Question

Does\(17\)divide each of these numbers?

a) \(68\)

b) \(84\)

c) \(357\)

d) \(1001\)

Short answer

(a) Yes

(b) No

(c) Yes

(d) No.

Step-by-step solution

Step 1

DEFINITIONS

a divides b if there exists an integer c such that \(ac = b\).

Notation: \(a|b\)

SOLUTION

a) \({\rm{17}}\)divides \(68\), because \(17 \cdot 4 = 68\).

Note: you can also notice this by using long division and noting that the remainder is \(0\).

\(1{\rm{ 7}}\mathop{\left){\vphantom{1\begin{array}{l}6{\rm{ 8}}\\{\rm{0}}\\\overline 6 {\rm{ 8}}\\6{\rm{ 8}}\\\overline {{\rm{ }}0} \end{array}}}\right.

\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}6{\rm{ 8}}\\{\rm{0}}\\\overline 6 {\rm{ 8}}\\6{\rm{ 8}}\\\overline {{\rm{ }}0} \end{array}}}}

\limits^{\displaystyle\,\,\, {0{\rm{ 4}}}}\)

Step 2

b) \(17 \cdot 4 = 68\)

\(17 \cdot 5 = 85\)

Thus if \(17 \cdot c = 84\), then c has to be between \(4{\rm{ and 5}}\). Thus c cannot be an integer and thus \(17\)does not divide \(84\).

Note: You can also notice this by using long division and noting that the remainder is not \(0\).

\(1{\rm{ 7}}\mathop{\left){\vphantom{1\begin{array}{l}{\rm{8 4}}\\{\rm{0}}\\\overline 8 {\rm{ 4}}\\6{\rm{ 8}}\\\overline {{\rm{1 6 }}} \end{array}}}\right.

\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}{\rm{8 4}}\\{\rm{0}}\\\overline 8 {\rm{ 4}}\\6{\rm{ 8}}\\\overline {{\rm{1 6 }}} \end{array}}}}

\limits^{\displaystyle\,\,\, {0{\rm{ 4}}}}\)

c) \(17\)divide \(17\), because \(17 \cdot 21 = 357\).

Note: You can also notice this by using long division and noting that the remainder is \(0\).

\(1{\rm{ 7}}\mathop{\left){\vphantom{1\begin{array}{l}{\rm{3 5 7}}\\{\rm{0}}\\\overline 3 {\rm{ 5}}\\{\rm{3 4}}\\\overline \begin{array}{l}{\rm{1 7}}\\{\rm{1 7}}\\\overline {{\rm{ }}0} {\rm{ }}\end{array} \end{array}}}\right.

\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}{\rm{3 5 7}}\\{\rm{0}}\\\overline 3 {\rm{ 5}}\\{\rm{3 4}}\\\overline \begin{array}{l}{\rm{1 7}}\\{\rm{1 7}}\\\overline {{\rm{ }}0} {\rm{ }}\end{array} \end{array}}}}

\limits^{\displaystyle\,\,\, {0{\rm{ 2 1}}}}\)

Step 3

d) \(17 \cdot 58 = 986\)

\(17 \cdot 59 = 1003\)

Thus if\(17 \cdot c = 1001c\), then c has to be between \(58{\rm{ and 59}}\). Thus c cannot be an integer and thus \(17\)does not divide \(1001\).

Note: You can also notice this by using long division and noting that the remainder is not \(0\).

\(1{\rm{ 7}}\mathop{\left){\vphantom{1\begin{array}{l}{\rm{1 0 0 1}}\\{\rm{0}}\\\overline 1 {\rm{ 0}}\\{\rm{ 0}}\\\overline \begin{array}{l}{\rm{1 0 0}}\\{\rm{ 8 5}}\\\overline \begin{array}{l}{\rm{ 1 5 1}}\\{\rm{ 1 3 6}}\\\overline {{\rm{ 1 5 }}} \end{array} {\rm{ }}\end{array} \end{array}}}\right.

\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}{\rm{1 0 0 1}}\\{\rm{0}}\\\overline 1 {\rm{ 0}}\\{\rm{ 0}}\\\overline \begin{array}{l}{\rm{1 0 0}}\\{\rm{ 8 5}}\\\overline \begin{array}{l}{\rm{ 1 5 1}}\\{\rm{ 1 3 6}}\\\overline {{\rm{ 1 5 }}} \end{array} {\rm{ }}\end{array} \end{array}}}}

\limits^{\displaystyle\,\,\, {0{\rm{ 0 5 8}}}}\)