What is the remainder when 47^51 is divided by 10?

We can use units digit cyclicity to find remainders when the divisor is 2, 5, or 10*
\(47^{51}\) has the same units digit as \(7^{51}\)

(1) Units digit cyclicity of 7 (multiply only the last digit by 7 to get the next power):
\(7^1 = 7\)
\(7^2 =\)__\(9\)
\(7^3=\) __\(3\)
\(7^4=\) __\(1\)
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\(7^5=\) __\(7\)
\(7^6=\) __\(9\)
\(7^7=\) __\(3\)
\(7^8=\) __\(1\)

Powers of 7 have a cyclicity of 4. (The units digits pattern repeats after 4 terms)

(2) Calculate where \(7^{51}\) "hits" in a cycle of 4
-- Start with a multiple of 4 that is close to 51
\(7^{48}\) has the same units digit as \(7^4\)
(48 has exactly 12 cycles of 4. The cycle now starts over.)
\(7^{49}\) has the same units digit as \(7^1\)
\(7^{50}\): same units digit as \(7^2\)
\(7^{51}\): same units digit as \(7^3\)

-- Or divide the exponent by cyclicity of 4
Any remainder, \(r\), tells us that the units digit of \(7^{51}\) is the same as that of \(7^{r}\).
Exponent/cyclicity: \(\frac{51}{4}=12\) plus \(r=3\)
\(7^{r}=7^{3}\)
\(7^{51}\) has the same units digit as \(7^3\)

(3) Use the units digit to find the remainder
Above: \(7^3\) has units digit of 3
\(\frac{3}{10}=0\) Remainder 3
The units digit of \(7^{51}\) is also 3

\(7^{51}\) divided by 10 thus also has a remainder of 3

Answer B

*Units digit cyclicity works for remainders when the divisor is 10 because any remainder IS the units digit. Whether the number is 3; or 33; or 823,543 = 7^7; or 343 = 7^3, when that number is divided by 10, the remainder is 3.