1955 AHSME Problems/Problem 37

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 $\boxed{\textbf{(B) } \text{9 and 5}}$ .

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1955 AHSME Problems/Problem 37

Solution

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