1955 AHSME Problems/Problem 37

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A three-digit number has, from left to right, the digits $h, t$, and $u$, with $h>u$. When the number with the digits reversed is subtracted from the original number, the units' digit in the difference is 4. The next two digits, from right to left, are:

$\textbf{(A)}\ \text{5 and 9}\qquad\textbf{(B)}\ \text{9 and 5}\qquad\textbf{(C)}\ \text{impossible to tell}\qquad\textbf{(D)}\ \text{5 and 4}\qquad\textbf{(E)}\ \text{4 and 5}$

Solution

We can set up the subtraction like this: \[\text{ h t u}\] \[- \text{u t h}\] \[-------\] \[\text{? ? 4}\] Since $u < h$, we need to borrow the one from the tens column. Since the result of the tens column is 0, the taken 1 would result in the tens digit being 9:

\[\text{ h t u}\] \[- \text{u t h}\] \[-------\] \[\text{? 9 4}\] We can assign a value for $u$ and $h$, since that doesn't impact the difference, so lets say that $u=3$ and $h=9$. \[\text{ 9 t 3}\] \[- \text{3 t 9}\] \[-------\] \[\text{? 9 4}\] Since $t < t+1$, we can subtract one from the hundred's digit: \[\text{ 9 t 3}\] \[- \text{3 t 9}\] \[-------\] \[\text{5 9 4}\] Out difference is $594$. Therefore, the next two digits, from right to left, are $\boxed{\textbf{(B) } \text{9 and 5}}$.

See Also

Go to the rest of the 1955 AHSME Problems

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