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Find the remainder when 6799 is divided by 7.

A. 4

B. 6

C. 1

D. 2

Answer: Option C

Solution(By Examveda Team)

$$\eqalign{ & {\text{Remainder of}}\frac{{{{67}^{99}}}}{7} \cr & {\text{or, }}R = \frac{{{{\left( {63 + 4} \right)}^{99}}}}{7} \cr} $$
63 is divisible by 7 for any power, so required remainder will depend on the power of 4
Required remainder
$$\eqalign{ & \frac{{{4^{99}}}}{7} = = R = = \frac{{{4^{\left( {96 + 3} \right)}}}}{7} \cr & \frac{{{4^3}}}{7} \Rightarrow \frac{{64}}{7} \Rightarrow \frac{{\left( {63 + 1} \right)}}{7} = = R \Rightarrow 1 \cr & \cr & {\text{Note}}: \cr & \frac{4}{7}{\text{remainder}} = 4 \cr & \frac{{\left( {4 \times 4} \right)}}{7} = \frac{{16}}{7}{\text{remainder}} = 2 \cr & \frac{{\left( {4 \times 4 \times 4} \right)}}{7} = \frac{{64}}{7} = 1 \cr & \frac{{\left( {4 \times 4 \times 4 \times 4} \right)}}{7} = \frac{{256}}{7}{\text{remainder}} = 4 \cr & \frac{{\left( {4 \times 4 \times 4 \times 4 \times 4} \right)}}{7} = 2 \cr} $$
If we check for more power we will find that the remainder start repeating themselves as 4, 2, 1, 4, 2, 1 and so on. So when we get A number having greater power and to be divided by the other number B, we will break power in (4n + x) and the final remainder will depend on x i.e. $$\frac{{{{\text{A}}^{\text{x}}}}}{{\text{B}}}$$

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Comments ( 1 )

  1. Palak Grover
    Palak Grover :
    5 years ago

    4^96*4^3/7=1*4^96
    since ,cyclicity of 4^n/7 is 3.
    so,at 4^3*32 0 i.e. 4^3*k 0
    now 4^3*k=1,
    hence,Rem(4^0/7)==1

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