If 1-2-22-23-21999 is divided by 5 - then the remainder is

The correct option is C 01+2+2^2 + 2^3+...+2^1999 = 1(2^2000 1)2 1 = 2^2000 1= 4^1000 1 = nbsp;(5 1)^1000 1 = ^1000C_05^1000 ^1000C_15^999+… ^1000C_99 ...

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Standard XII

Mathematics

Finding Remainder of Numbers of the Form a^b

If 1+2+22+23+...

Question

If $1 + 2 + 2^{2} + 2^{3} + \dots + 2^{1999}$ is divided by $5$ , then the remainder is

3

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0

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Solution

The correct option is C $0$
$1 + 2 + 2^{2} + 2^{3} + . . . + 2^{1999} = \frac{1 (2^{2000} - 1)}{2 - 1} = 2^{2000} - 1 = 4^{1000} - 1 = (5 - 1)^{1000} - 1 =^{1000} C_{0} 5^{1000} -^{1000} C_{1} 5^{999} + \dots -^{1000} C_{999} 5 + 1 - 1 = 5 k + 1 - 1 = 5 k$

Which is divisible by $5$ , so the remainder is $0$ .

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Finding Remainder of Numbers of the Form a^b

Standard XII Mathematics

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If 1+2+22+23+…+21999 is divided by 5, then the remainder is

The correct option is C 01+2+22+23+...+21999=1(22000−1)2−1=22000−1=41000−1=(5−1)1000−1= 1000C051000− 1000C15999+…− 1000C9995+1−1=5k+1−1=5k Which is divisible by 5, so the remainder is 0.

If $1 + 2 + 2^{2} + 2^{3} + \dots + 2^{1999}$ is divided by $5$ , then the remainder is