# My favorite Fermat Little Theorem with Pascal Triangle

Fermat Little Theorem: For any prime integer p, any integer m

$\boxed {m^{p} \equiv m \mod p}$

When m = 2,

$\boxed{2^{p} \equiv 2 \mod p}$

Note: 九章算数 Fermat Little Theorem (m=2)

$1 \: 1 \implies sum = 2 = 2^1 \equiv 2 \mod 1$

$1\: 2 \:1\implies sum = 4 = 2^2 \equiv 2 \mod 2 \;(\equiv 0 \mod 2)$

$1 \:3 \:3 \:1 \implies sum = 8= 2^3 \equiv 2 \mod 3$

1 4 6 4 1 => sum = 16= 2^4 (4 is non-prime)

$1 \:5 \:10\: 10\: 5\: 1 \implies sum = 32= 2^5 \equiv 2 \mod 5$

[PODCAST]

https://kpknudson.com/my-favorite-theorem/2017/9/13/episode-4-jordan-ellenberg

# The most addictive theorem in Applied mathematics

What is your favorite theorem ?

I have 2 theorems which trigger my love of Math :

1. Chinese Remainder Theorem: 韩信点兵, named after a 200 BCE Han dynasty genius general Han Xin （韩信） who applied this modern “Modular Arithmetic” in battle fields.
2. Fermat’s Last Theorem：The Math “prank” initiated by the 17CE amateur Mathematician Pierre de Fermat kept the world busy for 380 years until 1974 proved by Andrew Wiles.

Note 1: Lycée Pierre de Fermat (Classe Préparatoire) happens to be my alma mater named after this great Mathematician born in the same southern France “Airbus City” Toulouse.

Note 2: His another Fermat’s Little Theorem is used in modern computer cryptography.

China 南京航空航天大学 Nanjing University of Aeronautics and Astronautics set the WiFi password as the answer of this integral (first 6 digits).

Can you solve it?

(If can’t, please revise GCE “A-level” / Baccalaureate / 高考 Calculus 微积分)

Answer : Break the integral (I) into 2 parts:

I = A(x) + B(x)

$\displaystyle A(x) = \int_{-2 }^ {2} x^{3}. \cos \frac{x}{2}.\sqrt{4-x^2}dx$

$\displaystyle B(x) = \int_{-2 }^ {2} \frac{1}{2}\sqrt{4-x^2}dx$

A(x) = – A(-x) => Odd function
=> A(x) = 0 since its area canceled out over [-2, 2]

B(x) = B(-x) => Even function
$\displaystyle\implies B(x) = 2\int_{0 }^ {2} \frac{1}{2}\sqrt{4 - x^2}dx$
$\displaystyle\implies B(x) = \int_{0 }^ {2} \sqrt{4 - x^2}dx$

Let x = 2 sin t => dx = 2 cos t. dt

x = 2 = 2 sin t => sin t = 1 => t = π / 2

x = 0 = 2 sin t => sin t = 0 => t = 0

$\displaystyle B(x) = \int_{0 }^ {\pi/2} \sqrt{4 - 4.\sin^{2} {t} }. (2 \cos t. dt)$

$\displaystyle \implies B(x) = \int_{0 }^ {\pi/2} 2.\cos t. (2 \cos t. dt)$

$\displaystyle\implies B(x) = \int_{0 }^ {\pi/2} 4 \cos^{2} t. dt$

$\displaystyle \cos ^{2} t = \frac {1 + \cos 2t} {2 }$

$\displaystyle\implies B(x) = \int_{0 }^ {\pi/2} (2 + 2\cos 2t) . dt$

$\displaystyle\implies B(x) = (2 t) \Bigr|_{0 }^ {\pi/2} + (2. \frac{1}{2} \sin 2t) \Bigr|_{0 }^ {\pi/2}$

$\displaystyle\implies B(x) = (\pi ) + \sin \pi = \pi$

$\boxed{ I = \pi = 3.14159}$

# Spot The Mistake?

Solve:

$\displaystyle {\Bigl(\frac{2}{3}\Bigr)} ^{x} = {\Bigl(\frac{3}{2}\Bigr)}^{3}$

$\implies \displaystyle \frac {2^{x}} {3^{x}} = \frac {3^{3}} {2^{3}}$

$\implies \displaystyle{2^{3}}. {2^{x}} ={3^{3}}. {3^{x}}$

$\implies \displaystyle2^{3+x} =3^{3+x}$

Since the exponents (3+x) on both sides are equal,

=> 2 = 3

# Singapore PSLE Math baffled Anxious Parents

One afternoon 5 friends rented 3 bikes from 5 p.m. to 6:30 p.m. and took turns to ride on them. At any time, 3 of them cycled while the other 2 friends rested.

If each of them had the same amount of cycling time, how many minutes did each person ride on a bike?

Note: PSLE (Primary Schools Leaving Exams) is the Singapore National Exams for all 12 year-old pupils at Primary 6 year end. The result of which will determine which secondary school the pupil is qualified to enter the following year. Math subject, besides Science, English and mother tongue (Chinese or Malay or Tamil) are tested in PSLE.

[Answer] Try before you scroll down below ….

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Answer = each student rides 18 mins per bike (= 90 mins /5 ).

The “3” bikes are tricky “smokes” not relevant, it could be any “n” (<6) bikes , as long as total 90 mins, and each student rides same duration.