Elementary Number Theory Problems 4.3 Solution (David M. Burton's 7th Edition) - Q11

Background

All theorems, corollaries, and definitions listed in the book's order:

Theorems and Corollaries in Elementary Number Theory

All theorems and corollaries mentioned in David M. Burton’s Elementary Number Theory are listed by following the book’s order. (7th Edition) (Currently Ch 1 - 4)

RanblogRan

I will only use theorems or facts that are proved before this question. So, you will not see that I quote theorems or facts from the later chapters.

Question

Assuming that $495$ divides $273x49y5$, obtain the digits $x$ and $y$.

Solution

Let $n = 273x49y5$. We have $495 \mid 273x49y5$.

Divisibility by $9$

Since $495 = 9 \cdot 11 \cdot 5$, we have $9 \mid 495$. So, we have $9 \mid n$. Then the sum of the digits of $n$ must be divisible by $9$ according to Theorem 4.5:

Divisibility by $11$

Similarly, since $11 \mid 495$, the alternating sum of the digits of $n$ must be divisible by $11$ according to Theorem 4.6:

Constraints on $x$ and $y$

Since $x$ and $y$ are digits, $0 \leq x, y \leq 9$. So, we have

which means

Solving the System

Because $9 \mid (30 + x + y)$ and $30 \leq 30 + x + y \leq 48$, we have:

which implies that $x + y = 6$ or $x + y = 15$.

Because $11 \mid (12+x-y)$ and $3 \leq 12 + x - y \leq 21$, we have

which implies that $x - y = -1$.

$x - y = -1$ means that $x$ and $y$ are consecutive integers. Their sum must be an odd integer, which eliminates the situation for $x + y = 6$.

Conclusion

Combining $x + y = 15$ and $x - y = -1$, we get $x = 7$ and $y = 8$.

Read More: All My Solutions for This Book