3 - 722-2 - 1022 - 44 when divided by 18 leaves the remainder

3 × 7^22 + 2 × 10^22 44 = 3 × (1+ 6)^22 + 2 (1+9)^22 44 =3[^22C_0 + ^22C_1(6) + ^22C_2(6)^2 + ⋯ + ^22C_22(6)^22] + 2[^22C_0 + ^22C_1(9)+⋯+^22C_22(9)^22] ...

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Byju's Answer

Standard XI

Mathematics

Finding remainder of numbers of the form a^b

3 × 722+2 × 1...

Question

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

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Solution

$3 \times 7^{22} + 2 \times 10^{22} - 44$
$= 3 \times (1 + 6)^{22} + 2 (1 + 9)^{22} - 44$
$= 3 [^{22} C_{0} +^{22} C_{1} (6) +^{22} C_{2} (6)^{2} + \dots +^{22} C_{22} (6)^{22}] + 2 [^{22} C_{0} +^{22} C_{1} (9) + \dots +^{22} C_{22} (9)^{22}] - 44$
$= 3 \cdot^{22} C_{0} + 18 [^{22} C_{1} +^{22} C_{2} (6) + \dots +^{22} C_{22} (6)^{21}] + 2 \cdot^{22} C_{0} + 18 [^{22} C_{1} +^{22} C_{2} (9) + \dots +^{22} C_{22} (9)^{21}] - 44$
$= 3 + 18 k_{1} + 2 + 18 k_{2} - 44$
$= 18 (k_{1} + k_{2}) - 39$
$= 18 k_{3} - 54 + 15$
$= 18 (k_{3} - 3) + 15$
$= 18 k + 15$
$∴$ Remainder when divided by $18$ is $15.$

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3×722+2×1022−44 when divided by 18 leaves the remainder

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