How Many Zeros?

Problems: (Solutions below.)

#1. 2003 Chapter Team # 7--How many zeros are at the end of (100!)(200!)(300!) when multiplied out?

#2. How many zeros are at the end of 2013!? 

#3. How many zeros are at the end of 10! *9!*8!*7!*6!*5!*4!*3!*2!*1!*0!?

#4. What is the unit digit of 10! + 9! + 8! + 7! + 6! + 5! + 4! + 3! + 2! + 1! + 0!?

#5. The number \(3^{4}\times 4^{5}\times 5^{6}\) written out in full. How many zeros are there are at the end of the number?

#6. How many zeros are at the end of (31!)/(16!*8!*4!*2!*1!)

#7. 2009 National Sprint #18-- What is the largest integer n such that \(3^{n}\) is a factor of 1×3×5×…×97×99?

Solutions:

#1.For 100!, there are -- 100/5 = 20 , 20/5 = 4 (Stop when the quotient is not divisible by 5 and then add up all the quotients.), or 20 + 4 = 24 zeros.
For 200!, there are 200/5 =40 , 40/5 = 8, and  8/5 = 1, or total 40 + 8 + 1 = 49 zeros.
For 300!, there are 300/5 = 60, 60/5 = 12, and 12/5 = 2, or total 60 + 12 + 2 = 74 zeros.
Add all the quotients and you get 147 zeros. 

#2. Use the same method as #1 and the answer is 501 zeros.

#3:Starting at 5!, you have one "0", the same goes with 6!, 7!, 8!, and 9!
10! will give you 2 extra "0"s. Thus total 7 zeros.

#4: Since starting with 5! you have "0" for the unit digit, you only need to check 4! + 3! + 2! + 1! + 0!.
24 + 6 + 2 + 1 + 1 = 34 so the unit digit is 4.

#5: Make sure to prime factorize all the given number sequences, in this case, it's \(3^{4}\times 2^{10}\times 5^{6}\) after you do that.
2 * 5 = 10 will give you a zero since there are fewer 5s than 2s so the answer is 6 zeros.

#6: You need the same number of 2 and 5 multiple together to get a "0".
31! gives you 30/5 = 6, 6/5 = 1 or 6 + 1 = 7 multiples of 5
31! gives you 10//2 = 15...15/2 = 7...7/2 = 3...3/2 = 1 or 15 + 7 + 3 + 1 = 26 multiples of 2. 16!*8!*4!*2!*1! gives you 4 multiples of 5 and 8 + 4 + 2 + 1 (16!) + 4 + 2 + 1 (8!) + 2 + 1 (4!) + 1(2!)
= 26 multiples of 2.
Thus the multiples of 2s all cancel out, the answer is "0" zeros. 

#7: There are 3*1, 3*3, 3*5...3*33 or \(\dfrac {33-1} {2}+1=17\) multiples of 3.
There are 9*1, 9* 3...9*11 or \(\dfrac {11-1} {2}+1 = 6\) extra multiples of 3.
There are 27*1, 27*3 or 2 extra multiples of 3.
There is 81*1 or 1 extra multiples of 3.
Add them up and the answer is 26.

2014 AMC 8 Results , Problems, Answer key and Detailed Solutions.

E-mail me if you want to join my group lessons for Mathcounts/AMCs prep.
Different groups based on your current level of proficiency.

My students are amazing. They NEVER stop learning and they don't just do math.

Please don't contact me if you just want state/national questions since I'm extremely busy these days with the preps.
You can purchase Mathcounts National tests and other prep materials at Mathcounts store.

You can also use my online blog contents. If you really understand those concepts, I'm sure you can be placed in your state's top 10 in the above average states [Every year I'd received some online students'/parents' e-mails about their (their kids') state results], not the most competitive states, which are crapshoot for even the USAJMO qualifiers, that I know.

Take care and best of luck.

Have fun problem solving and good luck.

2014 AMC-8 problems and solutions from AoPS wiki

Comments :

#4 : We've been practicing similar problems to #4 so it should be a breeze if you see right away that the prime number "2" is involved. You'll get a virtual bump if you forgot about that again.

