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.

Sep. 4th, 2014 AMC-8 and Mathcounts State/National prep

To prepare for AMC-8 and Mathcounts simultaneously, it's a good idea to delete AMC-8 multiple choice options to make the test more like Mathcounts problems.

There are some tricky questions for AMC-8 test, so if you are at the Mathcounts state level in above average states, you might get better scores on AMC-10 tests than on AMC-8. (Sigh...)

That's what happened to quite a few of my students because their level is way up so they might not as focused as they worked on the more challenging, more interesting questions. Oh dear !!

Review the following very frequently tested concepts. You really don't need a lot of tools (formulas) but deeper understanding, tenacity and the love of thinking outside the box.

Review similar triangles

Review counting and probability

I'll put solutions to some of the "Mass Points" questions soon.

We all love "Mass Points".

Have fun problem solving. Cheers, Mrs. Lin

Find the area of the petal, or the football shape.

Find the area of the football shape, or the petal shape.
The below Mathcounts mini presents two methods.

Circle and area revisited from Mathcounts mini

The first question is exactly the same as this one.
Besides the two methods on the videos, you can also use the following methods.

Solution III:
You can also look at this as a Venn Diagram question.
One quarter circle is A and the other is B, and both are congruent. (center at opposite corner vertexes)

The overlapping part is C.

A + B - C = 6^2 so C = A + B - 36 or 18pi - 36





                                                                                               

Solution IV:

If you use the area of the rectangle,
which is 6 x 12 minus, the area of the half circle with a radius 6, you get the two white spots that are un-shaded.

Use the area of the square minus that will again give you the answer.
\(6^{2}-\left( 6*12-\dfrac {6^{2}\pi } {2}\right)\)
= 18pi - 36




Similar triangles and triangles that share the same vertexes/or/and trapezoid

Another link from my blog

Similar triangles, dimensional change questions are all over the place so make sure you really
understand them.

Take care and happy problem solving !!