Elementary Number Theory Problems 4.3 Solution (David M. Burton's 7th Edition) - Q6
My Solution for "Working modulo $9$ or $11$, find the missing digits in the calculations below: (a) $51840 \cdot 273581 = 1418243x040$. (b) $2x99561 = [3(523 + x)]^2$. (c) $2784x = x \cdot 5569$. (d) $512 \cdot 1x53125 = 1000000000$."
Background
All theorems, corollaries, and definitions listed in the book's order:
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
Working modulo $9$ or $11$, find the missing digits in the calculations below:
(a) $51840 \cdot 273581 = 1418243x040$.
(b) $2x99561 = [3(523 + x)]^2$.
(c) $2784x = x \cdot 5569$.
(d) $512 \cdot 1x53125 = 1000000000$.
Solution
(a)
Let us consider modulo $9$, and we will be using Theorem 4.5 throughout the calculations:
$$ \begin{equation} \begin{split} 51840 \cdot 273581 & \equiv 1418243x040 \\ (5 + 1 + 8 + 4 + 0) \cdot (2 + 7 + 3 + 5 + 8 + 1) & \equiv (1 + 4 + 1 + 8 + 2 + 4 + 3 + x + 0 + 4 + 0) \\ 18 \cdot 26 & \equiv 27 + x \\ (1 + 8) \cdot (2 + 6) & \equiv 2 + 7 + x \\ 72 & \equiv 9 + x \\ 7 + 2 & \equiv 9 + x \\ x & \equiv 0 \pmod {9} \end{split} \nonumber \end{equation} $$Then we consider modulo $11$, and we will be using Theorem 4.6:
$$ \begin{equation} \begin{split} 51840 \cdot 273581 & \equiv 1418243x040 \\ (0 - 4 + 8 - 1 + 5) \cdot (1 - 8 + 5 - 3 + 7 - 2) & \equiv (1 - 4 + 1 - 8 + 2 - 4 + 3 - x + 0 - 4 + 0) \\ 8 \cdot 0 & \equiv -13 - x \\ 0 & \equiv -13 - x \\ x & \equiv -13 \\ x & \equiv 9 \pmod {11} \end{split} \nonumber \end{equation} $$So, we know $0 \leq x \leq 9$, $x \equiv 0 \pmod {9}$ and $x \equiv 9 \pmod {11}$. Therefore, $x = 9$.
The rest is for Premium Members only
SubscribeAlready have an account? Log in
%20-%20Q6){kind=link}