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Find the lowest common multiple and highest common factors of given numbers. Also a question on identifying prime numbers.

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Two trains arrive at the same platform with different periods. Compute the LCM of the two periods to find the time they clash.

\n

This is a context question testing the student's ability to identify the lowest common multiple of two integer values which are not multiples of each other.

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Trains intended for platform A arrive at every multiple of $\\var{c}$ minutes after midday. Trains intended for platform B arrive at every multiple of $\\var{d}$ minutes after midday. There will be a clash when these times coincide.

\n

The first such time is the lowest common multiple of $\\var{c}$ and $\\var{d}$.

\n

To calculate the lowest common multiple of two numbers, you first need to calculate a list of common multiples for the individual numbers and then look for numbers that appear in both lists.

\n

Multiples of $\\var{c}$ are $\\var{c}, \\var{2c}, \\var{3c}, \\var{4c}, \\var{5c}$, $\\var{6c}$, $\\var{7c}$, $\\var{8c}$, $\\var{9c}$, $\\var{10c}$, $\\var{11c}$, $\\var{12c}$...

\n

Multiples of $\\var{d}$ are $\\var{d}, \\var{2d}, \\var{3d}, \\var{4d}, \\var{5d}$, $\\var{6d}$, $\\var{7d}$, $\\var{8d}$, $\\var{9d}$, $\\var{10d}$, $\\var{11d}$, $\\var{12d}$...

\n

The first number which appears in both lists is $\\var{lcm1}$. This is the lowest common multiple of $\\var{c}$ and $\\var{d}$.

\n

This means that there will be a clash in the arrival timetable at {newtime(lcm1)}, $\\var{lcm1}$ minutes after midday.

", "statement": "

A small train station hosts 2 north-bound platforms, platforms A and B.

\n

Scheduled trains arrive at platform A every $\\var{c}$ minutes, and at platform B every $\\var{d}$ minutes. The trains arrive and depart in less than a minute.

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At midday, trains arrive simultaneously at both platforms. Immediately after they depart, an electrical fault causes platform A to become unusable and the entirety of the arriving trains are diverted to arrive at platform B.

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How long after midday will it be before there is a clash in the arrival timetable?

\n

[[0]] minutes

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This is a simple question testing the student on their ability to calculate the lowest common multiple of two integers which are:

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Part a) - coprime;

\n

Part b) - where the greatest common divisor between the two integers is greater than one and not equal to either given number; and

\n

Part c) - where one of the integer is a multiple of the other.

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#### a)

\n

Here are the times tables for $\\var{a}$ and $\\var{b}$.

\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
 $\\var{a}$ $\\var{2a}$ $\\var{3a}$ $\\var{4a}$ $\\var{5a}$ $\\var{6a}$ $\\var{7a}$ $\\var{8a}$ $\\var{9a}$ $\\var{10a}$ $\\var{11a}$ $\\var{12a}$ $\\var{13a}$ $\\var{b}$ $\\var{2b}$ $\\var{3b}$ $\\var{4b}$ $\\var{5b}$ $\\var{6b}$ $\\var{7b}$ $\\var{8b}$ $\\var{9b}$ $\\var{10b}$ $\\var{11b}$ $\\var{12b}$ $\\var{13b}$
\n

The first number which appears in both lists is $\\var{a*b}$.

\n

Alternately, notice that $\\var{a}$ and $\\var{b}$ don't have any factors in common, so their greatest common divisor is $1$. So the lowest common multiple is just the product of the two numbers.

\n

\n

#### b)

\n

The lowest common multiple of $\\var{f}$ and $\\var{g}$ will be the product of the two numbers, divided by the greatest common divisor.

\n

The greatest common divisor of $\\var{f}$ and $\\var{g}$ is $\\var{gcd_fg}$.

\n

Therefore, the lowest common multiple will is

\n

\$\\frac{\\var{f}\\times\\var{g}}{\\var{gcd_fg}}=\\var{lcm_fg}\\text{.}\$

\n

#### c)

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$\\var{d}$ is a multiple of $\\var{c}$, as $\\var{d/c}\\times\\var{c}=\\var{d}.$

\n

The lowest common multiple of $\\var{c}$ and $\\var{d}$ will therefore be $\\var{d/c} \\times \\var{c} = 1 \\times \\var{d} = \\var{d}$.

