Difference between revisions of "1986 AJHSME Problems/Problem 8"

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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.
 
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, <math>2 \times B = 6</math>, <math>B = 3</math>
+
So, <math>2 \times \text{B}</math> has a units digit of <math>6</math>, so <math>\text{B}</math> is either <math>3</math> or <math>8</math>. If <math>\text{B}=3</math>, then the product is <math>32\times 73</math>, which is clearly too small, so <math>\text{B}=8</math>
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<math>\boxed{\text{E}}</math>
  
 
==See Also==
 
==See Also==
  
 
[[1986 AJHSME Problems]]
 
[[1986 AJHSME Problems]]

Revision as of 11:33, 24 January 2009

Problem

In the product shown, $\text{B}$ is a digit. The value of $\text{B}$ is

\[\begin{tabular}[t]{r}
\text{B}2 \\
\times 7\text{B} \\ \hline
6396 \\
\end{tabular}\] (Error compiling LaTeX. ! Missing $ inserted.)

$\text{(A)}\ 3 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$

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, $2 \times \text{B}$ has a units digit of $6$, so $\text{B}$ is either $3$ or $8$. If $\text{B}=3$, then the product is $32\times 73$, which is clearly too small, so $\text{B}=8$

$\boxed{\text{E}}$

See Also

1986 AJHSME Problems

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