@eye_kanyalak: More than a trip but I found you all Guys I really miss you 🥺

Eye Kanyalak Nookaew

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Thursday 13 August 2026 02:22:19 GMT

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urfav_Lowey :

why this makes me teary eyed!🥹💓

2026-08-13 22:36:50

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ຫມູບຸ້ງໆ🙇🏻‍♀️ :

2026-08-13 02:24:02

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Katrina Llegado :

🥹🥹🥹❤️‍🩹

2026-08-13 03:32:24

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Ponferrada Renato Jr :

@

2026-08-16 19:57:48

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Lên con mã đoạn rap Song: Đẹp Mã - Hùng Huỳnh Dc: me and @Thế Ngọc Trân #dance #trending #geminihunghuynh #depma #tiemgaudepmachallenge

Did you know this math secret? 🤯 Solve coprime questions in just 5 seconds using Euler’s Totient Formula! No more wasting time listing out numbers. Watch until the end and hit follow if you learned something new! 👇✨ Two numbers are coprime (or relatively prime) if their only shared factor is 1. Their greatest common divisor (GCD) or highest common factor (HCF) is 1. Euler's Totient Function phi(n) Objective To find how many integers from 1 to n are coprime to n (sharing no common factors except 1), assuming n has exactly two distinct prime factors, p and q. Step-by-Step Derivation Step 1: Start with the total pool There are exactly n integers from 1 to n. Step 2: Subtract the multiples of each prime factor We must remove any number that is divisible by p or divisible by q. The number of multiples of p is n/p The number of multiples of q is n/q Subtracting these from our total pool gives: n - n/p - n/q Step 3: Correct for double-counting (Inclusion-Exclusion) Numbers that are multiples of both p and q (multiples of pq) were included in both groups. Because they were subtracted twice, we must add them back exactly once to keep the count accurate. n - n/p - n/q + n/pq Step 4: Factor out the total n To simplify the expression, we factor n out of every term in the equation: n(1 - 1/p - 1/q + 1/pq) Step 5: Factor by grouping The algebraic expression inside the parentheses can be perfectly factored into a product of two distinct binomials: phi(n) = n(1-1/p)(1-1/q) #MathHacks #MathTricks #EulersTotient #SATMath #SmartThinking

Did you know this math secret? 🤯 Solve coprime questions in just 5 seconds using Euler’s Totient Formula! No more wasting time listing out numbers. Watch until the end and hit follow if you learned something new! 👇✨ Two numbers are coprime (or relatively prime) if their only shared factor is 1. Their greatest common divisor (GCD) or highest common factor (HCF) is 1. Euler's Totient Function phi(n) Objective To find how many integers from 1 to n are coprime to n (sharing no common factors except 1), assuming n has exactly two distinct prime factors, p and q. Step-by-Step Derivation Step 1: Start with the total pool There are exactly n integers from 1 to n. Step 2: Subtract the multiples of each prime factor We must remove any number that is divisible by p or divisible by q. The number of multiples of p is n/p The number of multiples of q is n/q Subtracting these from our total pool gives: n - n/p - n/q Step 3: Correct for double-counting (Inclusion-Exclusion) Numbers that are multiples of both p and q (multiples of pq) were included in both groups. Because they were subtracted twice, we must add them back exactly once to keep the count accurate. n - n/p - n/q + n/pq Step 4: Factor out the total n To simplify the expression, we factor n out of every term in the equation: n(1 - 1/p - 1/q + 1/pq) Step 5: Factor by grouping The algebraic expression inside the parentheses can be perfectly factored into a product of two distinct binomials: phi(n) = n(1-1/p)(1-1/q) #MathHacks #MathTricks #EulersTotient #SATMath #SmartThinking

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@marianafagundesoficial #marianafagundescy #mjmusiccy #SERTANEJO #SBT

@marianafagundesoficial #marianafagundescy #mjmusiccy #SERTANEJO #SBT

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