 #10 : Almost every test has this type of problem, inclusive, exclusive, between, calendar, space, terms, stages... It's very easy to twist the questions in the hope of confusing students, so slow down on this type of question or for the trickier ones, skip it first. You can always go back to it if you have time left after you get the much easier-to-score points.  (such as #12, 13, 18 -- if you were not trolled and others)

 #11 : Similar questions appear at AMC-10, Mathcounts.  For harder cases, complementary counting is easier.
This one, block walk is easier.

Counting Path on a Grid 

This one includes forbidden path finder

 #12: 1/ 3!

How about if there are 4 celebrities ? What is the probability that all the baby photos match with the celebrities ? only 1 baby photo matches,  only2 baby photos match, 3 baby photos match, or none matches ?

 #13: number theory

For sum of odd and /or even, it's equally likely --
odd + odd = even ; even + even = even
odd + even = odd ; even + odd = odd

For product of odd and/or even, it's not equally likely --
odd * even = even ; even * even = even
odd * odd = odd

For product probability questions, complementary counting with total minus the probability of getting odd product (all odds multiply)is much faster.

SAT/ACT has similar type of problems.

#15 : Central angle and inscribed angles --> Don't forget radius is of the same length.
Learn the basics from Regents prep 

#17: rate, time and distance could be tricky

Make sure to have the same units (hour, minutes or seconds) and it's a better idea writing down
R*T = D so you align the given infor. better.

Also, sometimes you can use direct/inverse relationship to solve seemingly harder problems in seconds.

 Check out the notes from my blog and see for yourself.

#18 : Trolled question. Oh dear !! 

1 4 6 4 1     , but it doesn't specify gender number(s) so 
(4 + 4)/2^4 is the most likely. 

#19 : more interesting painted cube problems --> one cube is completely hidden inside.

Painted cube animation from Fairy Math Tutors

Painted cube review   Use Lego or other plops to help you visualize how it's done.


#20 : Use 3.14 for pi and if you understand what shape is asked, it's not too bad.

#21: You can cross out right away multiples of three or sum of multiples of 3 by first glance.
For example 1345AA, you can cross out right away 3 and 45 (because 4 + 5 = 9, a multiple of 3).
You don't need to keep adding those numbers up. It's easier this way.

#22 : To set up two-digit numbers, you do 10x + y.
To set up three-digit numbers, you do 100x + 10y + z 

For those switching digits questions, sometimes faster way is to use random two or three digit numbers, not in this case, though.

#23 : This one is more like a comprehension question. Since it relates to birthday of the month, there are not many two digit primes you need to weight, so 11, 13 and 17. (19 + 17 would exceed any maximum days of the month). From there, read carefully and you should get the answer.

#24 : a more original question --

To maximize the median, which in this case is the average of the 50th and 51st term, you minimized the first 49 terms, so make them all 1s.
Don't forget the 51st term has to be equal or larger than the 50th term.

#25: The figure shown is just a partial highway image. 40 feet is the diameter and the driver's speed is 5 miles per hour, so units are not the same --> trap

I've found most students, when it comes to circular problems, tend to make careless mistakes because there are just too many variables.
Areas, circles, semi-circles, arch, wedge areas, and those Harvey like "think outside the box" fun problems.

Thus, it's a good idea to slow down for those circular questions. Easier said than remembered.

Happy Holiday !!

ACT worksheets

ACT worksheets

Let me know if you have any other questions. Best and take good care, Mrs. Lin 

Lesson   Equations of Circles

Practice : Practice with Equations of Circles

worksheet : 

http://www.jmap.org/StaticFiles/PDFFILES/WorksheetsByPI/Geometry/Coordinate_Geometry/Drills/PR_G.G.71.pdf

Similar triangles : 

http://www.regentsprep.org/regents/math/geometry/gp11/Lstrategy.htm

http://www.regentsprep.org/regents/math/geometry/gp11/PracSim.htm

worksheet : 

https://www.jmap.org/JMAP/RegentsExamsandQuestions/PDF/WorksheetsByPI-Topic/Geometry/Informal_and_Formal_Proofs/G.G.46.SideSplitterTheorem.pdf

2017 Mathcounts State Prep: Volume of a Regular Tetrahedron and Its Relationship with the Cube it's Embedded

How to find volume of a tetrahedron (right pyramid) with side length one.

