prove-that-the-sum-2-2-2-2-3-2-4-2-2019-2-2020-is-divisible-by-30-

Question Number 154401 by mathdanisur last updated on 18/Sep/21

prove that the sum

2 + 2^{2} + 2^{3} + 2^{4} + \dots + 2^{2019} + 2^{2020}

is divisible by 30

Answered by talminator2856791 last updated on 18/Sep/21

2^{k + 1} + 2^{k + 3} = 2^{k} (8 + 2)

= 10 \cdot 2^{k}

2^{k + 2} + 2^{k + 4} = 2^{k} (16 + 4)

= 20 \cdot 2^{k}

2^{k + 1} + 2^{k + 2} + 2^{k + 3} + 2^{k + 4} = 30 \cdot 2^{k}

S = 2 + 2^{2} + 2^{3} + 2^{4} + \dots . . + 2^{2019} + 2^{2020}

= (2 + 2^{2} + 2^{3} + 2^{4}) + (2^{5} + 2^{6} + 2^{7} + 2^{8}) + \dots . .

+ (2^{2017} + 2^{2018} + 2^{2019} + 2^{2020})

= (2 + 2^{2} + 2^{3} + 2^{4}) + 2^{4} (2 + 2^{2} + 2^{3} + 2^{4}) + \dots . .

+ 2^{2016} (2 + 2^{2} + 2^{3} + 2^{4})

= 30 + 2^{4} \cdot 30 + 2^{8} \cdot 30 + \dots . . + 2^{2016} \cdot 30

= 30 (1 + 2^{4} + \dots . . + 2^{2016})

\Rightarrow 30 ∣ S

Commented by mathdanisur last updated on 18/Sep/21

Very nice Ser thanks

Answered by Rasheed.Sindhi last updated on 18/Sep/21

\underset{―}{MOD 30}

2^{4} \equiv 16 \Rightarrow 2^{4 \times 1} \equiv 16

{(2^{4})}^{2} \equiv {(16)}^{2} \Rightarrow 2^{8} \equiv 16 \Rightarrow 2^{4 \times 2} \equiv 16

{(2^{8})}^{2} \equiv {(16)}^{2} \Rightarrow 2^{12} \equiv 16 \Rightarrow 2^{4 \times 3} \equiv 16

\dots \dots \dots \dots \dots

2^{4 \times k} \equiv 16

2^{4 k} . 2 \equiv 16.2 \equiv 2 \Rightarrow 2^{4 k + 1} \equiv 2

2^{4 k + 1} . 2 \equiv 2.2 \equiv 4 \Rightarrow 2^{4 k + 2} \equiv 4

2^{4 k + 2} . 2 \equiv 4.2 \equiv 8 \Rightarrow 2^{4 k + 3} \equiv 8

2^{4 k} \equiv 16 \Rightarrow 2^{4 k} \equiv 16

▸ S = 2 + 2^{2} + 2^{3} + 2^{4} + \dots + 2^{2019} + 2^{2020}

= \underset{k = 0}{\overset{505}{Σ}} (2^{4 k + 1} + 2^{4 k + 2} + 2^{4 k + 3} + 2^{4 k})

\equiv \underset{k = 1}{\overset{505}{Σ}} (2 + 4 + 8 + 16) . k = \underset{k = 1}{\overset{505}{Σ}} (30 k)

S = 30 \times \underset{k = 1}{\overset{505}{Σ}} k

∴ 30 ∣ S

Commented by mathdanisur last updated on 18/Sep/21

Very nice Ser thankyou

Answered by JDamian last updated on 18/Sep/21

S = 2 \frac{2^{2020} - 1}{2 - 1} = 2 (2^{2020} - 1)

S = 2^{2021} - 2

S m o d 30 = (2^{2021} - 2) m o d 30 =

= 2^{2021} m o d 30 - 2

∣ {(2^{5})}^{k} m o d 30 = {(30 + 2)}^{k} m o d 30 =

= 2^{k} m o d 30 ∣

2^{2021} m o d 30 = [2 \cdot {(2^{5})}^{404}] m o d 30 =

= (2 \cdot 2^{404}) m o d 30 = 2^{405} m o d 30 =

= {(2^{5})}^{81} m o d 30 = 2^{81} m o d 30 =

= 2 \cdot {(2^{5})}^{16} m o d 30 = 2^{17} m o d 30 =

= 2^{2} {(2^{5})}^{3} m o d 30 = 2^{5} m o d 30 = 2

S m o d 30 = 2 - 2 = 0

Commented by mathdanisur last updated on 18/Sep/21

Very nice Ser thankyou