", "statement": "

The lowest common multiple of two numbers is the first number which appears in both numbers' times tables.

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What is the lowest common multiple of $\\var{a}$ and $\\var{b}$?

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What is the lowest common multiple of $\\var{f}$ and $\\var{g}$?

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What is the lowest common multiple of $\\var{c}$ and $\\var{d}$?

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Prime

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Composite

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A number that only has two factors - itself and 1 - is known as a prime number.

\n

A number that can be divided without remainder by numbers other than itself and 1 is known as a composite number.

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$\\displaystyle\\var{b}$

", "

$\\displaystyle\\var{k}$

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$\\displaystyle\\var{f}$

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$\\displaystyle\\var{a}$

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$\\displaystyle\\var{d}$

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$\\displaystyle\\var{h}$

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$\\displaystyle\\var{j}$

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\n

The numbers and their factors are given below:

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 Number Prime/Composite Factors $\\var{b}$ Prime $1$, $\\var{b}$ $\\var{k}$ Prime $1$, $\\var{k}$ $\\var{f}$ Composite $1$, $2$, $\\var{f/2}$, $\\var{f}$ $\\var{a}$ Prime $1$, $\\var{a}$ $\\var{d}$ Composite $1$, $\\var{sqrtd}$, $\\var{d}$ $\\var{h}$ Composite $\\var{latex(hlist)}$ $\\var{j}$ Prime $1$, $\\var{j}$
\n

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Identify which of the following are prime numbers.

Sort a list of numbers into \"prime\" or \"composite\".

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This question tests the student's ability to identify the factors of some composite numbers and the highest common factors of two numbers.

#### a)

\n

i)

\n

$\\var{fourfac}$ has four factors: $1$, $3$, $\\var{fourfac/3}$ and $\\var{fourfac}$.

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It is possible to pair the factors up to prove that they are factors.

\n

\\\begin{align} 1\\times\\var{fourfac}&=\\var{fourfac}\\text{.}\\\\ 3\\times\\var{fourfac/3}&=\\var{fourfac}\\text{.}\\\\ \\end{align} \

\n

ii)

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$\\var{sixfac}$ has six factors: $1$, $2$, $3$, $\\var{sixfac/3}$, $\\var{sixfac/2}$ and $\\var{sixfac}$.

\n

Again, it is possible to pair the factors up to prove that they are factors.

\n

\\\begin{align} 1\\times\\var{sixfac}&=\\var{sixfac}\\text{.}\\\\ 2\\times\\var{sixfac/2}&=\\var{sixfac}\\text{.}\\\\ 3\\times\\var{sixfac/3}&=\\var{sixfac}\\text{.}\\\\ \\end{align} \

\n

\n\n

#### b)

\n

We now look for common factors between the two lists of factors, and the highest common factor will be the largest of these.

\n

\n

For $\\var{fourfac}$ and $\\var{sixfac}$, the highest common factor is $\\var{hc}$.

\n

#### c)

\n

Dividing both the numerator and denominator by the highest common factor gives:

\n

\$\\frac{\\var{sixfac}}{\\var{fourfac}} = \\frac{\\frac{\\var{sixfac}}{\\var{hc}}}{\\frac{\\var{fourfac}}{\\var{hc}}} = \\frac{\\var{sixfac/hc}}{\\var{fourfac/hc}}\\text{.}\$

\n

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Simplify $\\var{sixfac}/\\var{fourfac}$ by finding the highest common factor.

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Identify the factors of the following numbers in ascending order.

\n

i)     Factors of $\\var{fourfac}$:            $1$, [[4]], [[5]], $\\var{fourfac}$

\n

\n

ii)     Factors of $\\var{sixfac}$:          [[0]], $2$, [[1]], [[2]], $\\var{sixfac/2}$, [[3]]

What is the heighest common factor of $\\var{fourfac}$ and $\\var{sixfac}$?

\n

The heighest common factor is [[0]]

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If the same number appears in a list of factors for two numbers, it is a common factor. The largest of these common factors is the highest common factor.

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Use the result above to reduce the following fraction to its simplest form.

\n

$\\displaystyle \\frac{\\var{sixfac}}{\\var{fourfac}} =$ [[0]] [[1]]

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To simplify the fraction, divide both the numerator and denominator by the highest common factor.

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