Last Digit Problems

Sum of Digits of the Product

Question:  Determine the sum of the digits in the product of

18\times\underbrace{11...11}_{101 digits}\times\underbrace{33...33}_{101 digits}

Solution:

Last Two Digits of the Product 

Question: Find the last two digits of 3\times 7\times 11 \times 15 ...\times 95 \times 99

Solution:

Since 15 (3 x 5) and 35 (5 x 7) are in the product, therefore the product is divisible by 5 x 5 = 25.

Hence, the last 2 digits could be 25, 50, 75, 00.

Using modular arithmetic,

3,11,19,...,99\equiv 3\mod 4

\underbrace{3\times 11\times 19\times...\times99}_{25\quad numbers}\equiv (3)^{25}\mod 4\equiv (3)^{24}(3)\mod 4

\equiv (9)^{12}(3)\mod 4\equiv (1)^{12}(3)\mod 4\equiv 3\mod 4

Using the divisibility test of 4, we only need to consider the last 2 digits of the product.

Since only 75\equiv 3\mod 4 , then the last 2 digits of the product is 75.

Interesting Probability Questions

A trading firm asked me some interesting questions during rounds of phone interview. The rule was to think out loud, and no calculators and paper allowed! Yep, a lot of pressure in the interview. No background questions, just probability and brain teaser questions.

Suppose you were in a betting game. It is given that a treasure chest costs from $0 to $1000. You get the treasure chest if you bet an amount that is equal or higher than its actual value. You could sell the treasure chest to your friend for 50% more its actual value should you get it. How much would you bet?

Solution

The idea here is to maximize the profit. I think most people would think that betting $1000 is good since for sure the treasure chest would be gotten. However, suppose the actual value of the treasure chest is way smaller, then you would lose a lot of money even if you sell it to your friend!

So let’s use the concept of expectation value and probability to decide logically the amount that would optimize the profit.

Suppose I bet $p such that 0\le p\le 1000 . Then the probability that I’ll get the treasure chest is p/1000 .

Case 1: Actual value is less than p

Then the expectation value of the actual value is then the half of p, and the expected resell price for my friend is 1.5p/2. The expectation value of the profit is then 1.5p/2-p , which occurs p/1000 of the time.

Case 2: Actual value is more than p

Obviously, there would be no profit in this case and this happens 1 -p/1000 of the time.

Hence the overall expectation value of this game is

(p/1000)(1.5p/2 -p)+(1 - p/1000)(0) of the time , which is a negative value.

Therefore, this game would likely make someone lose money! So the best way to avoid losing money or maximize the gain is to not bet in this game. So my answer is $0 🙂

Next question is quite a classic brain teaser in probability. I think I’ve seen it when I was a teenager.

Roll a 12-sided die. You and your friend pick a unique whole number from 1 to 12. The player who picked a number closest to the number shown on the die will win. Suppose getting a 12 is 40% and the rest is a uniform distribution of 60%. What number would you pick? Would you choose to be the 1st or 2nd player? Why?

Solution

One of the misleading ways to solve this problem is to choose the whole number closest to the expectation value (8.4). The thing is, if you choose 8 and your friend choose 9, then the your chance of winning is 8(\frac{60\%}{11})=43\frac{7}{11}\% and your friend’s chance of winning is 100\%-43\frac{7}{11}\%=56\frac{4}{11}\% (higher than yours!). This is because 12 is more likely to be shown on the biased die. Thus choosing a number close to 12 is reasonable (i.e. choosing 9, 10 and 11 are worth to consider).

Another catch is to assume that the second player is also smart. If you choose to be the 1st player, your friend (2nd player) would also think of a way to maximize his chance of winning.

So the smart way is to choose to be the 1st player and then pick a right number such that your probability of winning is higher no matter what number the 2nd player choose (if possible).

Fortunately this is possible! By case-by-case trial, this number is 10.

