Factors and Multiples: Chapter Test

Prime and composite are settled by a count of DISTINCT factors: exactly two makes a number prime, more than two makes it composite.

$\text{factors of } 1: \; 1 \qquad \text{factors of } 2: \; 1, 2 \qquad \text{factors of } 9: \; 1, 3, 9$

The list for $1$ holds one number, which is neither exactly two nor more than two, so $1$ meets neither definition. The number $2$ has exactly two factors and is prime, its evenness costing it nothing because the factor $2$ is the number itself rather than a third factor. And $9$ is odd with three factors, so odd numbers can be composite.

Every place value is one more than a multiple of $9$, so each column hands over a multiple of $9$ and leaves one copy of its digit behind. Adding the digits collects those leftovers, and the verdict rests on that sum.

$5 + 3 + 8 + 2 = 18 = 9 \times 2$

So $5{,}382$ is divisible by $9$, and dividing confirms it: $5{,}382 = 9 \times 598$. The other three give digit sums $15$, $16$ and $17$, and none of those is a multiple of $9$.

Divide by the smallest prime that fits at each step. The number is even three times over.

$968 \div 2 = 484, \quad 484 \div 2 = 242, \quad 242 \div 2 = 121$

Now $121$ is odd and its digits add to $4$, so $2$ and $3$ are finished, and it does not end in $0$ or $5$. The next prime that fits is $11$, and $121 \div 11 = 11$, which is prime.

$968 = 2 \times 2 \times 2 \times 11 \times 11 = 2^3 \times 11^2$

Write each number in prime-power form and go prime by prime.

$75 = 3 \times 5^2, \qquad 120 = 2^3 \times 3 \times 5$

A common factor can carry no more copies of a prime than the number that owns fewer, so take the lowest power of each SHARED prime. The prime $2$ is missing from $75$ and drops out; the prime $3$ appears once in each; the prime $5$ appears twice in $75$ but once in $120$.

$\operatorname{GCF}(75, 120) = 3 \times 5 = 15$

As a check, $75 \div 15 = 5$ and $120 \div 15 = 8$, both whole.

The test for $4$ reads the number made by the last two digits, because $100 = 4 \times 25$; the test for $8$ reads the last three, because $1000 = 8 \times 125$.

$24 = 4 \times 6, \qquad 124 = (8 \times 15) + 4$

So $5{,}124$ clears the first test and fails the second. Of the others, $2{,}946$ fails the test for $4$ already, since $46$ is not a multiple of $4$, while $7{,}352$ and $6{,}488$ pass both tests, because $352 = 8 \times 44$ and $488 = 8 \times 61$.

Screen with the quick tests first, then divide.

$355 = 5 \times 71, \qquad 357 = 3 \times 119, \qquad 371 = 7 \times 53$

That leaves $349$. It is odd, its digit sum $16$ is not a multiple of $3$, and it does not end in $0$ or $5$. Dividing by $7$, $11$, $13$ and $17$ leaves a remainder every time, and the next candidate is $19$, where $19 \times 19 = 361$ has already passed $349$. There is no factor pair left to find, so $349$ is prime.

Every pack takes an equal share of each color, so the number of packs divides both totals, and as many packs as possible means the greatest of those common divisors.

$102 = 2 \times 3 \times 17, \qquad 170 = 2 \times 5 \times 17$

The primes shared by both are $2$ and $17$, each appearing once in each number.

$\operatorname{GCF}(102, 170) = 2 \times 17 = 34$

So there are $34$ packs, each holding $102 \div 34 = 3$ badges of the first color and $170 \div 34 = 5$ of the second.

A tree is finished exactly when every leaf is prime. In the second list the leaf $9$ is not: its digits add to $9$, so $3$ divides it.

$9 = 3 \times 3, \qquad 2 \times 9 \times 2 \times 13 = 468$

Splitting that leaf turns the second list into $2$, $3$, $3$, $2$, $13$, which is the first list in a different order.

$468 = 2^2 \times 3^2 \times 13$

A whole number greater than $1$ has one prime factorization apart from the order of its factors, so two finished trees can never disagree. Where a mismatch shows up, one of the trees has been stopped a branch early, which is exactly what the leaf $9$ records here.

Split the number at the thousands column, into the part built from thousands and above and the number made by the last three digits.

$N = (1000 \times m) + L, \qquad 1000 = 8 \times 125$

The first piece is $8 \times 125 \times m$ whatever $m$ is, so it leaves remainder $0$ and cannot affect the verdict. The whole question therefore rests on $L$, the last three digits. The same split one column earlier, where $100 = 4 \times 25$, is why the test for $4$ needs only two digits.

For two numbers, the greatest common factor and the least common multiple take the lower and the higher power of every prime, which is that pair of counts over again, so together they multiply back to the numbers multiplied together.

$\operatorname{GCF}(56, 70) \times \operatorname{LCM}(56, 70) = 56 \times 70 = 3{,}920$

Dividing the product by the greatest common factor leaves the least common multiple.

$\operatorname{LCM}(56, 70) = 3{,}920 \div 14 = 280$

The prime powers agree: $56 = 2^3 \times 7$ and $70 = 2 \times 5 \times 7$, so the highest powers give $2^3 \times 5 \times 7 = 280$.

Write all three in prime-power form and take the highest power of every prime that appears anywhere, because a common multiple must hold enough copies for each number on its own.

