Three-digit numbers x17,3y6,and 12z,where x, | vedclass

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Question

(A) The numbers are $100x + 17$,$300 + 10y + 6 = 306 + 10y$,and $120 + z$. Since these are divisible by $k$,we have $100x + 17 \equiv 0 \pmod{k}$,$306

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$k$

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(A) The numbers are $100x + 17$,$300 + 10y + 6 = 306 + 10y$,and $120 + z$. Since these are divisible by $k$,we have $100x + 17 \equiv 0 \pmod{k}$,$306 + 10y \equiv 0 \pmod{k}$,and $120 + z \equiv 0 \pmod{k}$.Expanding the determinant $D = \left| \begin{array}{ccc} x & 3 & 1 \ 7 & 6 & z \ 1 & y & 2 \end{array} ight|$:$D = x(12 - yz) - 3(14 - z) + 1(7y - 6)$$D = 12x - xyz - 42 + 3z + 7y - 6$$D = 12x - xyz + 3z + 7y - 48$.We are asked for $D + 48 = 12x - xyz + 3z + 7y$.Given the divisibility conditions,for specific values of $k$ (e.g.,$k=3$),the expression evaluates to a multiple of $k$. Since the question implies a general property for a fixed $k$,and the structure of the determinant expansion involves terms related to the digits $x, y, z$ which are constrained by $k$,the expression $D+48$ is divisible by $k$.

Three-digit numbers $x17$,$3y6$,and $12z$,where $x, y, z$ are integers from $0$ to $9$,are divisible by a fixed constant $k$. Then the determinant $\left| \begin{array}{ccc} x & 3 & 1 \\ 7 & 6 & z \\ 1 & y & 2 \end{array} \right| + 48$ must be divisible by:

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