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$\\var{numbers[0]} =$ [[0]]

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$\\var{numbers[1]} =$ [[0]]

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$\\var{numbers[2]} =$ [[0]]

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$\\var{numbers[3]} =$ [[0]]

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Factorize completely the following numbers.

\n

For example if you are factorizing $1998$ then we have $1998 = 2 \\times 3^3 \\times 37$ and you would enter 2 * 3^3 * 37.

", "tags": ["checked2015"], "rulesets": {}, "type": "question", "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Pick four numbers from $1900\\dots 2015$ and ask the student to factorise them.

\n

Custom marking scripts make sure the student has entered a complete factorisation.

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Split into factors, each factor a power of a prime number and include the multiplication sign * between the factors

", "showStrings": false, "partialCredit": 0, "strings": ["*", "^"]}, "vsetrange": [0, 1], "scripts": {"constructor": {"script": "question.createFactorisePart(this);", "order": "after"}, "mark": {"script": "question.markFactorisePart(this);", "order": "instead"}}, "answersimplification": "unitPower,zeroPower,unitFactor", "expectedvariablenames": [], "showpreview": true, "checkingtype": "absdiff", "checkingaccuracy": 0.001, "minlength": {"length": 7, "message": "", "partialCredit": 0}, "type": "jme", "checkvariablenames": false, "showCorrectAnswer": true, "marks": 3, "vsetrangepoints": 5}], "type": "gapfill", "prompt": "\n\n\n

Factorize completely $\\var{ntbf}$.

\n\n\n\n

Input your answer in the form p^r * q^s * ... where $p, q, \\dots$ are distinct primes and $r, s, \\dots$ are their powers.

\n\n\n\n

$\\var{ntbf}=\\;\\;$[[0]]

\n\n\n\n

(There is a Maple function $\\mathrm{ifactor}(n)$ which factorizes integers.)

\n\n", "showCorrectAnswer": true, "marks": 0}], "variablesTest": {"condition": "", "maxRuns": 100}, "statement": "", "tags": ["Arithmetic", "arithmetic", "checked2015", "factorisation", "factorization", "integers", "MAS3214", "number theory", "prime factorisation", "prime factorization", "prime factors", "prime numbers", "prime powers"], "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "preamble": {"css": "", "js": "question.createFactorisePart = function(p) {\n var n = parseInt(p.settings.correctAnswer);\n var factors = Numbas.math.factorise(n);\n\n p.settings.correctAnswer = p.display.correctAnswer = factors.map(function(pow,i){\n if(pow==1) {\n return Numbas.math.primes[i];\n } else if(pow>0){\n return Numbas.math.primes[i]+'^'+pow;\n } else {\n return '';\n }\n }).filter(function(x){return x}).join(' * ');\n\n p.display.correctAnswerLaTeX = factors.map(function(pow,i){\n if(pow==1) {\n return Numbas.math.primes[i];\n } else if(pow>0){\n return Numbas.math.primes[i]+'^{'+pow+'}';\n } else {\n return '';\n }\n }).filter(function(x){return x}).join(' \\\\times ');\n}\n\nquestion.markFactorisePart = function(p) {\n var m = Numbas.jme.display.matchExpression('m_all(m_any(m_number,m_number^m_number));factors*??',p.studentAnswer,true);\n\n function onlyPrimes(tree) {\n var number;\n if(tree.tok.type=='op' && tree.tok.name=='*') {\n return onlyPrimes(tree.args[0]) && onlyPrimes(tree.args[1]);\n } else if(tree.tok.type=='op' && tree.tok.name=='^') {\n number = tree.args[0].tok.value;\n } else {\n number = tree.tok.value;\n }\n return Numbas.math.primes.contains(number);\n }\n\n if(!onlyPrimes(m.factors)) {\n p.setCredit(0,'Your answer is not completely factorised.');\n return ;\n }\n JMEPart.prototype.mark.apply(p);\n}"}, "type": "question", "metadata": {"notes": "

16/08/2012:

\n

Added tags.

\n

Added description.

\n

No advice given.

", "licence": "Creative Commons Attribution 4.0 International", "description": "

Factorising 5 to 7 digit numbers into a product of prime powers.

\n

Uses the marking algorithms from question 1 of this CBA

"}, "advice": "\n\n\n\n\n"}, {"name": "Blathnaid's copy of Perform arithmetic in Z_n", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Blathnaid Sheridan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/447/"}], "functions": {}, "ungrouped_variables": ["a1", "g5", "ans3", "g3", "ans1", "ans2", "t5", "ans4", "ans5", "s5", "h5", "b4", "b5", "a3", "a2", "a5", "b1", "b2", "b3", "a4"], "tags": ["checked2015", "MAS3214", "Modular arithmetic", "modular arithmetic"], "advice": "

In the the last part, working out $(\\var{a5}+\\var{b5})\\times (\\var{g5}+\\var{h5}) \\bmod{X}$, it is sometimes easier to work out $(\\var{a5}+\\var{b5}) \\bmod{X}$ and $(\\var{g5}+\\var{h5}) \\bmod{X}$ separately, giving two numbers in the range $[0 \\dots X-1]$, and then to multiply them together.

