Question

The sum of last three digit if the number N = $7^{100}-3^{100}$ is

• A. 0
• B. 1
• C. 3
• D. 4

Answer 1

Answer: (A)

0

Answer 2

To find the sum of the last three digits of the number $N = 7^{100} - 3^{100}$, we need to find the value of $N$ modulo 1000. The last three digits of a number are represented by its remainder when divided by 1000.

We can use the Carmichael function, $\lambda(n)$, which is the smallest positive integer such that $a^{\lambda(n)} \equiv 1 \pmod{n}$ for all integers $a$ coprime to $n$.

First, find the prime factorization of 1000: $1000 = 10^3 = (2 \times 5)^3 = 2^3 \times 5^3$.

Next, calculate $\lambda(1000)$: $\lambda(1000) = \text{lcm}(\lambda(2^3), \lambda(5^3))$.

For a prime $p$ and positive integer $k$:

• If $p$ is an odd prime, $\lambda(p^k) = \phi(p^k) = p^k - p^{k-1}$.
• For $p=2$, $\lambda(2) = 1$, $\lambda(4) = 2$, and $\lambda(2^k) = 2^{k-2}$ for $k \ge 3$.

Let's calculate the individual Carmichael values:

1. For $2^3$: $\lambda(2^3) = \lambda(8) = 2^{3-2} = 2^1 = 2$.

2. For $5^3$: $\lambda(5^3) = \phi(5^3) = 5^3 - 5^{3-1} = 125 - 25 = 100$.

Now, calculate $\lambda(1000)$: $\lambda(1000) = \text{lcm}(2, 100) = 100$.

According to Carmichael's theorem, if $\gcd(a, n) = 1$, then $a^{\lambda(n)} \equiv 1 \pmod{n}$. In our case, $n=1000$ and $\lambda(1000) = 100$.

1. For $7^{100} \pmod{1000}$: Since $\gcd(7, 1000) = 1$, and the exponent is 100, which is equal to $\lambda(1000)$: $7^{100} \equiv 1 \pmod{1000}$.

2. For $3^{100} \pmod{1000}$: Since $\gcd(3, 1000) = 1$, and the exponent is 100, which is equal to $\lambda(1000)$: $3^{100} \equiv 1 \pmod{1000}$.

Now, substitute these results back into the expression for $N$: $N = 7^{100} - 3^{100} \pmod{1000}$ $N \equiv 1 - 1 \pmod{1000}$ $N \equiv 0 \pmod{1000}$.

This means that $N$ is a multiple of 1000. Any number that is a multiple of 1000 has its last three digits as 000. For example, $N = \dots000$. The last three digits are 0, 0, and 0.

The sum of the last three digits is $0 + 0 + 0 = 0$.