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

(→‎SOLUTION 2)
(→‎SOLUTION 2)
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==SOLUTION 2==
 
==SOLUTION 2==
  
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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>.
+
* <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>.
 +
 
 +
Let's define <math>U</math>(<math>x</math>) as units digit funtion of <math>x</math>.
  
−
Let's define <math>U(</math>x<math>)</math> as units digit funtion of <math>x</math>.
 
 
We can clearly observe that,
 
We can clearly observe that,
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<math>U(</math>7^1<math>)</math>= <math>7</math>
 
−
    <math>.
 
−
    .
 
−
    .</math>
 
−
<math>U(</math>7^4)<math>= </math>1<math>
 
  
−
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>.
+
<math>U</math>(<math>7^1</math>)= <math>7</math>
 +
    .      .
 +
    .      .
 +
    .      .
 +
<math>U</math>(<math>7^4)= </math>1<math>
 +
 
 +
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<math> + </math>1<math>  \Rightarrow </math>U(<math>7</math><math>^753</math>)<math> = </math>7<math>.
 
   
 
   
 
</math> ~GEOMETRY-WIZARD $
 
</math> ~GEOMETRY-WIZARD $

Revision as of 07:52, 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 $p$ 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)=$1$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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