\n

For example, working $\\bmod{9}$ we have:

\n

\\begin{align}
\\var{a5}+\\var{b5}&\\equiv \\var{mod(a5+b5,9)} \\bmod{9}, \\\\
\\var{g5}+\\var{h5}&\\equiv \\var{mod(g5+h5,9)} \\bmod{9}. \\\\ \\\\
(\\var{a5}+\\var{b5})\\times (\\var{g5}+\\var{h5}) &\\equiv \\var{s5} \\times \\var{t5} \\bmod{9} \\\\
&\\equiv \\var{mod(ans5,9)} \\bmod{9}
\\end{align}

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Perform the following calculations in $\\mathbb{Z}_{2},\\;\\;\\mathbb{Z}_{9},\\;\\;\\mathbb{Z}_{10}$.

\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n
$\\mathbb{Z}_{2}$$\\mathbb{Z}_{9}$$\\mathbb{Z}_{10}$
$\\var{a1}+\\var{b1}$[[0]][[1]][[2]]
$\\var{a2}\\times\\var{b2}$[[3]][[4]][[5]]
$\\var{a3}\\times(\\var{b3}+\\var{g3})$[[6]][[7]][[8]]
$\\var{a4}\\times\\var{b4}$[[9]][[10]][[11]]
$(\\var{a5}+\\var{b5})\\times (\\var{g5}+\\var{h5})$[[12]][[13]][[14]]
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0.2, "type": "numberentry", "showPrecisionHint": false}, {"allowFractions": false, "scripts": {}, "maxValue": "{mod(ans4,2)}", "minValue": "{mod(ans4,2)}", "correctAnswerFraction": false, "showCorrectAnswer": true, "marks": 0.2, "type": "numberentry", "showPrecisionHint": false}, {"allowFractions": false, "scripts": {}, "maxValue": "{mod(ans4,9)}", "minValue": "{mod(ans4,9)}", "correctAnswerFraction": false, "showCorrectAnswer": true, "marks": 0.2, "type": "numberentry", "showPrecisionHint": false}, {"allowFractions": false, "scripts": {}, "maxValue": "{mod(ans4,10)}", "minValue": "{mod(ans4,10)}", "correctAnswerFraction": false, "showCorrectAnswer": true, "marks": 0.2, "type": "numberentry", "showPrecisionHint": false}, {"allowFractions": false, "scripts": {}, "maxValue": "{mod(ans5,2)}", "minValue": "{mod(ans5,2)}", "correctAnswerFraction": false, "showCorrectAnswer": true, "marks": 0.2, "type": "numberentry", "showPrecisionHint": false}, {"allowFractions": false, "scripts": {}, 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"description": ""}, "g3": {"definition": "random(3..6)", "templateType": "anything", "group": "Ungrouped variables", "name": "g3", "description": ""}, "ans1": {"definition": "a1+b1", "templateType": "anything", "group": "Ungrouped variables", "name": "ans1", "description": ""}, "ans2": {"definition": "a2*b2", "templateType": "anything", "group": "Ungrouped variables", "name": "ans2", "description": ""}, "t5": {"definition": "mod(g5+h5,9)", "templateType": "anything", "group": "Ungrouped variables", "name": "t5", "description": ""}, "ans4": {"definition": "a4*b4", "templateType": "anything", "group": "Ungrouped variables", "name": "ans4", "description": ""}, "ans5": {"definition": "(a5+b5)*(g5+h5)", "templateType": "anything", "group": "Ungrouped variables", "name": "ans5", "description": ""}, "s5": {"definition": "mod(a5+b5,9)", "templateType": "anything", "group": "Ungrouped variables", "name": "s5", "description": ""}, "h5": {"definition": "random(5..9)", "templateType": "anything", 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{"definition": "random(3..6)", "templateType": "anything", "group": "Ungrouped variables", "name": "b3", "description": ""}, "a4": {"definition": "random(6..9)", "templateType": "anything", "group": "Ungrouped variables", "name": "a4", "description": ""}}, "metadata": {"notes": "

16/08/2012:

\n


Added tags.

\n

Added description.    

", "description": "

Calculations in $\\mathbb{Z_n}$ for three values of $n$.     

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\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
 $\\mathbb{Z}_{3}$$\\mathbb{Z}_{5}$$\\mathbb{Z}_{7}$$\\mathbb{Z}_{11}$
$\\simplify[!basic]{{a1}*{b1}+{c1}*{d1}-{f1}}$[[0]][[1]][[2]][[3]]
$\\var{a2}$[[4]][[5]][[6]][[7]]
$ \\displaystyle{ \\frac{1}{\\var{b3}} } $[[8]][[9]][[10]][[11]]
$ \\displaystyle{ \\frac{\\var{c3}}{\\var{b3}} } $[[12]][[13]][[14]][[15]]
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Simplify the following in each of $\\mathbb{Z}_{3}, \\; \\mathbb{Z}_{5}, \\; \\mathbb{Z}_{7}$ and $\\mathbb{Z}_{11}$.

