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

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== Problem 28==
 
== Problem 28==
  
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If <math>2137^{753}</math> is multiplied out, the units' digit in the final product in the final product is:
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If <math>2137^{753}</math> is multiplied out, the units' digit in the final product is:
  
 
<math>\textbf{(A)}\ 1\qquad
 
<math>\textbf{(A)}\ 1\qquad
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\textbf{(C)}\ 5\qquad
 
\textbf{(C)}\ 5\qquad
 
\textbf{(D)}\ 7\qquad
 
\textbf{(D)}\ 7\qquad
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\textbf{(E)}\ 9</math>  
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\textbf{(E)}\ 9</math>
  
 
==Solution==
 
==Solution==
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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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==Alternate Solution==
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* <math>Lemma</math> (<math>Fermat's</math> <math>Theorem</math>): If <math>p</math> is a prime and <math>a</math> is an integer prime to <math>p</math> then we have <math>a^{p-1} \equiv 1\ (\textrm{mod}\ p)</math>.
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*Let's define <math>U</math>(<math>x</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>(<math>7^1</math>)= <math>7</math>
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  .      .
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  .      .
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  .      .
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  <math>U</math>(<math>7^4)= </math>1<math></math>
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and we can see by Fermat's Theorem that this cycle repeats with the cyclicity of <math>4</math> .                                                   
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Now <math>753</math> = <math>4k</math> + <math>1</math> <math> =></math> <math>U(</math>7<math>^{753}</math>)<math> = </math>7<math>.
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</math> ~GEOMETRY-WIZARD $
  
 
==See Also==
 
==See Also==

Latest revision as of 08:08, 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 $\boxed{\textbf{(D)}}$.

Alternate Solution

  • $Lemma$ ($Fermat's$ $Theorem$): If $p$ is a prime and $a$ is an integer prime to $p$ then we have $a^{p-1} \equiv 1\ (\textrm{mod}\ p)$.
  • Let's define $U$($x$) as units digit funtion of $x$.

We can clearly observe that, 

 $U$($7^1$)= $7$
  .      .
  .      .
  .      .
 $U$($7^4)=$1$$ (Error compiling LaTeX. Unknown error_msg)

and we can see by Fermat's Theorem that this cycle repeats with the cyclicity of $4$ . Now $753$ = $4k$ + $1$ $=>$ $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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