Art of Problem Solving

1961 AHSME Problems/Problem 28: Difference between revisions

Rockmanex3 (talk | contribs)
Solution to Problem 28
 
Line 1: Line 1:
== Problem 28==
== Problem 28==


If <math>2137^{753}</math> is multiplied out, the units' digit in the final product in the final product is:
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
Line 7: Line 7:
\textbf{(C)}\ 5\qquad
\textbf{(C)}\ 5\qquad
\textbf{(D)}\ 7\qquad
\textbf{(D)}\ 7\qquad
\textbf{(E)}\ 9</math>  
\textbf{(E)}\ 9</math>


==Solution==
==Solution==

Revision as of 03:51, 17 March 2020

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 $\boxed{\textbf{(D)}}$.

See Also

1961 AHSC (ProblemsAnswer KeyResources)
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

These problems are copyrighted © by the Mathematical Association of America.