The last three digits of the number N = 7^{10 | vedclass

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(D) We need to find $N \pmod{1000}$,where $N = 7^{100} - 3^{100}$.Note that $7^2 = 49$ and $3^2 = 9$. Thus $7^4 = 2401 \equiv 1 \pmod{1000}$ is not di

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$000$

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(D) We need to find $N \pmod{1000}$,where $N = 7^{100} - 3^{100}$.Note that $7^2 = 49$ and $3^2 = 9$. Thus $7^4 = 2401 \equiv 1 \pmod{1000}$ is not directly helpful,but we can use the binomial expansion.$N = (5+2)^{100} - (5-2)^{100}$.Using the binomial expansion: $(x+y)^n - (x-y)^n = 2 \sum_{k 	ext{ odd}} \binom{n}{k} x^{n-k} y^k$.Here $n=100, x=5, y=2$.$N = 2 [ \binom{100}{1} 5^{99} \cdot 2^1 + \binom{100}{3} 5^{97} \cdot 2^3 + \dots + \binom{100}{99} 5^1 \cdot 2^{99} ]$.Each term contains a factor of $5^k$ where $k \ge 1$. Specifically,for $k \ge 3$,$5^k$ is a multiple of $125$. Combined with the factor of $2^3=8$,these terms are multiples of $1000$.The first term is $2 \cdot 100 \cdot 5^{99} \cdot 2 = 400 \cdot 5^{99} = 400 \cdot 5^3 \cdot 5^{96} = 50000 \cdot 5^{96}$,which is a multiple of $1000$.Thus,$N \equiv 0 \pmod{1000}$,meaning the last three digits are $000$.

The last three digits of the number $N = 7^{100} - 3^{100}$ are

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