When 3^{2023} is divided by 16,the remainder | vedclass

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(B) We need to find the remainder when $3^{2023}$ is divided by $16$. $3^{2023} = 3 \cdot (3^2)^{1011} = 3 \cdot (9)^{1011} = 3 \cdot (8+1)^{1011}$. U

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(B) We need to find the remainder when $3^{2023}$ is divided by $16$. $3^{2023} = 3 \cdot (3^2)^{1011} = 3 \cdot (9)^{1011} = 3 \cdot (8+1)^{1011}$. Using the Binomial Theorem,$(1+x)^n = 1 + nx + \frac{n(n-1)}{2}x^2 + \dots + x^n$. $3(1+8)^{1011} = 3 \left[ 1 + 1011 \cdot 8 + \binom{1011}{2} 8^2 + \dots + 8^{1011} ight]$. Since $8^2 = 64$,all terms from the third term onwards are divisible by $16$. $3^{2023} = 3(1 + 8088 + 16k)$ for some integer $k$. $3^{2023} = 3(8089 + 16k) = 24267 + 48k$. Now,divide $24267$ by $16$: $24267 = 16 	imes 1516 + 11$. Thus,the remainder is $11$.

When $3^{2023}$ is divided by $16$,the remainder obtained is

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