It’s because of the following:

Case 1: 2nd player chose 11
1st player’s chance: 10(\frac{60\%}{11}) = 54 \frac{6}{11}\%%
2nd player’s chance: 100\%-54\frac{6}{11}\%=45\frac{5}{11}\%

Case 2: 2nd player chose 9
1st player’s chance: 100\%-49\frac{1}{11}\%=50\frac{10}{11}\%
2nd player’s chance = 9(\frac{60\%}{11}) = 49 \frac{1}{11}\%%

In either case, 1st player’s chance of winning is higher than the 2nd player.

So the being the 1st player and choosing 10 will maximize the chance of winning.

Dior Cruise 2011

I’m always a fan of barbie-look. Dior (John Galliano) did a runway debut in Shanghai back in 2010 with the following lovely collection. They are too feminine and cute, that’s why I love them! I guess the cuts were made to cater Asian style and imitate 60s inspired looks.

Here’s the video:

Here are my favorites:

The model is Frida Gustavsson and she was so stunning in the show.

Karlie Kloss opened the show. She must be! Her catwalk is gorgeous.

AIMO Geometry 2017

This year’s Australian Intermediate Mathematics Olympiad (AIMO) questions are more about enumeration of cases (quite boring paper).

The only problems I find interesting are the following:

1. Triangle ABC has AB = 90, BC = 50 and CA = 70. A circle is drawn with center P on AB such that CA and CB are tangents to the circle. Find 2AP.

2. In quadrilateral PQRS, PS = 5, SR = 6, RQ = 4 and \angle P = \angle Q = 60^0. Given that 2PQ=a+\sqrt{b}, where a and b are unique positive integers, find the value of a+b.

Here are my solutions:
(Feel free to comment if you find another way to solve the problems)

Triangle ABC has AB = 90, BC = 50 and CA = 70. A circle is drawn with center P on AB such that CA and CB are tangents to the circle. Find 2AP.

Short solution:

Let R and Q be the points of tangency.

In the diagram, triangle CPA and CPB have the same height, which is the radius of the circle, if we take AC and BC as the bases.
So the ratio of their areas is the same as the ratio of their bases, i.e. \frac{[CPA]}{[CPB]}=\frac{70}{50}=\frac{7}{5}

Also, AP and and PB can be taken as the base of \Delta CPA and \Delta CPB , respectively.
Now, dropping a perpendicular line segment from C to AB would form the height of both of these triangles. Hence, the triangles would have the same height if we take AP and PB as the bases.

It follows that \frac{[CPA]}{[CPB]}=\frac{AP}{PB}=\frac{AP}{90-AP} , and therefore \frac{AP}{90-AP}=\frac{7}{5} .

Solving this gives 2AP=105 .

Comment: This is a type of problem that could take some time to solve since determining the right direction is not obvious, especially if a student is not used to the ratio method above. Note that I didn’t use other theorems besides from the radius-tangent theorem. A lot of students were able to answer the other AIMO questions and just left this question unanswered. Some students wrote solutions that contain cumbersome algebra because they used Pythagoras theorem and other theorems involving circles. I find this question interesting since it could take some time to figure out how to start solving the question, yet the solution is simple and short 🙂

 

 

In quadrilateral PQRS, PS = 5, SR = 6, RQ = 4 and \angle P = \angle Q = 60^0. Given that 2PQ=a+\sqrt{b}, where a and b are unique positive integers, find the value of a+b.

Solution:

Since there are 60 degrees angles, then expect that the solution would involve 30-60-90 triangles.

Then we draw assisting lines SB and RC such that B and C are on PQ and SB\perp PQ and RC\perp PQ. It would be easier as well if we have isosceles trapezoid, so AR is constructed below such that A is on PS and PA = QR. Let the intersection of SB and AR be point D.

It follows that PQRA is isosceles trapezoid with PQ\parallel AR. So, ADS is a 30-60-90 triangle as well and \Delta ADS\sim\Delta PBS .

In PBS, PB = 5/2, whereas in RCQ, CQ = 2.

In ADS, DS=\frac{\sqrt{3}}{2} , then DR=\frac{\sqrt{141}}{2} by Pythagorean theorem.

As BCRD is a rectangle, BC=DR=\frac{\sqrt{141}}{2} .

