Find the remainder when 3 times 7^{22} + 2 ti | vedclass

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(C) We need to find the remainder of $3 	imes 7^{22} + 2 	imes 10^{22} - 44$ when divided by $18$. Note that $7 = 18 - 11$ or $7 \equiv -11 \pmod{18

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(C) We need to find the remainder of $3 	imes 7^{22} + 2 	imes 10^{22} - 44$ when divided by $18$. Note that $7 = 18 - 11$ or $7 \equiv -11 \pmod{18}$,but it is easier to write $7 = 1 + 6$. Using the Binomial Theorem: $7^{22} = (1 + 6)^{22} = 1 + 22 	imes 6 + \binom{22}{2} 	imes 6^2 + \dots = 1 + 132 + 18k = 133 + 18k \equiv 7 \pmod{18}$. Alternatively,consider modulo $18$: $3 	imes 7^{22} + 2 	imes 10^{22} - 44 \pmod{18}$. Since $7^2 = 49 = 2 	imes 18 + 13 \equiv -5 \pmod{18}$,this is not ideal. Let's use $7^2 = 49 \equiv 13 \equiv -5 \pmod{18}$. $7^{22} = (7^2)^{11} \equiv (-5)^{11} \pmod{18}$. $10^2 = 100 = 5 	imes 18 + 10 \equiv 10 \pmod{18}$. $10^{22} = (10^2)^{11} \equiv 10^{11} \pmod{18}$. Actually,using $7 = 1 + 6$: $3(1+6)^{22} + 2(1+9)^{22} - 44 \equiv 3(1 + 22 	imes 6) + 2(1 + 22 	imes 9) - 44 \pmod{18}$ $\equiv 3(1 + 132) + 2(1 + 198) - 44 \pmod{18}$ $\equiv 3(1 + 6) + 2(1 + 0) - 44 \pmod{18}$ $\equiv 3(7) + 2(1) - 44 \pmod{18}$ $\equiv 21 + 2 - 44 \pmod{18}$ $\equiv 23 - 44 \pmod{18}$ $\equiv -21 \pmod{18}$ $\equiv -21 + 36 \pmod{18}$ $\equiv 15 \pmod{18}$. Thus,the remainder is $15$.

Find the remainder when $3 \times 7^{22} + 2 \times 10^{22} - 44$ is divided by $18$.

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