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Difference between revisions of "1961 AHSME Problems/Problem 28"

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Notice that the unit digit eventually cycles to itself when the exponent is increased by <math>4</math>.  It also does not matter what the other digits are in the base because the units digit is found by multiplying by only the units digit.  Since <math>753</math> leaves a remainder of <math>1</math> after being divided by <math>4</math>, the units digit of <math>2137^{753}</math> is <math>7</math>, which is answer choice <math>\boxed{\textbf{(D)}}</math>.
 
Notice that the unit digit eventually cycles to itself when the exponent is increased by <math>4</math>.  It also does not matter what the other digits are in the base because the units digit is found by multiplying by only the units digit.  Since <math>753</math> leaves a remainder of <math>1</math> after being divided by <math>4</math>, the units digit of <math>2137^{753}</math> is <math>7</math>, which is answer choice <math>\boxed{\textbf{(D)}}</math>.
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==SOLUTION 2==
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* <math>Lemma (Fermat's Theorem)</math>: If <math>a</math> is an integer and <math>p</math> is a prime that is prime to a the we have <math>a^p-1</math> \equiv 1 (<math>mod</math> <math>p</math>).
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Let's define <math>U(</math>x<math>)</math> as units digit funtion of <math>x</math>.
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We can clearly observe that,
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<math>U(</math>7^1<math>)</math>= <math>7</math>
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    <math>.</math>
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    <math>.</math>
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    <math>.</math>
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<math>U(</math>7^4)<math>= </math>1<math>
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and we can see by Fermat's Theorem that this cycle repeats with the cyclicity of </math>4<math>. Now </math>753<math> = </math>4k + 1<math>  \Rightarrow </math>U(<math>7^753</math>)<math> =</math>7<math>.
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  </math> ~GEOMETRY-WIZARD $
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==See Also==
 
==See Also==

Revision as of 07:43, 31 December 2023

Problem 28

If $2137^{753}$ is multiplied out, the units' digit in the final product is:

$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 9$

Solution

$7^1$ has a unit digit of $7$. $7^2$ has a unit digit of $9$. $7^3$ has a unit digit of $3$. $7^4$ has a unit digit of $1$. $7^5$ has a unit digit of $7$.

Notice that the unit digit eventually cycles to itself when the exponent is increased by $4$. It also does not matter what the other digits are in the base because the units digit is found by multiplying by only the units digit. Since $753$ leaves a remainder of $1$ after being divided by $4$, the units digit of $2137^{753}$ is $7$, which is answer choice $\boxed{\textbf{(D)}}$.

SOLUTION 2

  • $Lemma (Fermat's Theorem)$: If $a$ is an integer and $p$ is a prime that is prime to a the we have $a^p-1$ \equiv 1 ($mod$ $p$).

Let's define $U($x$)$ as units digit funtion of $x$. We can clearly observe that, $U($7^1$)$= $7$ 

   $.$
   $.$
   $.$

$U($7^4)$=$1$and we can see by Fermat's Theorem that this cycle repeats with the cyclicity of$4$. Now$753$=$4k + 1$\Rightarrow$U($7^753$)$=$7$.$ ~GEOMETRY-WIZARD $


See Also

1961 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 27 		Followed by
Problem 29
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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