$18 = 2 \times 3^2, \qquad 30 = 2 \times 3 \times 5, \qquad 50 = 2 \times 5^2$

The most copies of $2$ any one of them holds is one, the most copies of $3$ is two, in $18$, and the most copies of $5$ is two, in $50$.

$\operatorname{LCM}(18, 30, 50) = 2 \times 3^2 \times 5^2 = 450$

Each number divides it: $450 \div 18 = 25$, $450 \div 30 = 15$ and $450 \div 50 = 9$.

Passing two tests puts both divisors inside the number, but it delivers their product only when the two share no factor above $1$. Otherwise the shared part is counted twice and the claim overshoots.

$5 \times 9 = 45, \qquad 5 \text{ and } 9 \text{ share no factor above } 1$

So a number carrying a factor $5$ and a factor $9$ carries them without overlap and is a multiple of $45$, and every multiple of $45$ passes both tests, which is the other direction. Each of the other three pairings shares a factor: $24$ passes the tests for $2$ and $8$, $12$ passes the tests for $3$ and $6$, and $20$ passes the tests for $4$ and $10$, yet none of the three is a multiple of the product claimed.

The failed tests rule out the divisors they name and nothing more. The search still owes every prime whose square does not pass $493$, which runs $7$, $11$, $13$, $17$ and $19$, since $19 \times 19 = 361$ is below $493$ while $23 \times 23 = 529$ is above it.

$493 \div 17 = 29, \qquad \text{so } 493 = 17 \times 29$

The divisions by $7$, $11$ and $13$ each leave a remainder, but $17$ goes in exactly, so a third factor exists and the number is composite. Failing every taught test is not the same as being prime: those tests reach only as far as $10$.

Each job falls on the multiples of its own cycle, so they coincide on a common multiple and NEXT coincide on the smallest one.

$15 = 3 \times 5, \qquad 40 = 2^3 \times 5$

Take the highest power of every prime that appears in either.

$\operatorname{LCM}(15, 40) = 2^3 \times 3 \times 5 = 120$

So $120$ days pass, which is $120 \div 15 = 8$ deep cleans and $120 \div 40 = 3$ stock counts, both whole counts as they must be.

Divide by the smallest prime that fits, over and over, until the quotient reaches $1$.

$1{,}176 \div 2 = 588, \quad 588 \div 2 = 294, \quad 294 \div 2 = 147$

Now $147$ is odd, so $2$ is finished, and its digits add to $12$, so $3$ fits: $147 \div 3 = 49$. The quotient $49$ is $7 \times 7$, and dividing twice by $7$ reaches $1$.

$1{,}176 = 2^3 \times 3 \times 7^2$

Multiplying back, $8 \times 3 \times 49 = 1{,}176$.

Take the lowest power of each shared prime for one answer and the highest power of every prime for the other.

$\operatorname{GCF}(a, b) = 2^2 \times 5 = 20, \qquad \operatorname{LCM}(a, b) = 2^3 \times 5^2 \times 7 = 1{,}400$

Multiplying them gives $20 \times 1{,}400 = 28{,}000$. Between them the two answers take the lower and the higher count of every prime, which is the pair of counts over again, so the result has to be $a$ times $b$.

$a \times b = 100 \times 280 = 28{,}000$

Every place value sits one above a string of nines.

$10 = 9 + 1, \qquad 100 = 99 + 1, \qquad 1000 = 999 + 1$

A digit $b$ in the tens column therefore contributes $b \times 10 = (b \times 9) + b$: a multiple of $9$, and one loose copy of $b$. Every column behaves that way, so the number is a multiple of $9$ plus the sum of its digits, and only that sum can decide the verdict.

$726 = (7 \times 99) + (2 \times 9) + (7 + 2 + 6)$

Here the digit sum is $15$, which is not a multiple of $9$, so $726$ is not either, although it is a multiple of $3$.

Every whole number greater than $1$ has exactly one prime factorization, apart from the order in which the factors are written. Two numbers with the same factorization are therefore the same product, and so the same number.

$2^2 \times 5 = 20, \qquad 2 \times 5^2 = 50$

Those two are built from the same primes and are not equal, so sharing the primes is not enough: the COUNTS have to match as well. Order, by contrast, is set aside by the theorem, so $2 \times 3 \times 7$ and $7 \times 3 \times 2$ are one factorization. And the theorem starts at $2$, because $1$ is not prime and so is not a product of primes at all.

The rule for two numbers works because a pair of counts holds nothing but a lower one and a higher one, so the two answers use the pair up. A third number brings a middle count that neither answer ever reaches for, so the rule survives only when there is nothing in the middle to lose, and that happens when no two of the three share a prime.

$4 = 2^2, \qquad 9 = 3^2, \qquad 25 = 5^2$

No prime appears in more than one of them, so the greatest common factor is $1$ and the least common multiple is everything at once.

$1 \times (2^2 \times 3^2 \times 5^2) = 900 = 4 \times 9 \times 25$

In each of the other three sets some prime is held by two of the numbers, and the copies in the middle go uncounted.

Peel off the small primes first. The number is even, and even again once halved, and the next quotient ends in $5$.

$2{,}860 \div 2 = 1{,}430, \quad 1{,}430 \div 2 = 715, \quad 715 \div 5 = 143$

Now $143$ is odd, its digits add to $8$ and it does not end in $0$ or $5$, so those tests all fail and the search moves up. Dividing by $7$ leaves a remainder, and dividing by $11$ does not: $143 \div 11 = 13$, which is prime.

$2{,}860 = 2^2 \times 5 \times 11 \times 13$