Fedemaravilla wrote:

What is the remainder when \(47^{51}\) is divided by 10?

(A) 1

(B) 3

(C) 5

(D) 7

(E) 9

\(7^1 = 7\)

\(7^2 =\)__\(9\)

\(7^3=\) __\(3\)

\(7^4=\) __\(1\)

\(7^5=\) __\(7\)

\(7^6=\) __\(9\)

- cyclicity is 4 (starts over again after 4)

You can split the exponents any way you want:

1)\(47^{51} = 47^{1}47^{50}\)

Units digit for \(7^1\)= 7

Units digit for \(7^{50}\):

Divide the exponent by the cyclicity of 4.

\(\frac{50}{4}\) = 12 + remainder 2. The remainder, 2, tells us to use the units digit for \(7^2\).

\(7^{50}\) has the same units digit as \(7^2\): 9

Now we have (_ _ _ 9) * (_7) = _3

Any number divided by 10 will have, as its remainder, the number's last digit.

Remainder 3

Answer B

2) \(47^{51} = 47^347^{48}\)

Units digit of \(47^3 = 7^3 = 3\)

Units digit of \(47^{48}\) (divide by cyclicity of 4):

\(\frac{48}{4} = 12\), no remainder.

When there is no remainder, the units digit is the same as 7's cyclicity; \(7^4\)

\(47^{48}\) has the same units digit as \(7^4\): 1

We have (_ _3) * (_ _ _1) = 3.

A number divided by 10 has, as its remainder, the number's last digit. \(\frac{3}{10}\) = remainder 3.

Answer B

_________________

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that within me there lay an invincible summer.

-- Albert Camus, "Return to Tipasa"