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a)

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We seek a number

\n

\\[r=\\var{a} \\bmod \\var{n} \\implies \\var{a} = m\\var{n}+r \\]

\n

where $m$ is a positive integer and $0 \\le r \\le \\var{n-1}$.

\n

This is achieved by dividing $\\var{a}$ by $\\var{n}$ and we find that:

\n

\\begin{align}
\\frac{\\var{a}}{\\var{n}} &= \\var{m}+\\frac{\\var{r}}{\\var{n}} \\\\
\\implies \\var{a} &= \\var{m}\\times \\var{n}+\\var{r}
\\end{align}

\n

on multiplying through by $\\var{n}$.

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Hence $r=\\var{r}$ and $\\var{a} \\bmod \\var{n}=\\var{r}$.

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You can use your calculator to find this as follows:

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Divide $\\var{a}$ by $\\var{n}$ to get

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\\[ \\frac{\\var{a}}{\\var{n}}=\\var{a/n}=\\var{m}+\\var{a/n-m} \\]

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Then multiplying $\\var{a/n-m}$ by $\\var{n}$ gives the remainder, i.e.

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\\[ \\var{a/n-m}\\times \\var{n} = \\var{r} \\]

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This must be an integer if no error is introduced, but the calculator result will be either an integer or very close to an integer – so you need to round to the integer in that case.

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b)

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As in a),

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\\[\\frac{\\var{a1}}{\\var{n1}}=\\var{m1}+\\frac{\\var{r1}}{\\var{n1}}\\]

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So $\\var{a1} \\bmod \\var{n1} = \\var{r1}$

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c)

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\\[\\frac{\\var{a2}}{\\var{n2}}=\\var{m2}+\\frac{\\var{r2}}{\\var{n2}}\\]

\n

So $\\var{a2} \\bmod \\var{n2} = \\var{r2}$

", "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "parts": [{"prompt": "
    \n
  1. $\\var{a} \\bmod \\var{n}=$ [[0]]
    Input your answer as an integer $a$, $0 \\le a \\le \\var{n-1}$.
  2. \n
  3. $\\var{a1} \\bmod \\var{n1}=$ [[1]]
    Input your answer as an integer $b$, $0 \\le b \\le \\var{n1-1}$.
  4. \n
  5. $\\var{a2} \\bmod \\var{n2}=$ [[2]]
    Input your answer as an integer $c$, $0 \\le c \\le \\var{n2-1}$.
  6. \n
\n

 

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Find the following numbers modulo the given number $n$.

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Modular arithmetic. Find the following numbers modulo the given number $n$. Three examples to do. Courtesy of Newcastle University

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