Difference between revisions of "1986 AJHSME Problems/Problem 8"
Aquakitty11 (talk | contribs) |
Mathgeek2006 (talk | contribs) m (→Problem) |
||
| Line 3: | Line 3: | ||
In the product shown, <math>\text{B}</math> is a digit. The value of <math>\text{B}</math> is | In the product shown, <math>\text{B}</math> is a digit. The value of <math>\text{B}</math> is | ||
| − | <cmath>\begin{ | + | <cmath>\begin{array}{rr} |
| − | \text{B}2 \\ | + | &\text{B}2 \\ |
| − | \times 7\text{B} \\ \hline | + | \times& 7\text{B} \\ \hline |
| − | 6396 \\ | + | &6396 \\ |
| − | \end{ | + | \end{array}</cmath> |
<math>\text{(A)}\ 3 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8</math> | <math>\text{(A)}\ 3 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8</math> | ||
Latest revision as of 18:35, 10 March 2015
Problem
In the product shown, 
is a digit. The value of is
![\[\begin{array}{rr} &\text{B}2 \\ \times& 7\text{B} \\ \hline &6396 \\ \end{array}\]](http://latex.artofproblemsolving.com/1/e/4/1e4699049e51f4d0f4365532c7936d2d0a0b448a.png)
Solution
Note that in any multiplication problem, the only 2 digits that will influence the last digit of the number will be the last digits of each number being multiplied.
So, 
has a units digit of 
, so is either 
or 
. If 
, then the product is 
, which is clearly too small, so 
See Also
| 1986 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