Therefore, PQ=PB+BC+CQ=\frac{9+\sqrt{141}}{2} , and hence a+b=9+141=150 .

Comment: Assisting lines such as SB, RC helped us determine the size of some parts of PQ. The catch is creating an isosceles trapezoid would make the problem easy to handle 😛

Cylic Trapezoid in a Triangle

In \Delta ABC, P, Q and R are the midpoints of AC, AB and BC respectively. Lines PR and QS intersect at T. If BS\perp AC , prove that TS=TP .

Solution

Connect points Q and R. Then by midpoint theorem, AC\parallel QR and 2QR=AC.

Let M be the intersection of SB and QR.

Since SB is the height of \Delta ABC , then MB is the height of \Delta QBR (i.e. QR\perp MB ). It follows that M is the midpoint of SB and \Delta SRM\cong \Delta BRM by SAS postulate. Hence, SR=BR . As R is the midpoint of CB, then SR=CR, and so \angle RCS\cong\angle RSC . Also, \angle RSP=180-\angle RSC=180-\angle RCS .

Due to the midpoint theorem, PQRC is a parallelogram, which implies that \angle RCS\cong\angle PQR .

In trapezoid PQRS, opposite angles \angle RSP and \angle PQR add up to 180. Therefore PQRS is a cyclic quadrilateral.

Now \angle PSQ\cong\angle PRQ (angle subtended by arc PQ). SP\parallel RQ , then \angle PRQ\cong\angle SPR. Then \angle PSQ\cong\angle SPR . Therefore, TS=TP .

Addendum:
Clearly, a cyclic trapezoid would be formed also by connecting the midpoint of each side and the perpendicular foot on another side.

Theorem: The quadrilateral formed by the midpoint of each side and a perpendicular foot of a triangle is a cyclic trapezoid.

Triangle Altitudes Inequality

In \Delta XYZ , the lengths of two altitudes are a and b , where a>b . Show that the third side altitude is of range \left(\frac{ab}{a+b},\frac{ab}{a-b}\right) .

Solution:
Let the sides of \Delta XYZ be s_1, s_2, s_3 , and the corresponding altitudes be a,b,c, respectively.

Assuming a>b implies that s_2 > s_1.

We then have as_1=bs_2=cs_3 \Rightarrow c=\frac{as_1}{s_3}=\frac{bs_2}{s_3} .

By triangle inequality, s_2-s_1<s_3<s_2+s_1 , and thus \frac{1}{s_2+s_1}<\frac{1}{s_3}<\frac{1}{s_2-s_1} .

 

Writing \frac{1}{s_2+s_1}<\frac{1}{s_3} in terms of the altitudes, we obtain

\frac{bs_2}{s_2+s_1}<\frac{bs_2}{s_3} \Rightarrow \frac{b}{1+\frac{s_1}{s_2}}<c \Rightarrow \frac{b}{1+\frac{b}{a}} <c \Rightarrow \frac{ab}{a+b}<c .

In a similar manner, it can be shown that c< \frac{ab}{a-b} from \frac{1}{s_3}<\frac{1}{s_2-s_1} .

Therefore, \frac{ab}{a+b}<c<\frac{ab}{a-b}  (Triangle Altitudes Inequality).

Amazing Yanina

Wow.. jaw dropping haute couture. This is so far the most amazing haute couture I’ve seen this year!
I love the sophistication, elegance and ‘femininity’ with mix of sweetness and sexiness

UAlberta Style on Campus #1

I had a chance to talk to this gorgeous fashionista in campus. She’s the first one to be featured this winter term 2015 on UAlberta Style on Campus.

“Dress for Success”

“I am in my last semester of Sociology. A lot of my friends say I overdress for school and look like a business woman, but I have been at this school for 6 years. The only thing that makes me feel better about my life sentence at the University is dressing up sometimes.”

Shiva

Galaxy Pants

I went shopping 2 weeks ago. I saw these unusual designs on pants in some stores in mall. Space is really amazing, and beautiful at the same time! What can you say about the galaxy pants? 😉

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