\n

For the last two questions, recall that for a prime $p$, $\\displaystyle{\\frac{1}{b} \\bmod{p}}$ is the unique solution to $bx \\equiv 1 \\bmod{p}$.

\n

You must input all values as positive integers; negative integers are not accepted.

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Express various integers and rationals mod $\\mathbb{Z}_3, \\;\\mathbb{Z}_5,\\;\\mathbb{Z}_7,\\;\\mathbb{Z}_{11}$

"}, "advice": "

For the first row, it may help to simplify each of the numbers modulo the number corresponding to each column, and then perform the calculation.

\n

For example,

\n

\\begin{align}
\\simplify[!basic]{{a1}*{b1}+{c1}*{d1}-{f1}} &\\equiv \\simplify[!basic]{{mod(a1,3)}*{mod(b3,3)}+{mod(c1,3)}*{mod(d1,3)}-{mod(f1,3)}}  \\mod{3} \\\\
&\\equiv \\var{ans13} \\mod{3}
\\end{align}

\n

In the second row, you just have to work out the remainder when dividing $\\var{a2}$ by each of $3$, $5$, $7$ and $11$.

\n

In the third row we have to find the inverse of $\\var{b3}$ in each of $\\mathbb{Z}_{3}$, $\\mathbb{Z}_{5}$, $\\mathbb{Z}_{7}$ and $\\mathbb{Z}_{11}$.

\n

It should be clear to you that $\\var{b3}$ is coprime to each of $3$, $5$, $7$ and $11$, hence we can find an inverse in each of $\\mathbb{Z}_{3}$, $\\mathbb{Z}_{5}$, $\\mathbb{Z}_{7}$ and $\\mathbb{Z}_{11}$.

\n

You can do this in $\\mathbb{Z}_{3}$ by finding $x$ such that $\\var{b3}x \\equiv 1 \\bmod{3}$, similarly for $5$, $7$, $11$.

\n

You can do this by trying values to find one that works; this is easy for $\\mathbb{Z}_{3}$ since there are only two cases to check, and you will find that

\n

$b = \\var{inv3}$ satisfies $\\var{inv3} \\times \\var{b3} = \\var{inv3*b3} \\equiv 1 \\bmod{3}.$

\n

Inverse in $\\mathbb{Z}_{5}$

\n

If you cannot immediately find an inverse in $\\mathbb{Z}_{5}$, you can use the Euclidean algorithm to find $a$ and $b$ such that

\n

\\[\\var{b3}b +5a = \\operatorname{gcd}(\\var{b3},5) = 1\\]

\n

as $\\var{b3}$ and $5$ are coprime.

\n

It follows that $\\var{b3}b \\equiv 1 \\bmod{5}$, and hence $b \\bmod{5}$ is the inverse of $\\var{b3}$ in $\\mathbb{Z}_{5}$.

\n

Using the Euclidean algorithm I find that

\n

\\[\\simplify[!basic]{ {max(5,b3)}*{p5} + {min(5,b3)}*{q5} = 1 }.\\]

\n

Hence the inverse is

\n

\\[ b = \\var{rinv5} \\equiv \\var{inv5} \\bmod{5}.\\]

\n

Inverse in $\\mathbb{Z}_{7}$

\n

Once again if you cannot spot the solution:

\n

Using the Euclidean algorithm I find that

\n

\\[\\simplify[!basic]{ {max(7,b3)}*{p7} + {min(7,b3)}*{q7} = 1 }.\\]

\n

Hence the inverse of $\\var{b3}$ in $\\mathbb{Z}_{7}$ is

\n

\\[b = \\var{rinv7} \\equiv \\var{inv7} \\bmod{7}.\\]

\n

Inverse in $\\mathbb{Z}_{11}$

\n

Using the Euclidean algorithm I find that

\n

\\[\\simplify[!basic]{ {max(11,b3)}*{p11} + {min(11,b3)}*{q11} = 1}. \\]

\n

Hence the inverse of $\\var{b3}$ in $\\mathbb{Z}_{11}$ is

\n

\\[ b = \\var{rinv11} \\equiv \\var{inv11} \\bmod{11}.\\]

\n

Last Question

\n

The last question's answers are found by, for example in $\\mathbb{Z}_{7}$, considering (all $\\bmod 7$)

\n

\\[\\frac{\\var{c3}}{\\var{b3}} = \\var{c3} \\times \\frac{1}{\\var{b3}} \\equiv \\var{c3} \\times \\var{inv7} \\equiv \\var{c3*inv7} \\equiv \\var{minv7} \\bmod{7}.\\]

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