If 540 is divided by 11- then remainder is - and if 22003 is divided by 17- then remainder is - Then the value of - is

5^40=(5^2 )^20=(22+3 )^20=22λ+3^20,λ∈ℕ Also, 3^20=(3^2 )^10=(11 2 )^10=11μ+2^10, μ∈ℕ Now, 2^10=1024=11× 93+1 ∴ Remainder =1 i.e., α=1 2^2003=2^3· 2^2000=8( ...

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Byju's Answer

Standard XII

Mathematics

Cube Root of a Complex Number

If 540 is div...

Question

If $5^{40}$ is divided by $11,$ then remainder is $α$ and if $2^{2003}$ is divided by $17,$ then remainder is $β .$ Then the value of $(β - α)$ is

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Solution

The correct option is C $7$
$5^{40} = {(5^{2})}^{20} = {(22 + 3)}^{20} = 22 λ + 3^{20}, λ \in N$
Also, $3^{20} = {(3^{2})}^{10} = {(11 - 2)}^{10} = 11 μ + 2^{10}, μ \in N$
Now, $2^{10} = 1024 = 11 \times 93 + 1$
$∴$ Remainder $= 1$ i.e., $α = 1$

$2^{2003} = 2^{3} \cdot 2^{2000} = 8 {(2^{4})}^{500} = 8 {(16)}^{500}$
$= 8 {(17 - 1)}^{500} = 8 (17 v + 1), v \in N$
$= 8 \times 17 v + 8$
$∴$ Remainder $= 8$ i.e., $β = 8$
$\Rightarrow β - α = 8 - 1 = 7$

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If 540 is divided by 11, then remainder is α and if 22003 is divided by 17, then remainder is β. Then the value of (β−α) is

The correct option is C 7540=(52)20=(22+3)20=22λ+320,λ∈N Also, 320=(32)10=(11−2)10=11μ+210,μ∈N Now, 210=1024=11×93+1 ∴ Remainder =1 i.e., α=1 22003=23⋅22000=8(24)500=8(16)500 =8(17−1)500=8(17v+1),v∈N =8×17v+8 ∴ Remainder =8 i.e., β=8 ⇒β−α=8−1=7

If $5^{40}$ is divided by $11,$ then remainder is $α$ and if $2^{2003}$ is divided by $17,$ then remainder is $β .$ Then the value of $(β - α)$ is