Last two and three digits of $7^{754}$

Question

How can we find the last two and three digits of $7^{754}$?

I tried with the binomial expansion and all but i don't seem to reach anywhere with it. Is there any other way too?

Answer 1

Since you mentioned the Binomial Theorem, let us use it. Note that we are finding the last three digits of $49^{377}$, that is, of $$(-1+50)^{377}.$$ Expand. We get $$(-1)^{377} +(377)(-1)^{376}(50) +\frac{(377)(376)}{2}(-1)^{375}(50)^2+\text{terms divisible by $1000$}.$$ Actually, $\frac{(377)(376)}{2}(-1)^{375}(50)^2$ is also divisible by $1000$, so we can forget about it.

Finally calculate the remainder when $(377)(50)-1$ is divided by $1000$. This can be done in one's head.

Remark:: The calculation took advantage of the fact that $7^2$, from a decimal point of view, is "nice." The standard general procedure is to use a combination of the binary method of modular exponentiation, combined with number-theoretic results such as Fermat's Theorem and Euler's Theorem.

Answer 2

I don't think that binomial expansion is the way to go. Perhaps for 2 digits try mod 4 and mod 25 separately. For the 3 digits one try mod 8 and mod 125 separately.

Answer 3

For the last digit, you need to find the remainder of $7^{754}$ when divided by $10$.

Take a look at the remainders of $7^n$ for small values of $n$:

• $7^1 \to 7$
• $7^2 \to 9$
• $7^3 \to 3$
• $7^4 \to 1$
• $7^5 \to 7$
• $7^6 \to 9$

Can you see a pattern?

You can do the same for two digits, the pattern is still there.

Answer 4

The last two digits of powers of $7$ go in a cycle -$ 07, 49, 43, 01.$ So if raised to power $754$ it has remainder $2$ when divided by $4$ so last two digits are 49. Similarly go for three digits