = 900 divided by 12 = 75
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