The above link gives you a visual interpretation of the relationship of a regular tetrahedron, its
relationship with the cube that it is embedded and the other kind of tetrahedron (right angle pyramid).

The side of the cube is \(\dfrac {S} {\sqrt {2}}\) so the volume of the regular embedded tetrahedron is
\(\dfrac {1} {3}\times \left( \dfrac {S} {\sqrt {2}}\right) ^{3}\)=\(\dfrac {1} {3}*\dfrac {s^{3}} {2\sqrt {2}}\)= \(\dfrac {\sqrt {2}S^{3}} {12}\).

You can also fine the height of the tetrahedron and then \(\dfrac {1} {3}\)*base*space height to get the volume.
Using Pythagorean theory, the hypotenuse S and one leg which is \(\dfrac {2} {3}\) of the height of the equilateral triangle base, you'll get the space height.

2016 AMC-8 Prep

Interesting articles on math for this week:

How to Fall in Love With Math by Manil Suri from the New York Times

The Simpsons' secret formula : it's written by math geeks by Simon Singh from the Guardian

For this week's self studies (part I work):
Review:

2009 #25 : Review using the Harvey method. :D

2008 #25   Don't use the method on the link. Use the much faster method we talked about at our lesson.

2008 #24  Make a chart. Slow down on similar question such as this one. 
This type of problem is very easy to make mistake on under or overcounting. 
Skip first and definitely slow down and double, triple check. 

2007 #25 Read the solution if you don't get the method we talked about at our lesson.
It takes time to develop insights so you need to be patient.
If you understand the method, this question will be easy, right ?
Stay with this question longer.

2007 #24 
Aayush's method is faster.
(To get the sum of three digits that is a multiple of 3, you either get rid of 1 or 4 [do you see it jumps by 3, why?] ) , so the answer is 1/2.

2006 #25
I've seen other problems (AMC-10s) using the oddest prime, which is "2", the only prime number that is even.
Thus, make sure you understand this question.

2006#24  Taking out the factor question.
Also, learn "1001 = 7 x 11 x 13"
"23 x 29 = 667"

2005 #25 Venn diagram is your friend.

2005 #24 Working backwards is the way to go.

For part II work :
This week, work on the last 5 problems from AMC-8 year 2010, 2011, 2012, 1999 and 1998.
Here is the link from AoPs.

Sam' original question:
David has a bag of 8 different-colored six-sided dice. Their colors are red, blue, yellow, green, purple, orange, black, and white. What is the probability that David takes out a red die, rolls a 6, then takes out a purple die, and rolls another 6 without replacement?

Solution:
The probability of rolling a 6 on a red die is 1/8 * 1/6 = 1/48. The probability of rolling a purple die and rolling a 6 after that, without replacement, is 1/7 * 1/6 = 1/42. Therefore, to get both events, 1/48 * 1/42 = 1/2016.

Evan's compiled question:
\(\sqrt {18+8\sqrt {2}}=a+b\sqrt {c}\)
a, b and c are positive integers. Find a + b + c.
Solution:
Square both sides and you have \(18+8\sqrt{2}\) = \( a^2 + 2ab\sqrt {2} + b^2c\)
You can see ab = 4 = 4 x 1 and c = 2
a = 4, b = 1 and c = 2 so the sum is 7.

Sounak's problem:
A rhombus with sides 4 is drawn. It has an angle of 60 degrees. What is length of the longer diagonal?

Solution:
Well first you have to draw the rhombus's height .The resulting triangle will be 30,60, 90 triangle.
We know the hypotenuse is 4 so now we know the rest of the sides are \(2\sqrt {3}\) and 2.
Now if we draw the diagonal we see that it makes another right angle triangle.
We know the legs of this triangle are the same as the previous lengths so then we know the diagonal is \(4\sqrt {3}\).