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Divisibility Rules: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 The counter setting

A counter reads $6{,}478$ . What is the smallest positive whole number that can be added to make its reading divisible by $10$ ?
1. Hint 1
  Look at what the ones digit must be when a number is a multiple of $10$ .
2. Hint 2
  Count forward from the current reading to the next number ending in $0$ .
Answer

$2$ .

Full solution

A multiple of $10$ ends in $0$ .
The readings immediately after $6{,}478$ are $6{,}479$ and $6{,}480$ , so the second is the first suitable reading.
$6{,}480-6{,}478=2$

Adding just $1$ leaves a last digit of $9$ , so a smaller positive increase does not work.

Answer

$2$ .

Key idea

The ones digit alone decides how much must be added to reach a multiple of ten.
Problem 1: Mark complete
Problem 2 A repeated digit

A five-digit whole number has $7$ in every place. Is it divisible by $3$ ?
1. Hint 1
  Divisibility by $3$ depends on the sum of all the digits.
2. Hint 2
  Add five copies of $7$ , then check whether that sum is a multiple of $3$ .
Answer

No.

Full solution

The five digits contribute the following sum.
$5\times7=35$

The sum is not divisible by $3$ .
$35=3\times11+2$
Therefore the original number, $77{,}777$ , is not divisible by $3$ .

Answer

No.

Key idea

Repeated digits contribute repeated copies to the digit sum.
Problem 2: Mark complete
Problem 3 A repeating label

A label is made by writing the digits $2$ , $4$ , $8$ , and then $2$ , $4$ , $8$ again, in that order. Is the resulting whole number divisible by $8$ ?
1. Hint 1
  Only the number formed by the last three digits matters for divisibility by $8$ .
2. Hint 2
  Compare that final block with $8\times30$ and account for what remains.
Answer

Yes.

Full solution

The label is $248{,}248$ , whose last three digits form $248$ .
$248=8\times31$
Because that block is a multiple of $8$ , the whole number is also a multiple of $8$ .

The thousands part contributes no remainder.
$248{,}000=8\times31{,}000$

Answer

Yes.

Key idea

For divisibility by eight, a long number can be reduced to its final three-digit block.
Problem 3: Mark complete
Problem 4 The tray order

A workshop has $59$ trays with $37$ plugs on each tray. What is the smallest number of extra plugs needed so that all the plugs can be packed into bags of $6$ with none left over?
1. Hint 1
  Find the total first, then look for the next total that can fill bags of $6$ .
2. Hint 2
  A suitable total must be even and have a digit sum divisible by $3$ .
Answer

$1$ plug.

Full solution

The trays contain this many plugs.
$59\times37=2{,}183$
This number is odd, so it cannot be a multiple of $6$ .

Try adding one plug.
$2{,}183+1=2{,}184$
The new number is even, and its digit sum is $15$ , a multiple of $3$ .
Passing both tests gives a multiple of $6$ , since $2$ and $3$ share no factor above $1$ .

The bags can therefore be filled exactly.
$2{,}184\div6=364$
No extra plugs fails, and one works, so one is the minimum.

Answer

$1$ plug.

Key idea

A multiple of six must pass both the even-number test and the digit-sum test for three.
Problem 4: Mark complete
Problem 5 The two deliveries

Two deliveries contain $1{,}460$ bolts and $2{,}340$ bolts, to be packed in bags of $8$ . Can either delivery be packed with none left over on its own? Can the two together, and if so, into how many bags? Is $8$ a factor of the combined total?
1. Hint 1
  Check the final three digits of each delivery, then check the combined total.
2. Hint 2
  When adding the deliveries, their leftovers may complete another bag.
Answer

Neither on its own; together, yes, in $475$ bags; yes, $8$ is a factor of $3{,}800$ .

Full solution

Each separate delivery fails the test for $8$ .
Its last three digits leave $4$ over.
$460=8\times57+4$ $340=8\times42+4$
The thousands parts are divisible by $8$ , so neither whole delivery fills bags exactly.

Combine the deliveries.
$1{,}460+2{,}340=3{,}800$ $3{,}800\div8=475$
The two leftovers of $4$ together fill a bag, and the combined total fills $475$ bags exactly.

Check the count.
$475\times8=3{,}800$

The combined division leaves remainder $0$ , which is what divisible means.
So $8$ is a factor of the combined total $3{,}800$ , and the same fact says $3{,}800$ is a multiple of $8$ .

Answer

Neither on its own; together, yes, in $475$ bags; yes, $8$ is a factor of $3{,}800$ .

Key idea

Two totals that each leave something over can together form an exact multiple.
Problem 5: Mark complete
Problem 6 The number cards

Four cards show $7$ , $4$ , $1$ , and $6$ . Use every card exactly once to make the greatest four-digit number divisible by $4$ . What number do you make?
1. Hint 1
  The final two cards must form a multiple of $4$ .
2. Hint 2
  The last card must be $4$ or $6$ ; list the endings before arranging the rest.
Answer

$7{,}416$ .

Full solution

The last card must be even, so it is $4$ or $6$ .
With these cards, the endings divisible by $4$ are $16$ , $64$ and $76$ .
The endings $14$ , $74$ and $46$ are even, but each leaves remainder $2$ , so they fail.

For each passing ending, put the other two cards in decreasing order in front.
The ending $16$ gives $7{,}416$ , the ending $64$ gives $7{,}164$ , and the ending $76$ gives $4{,}176$ .
The greatest of the three is $7{,}416$ .

The three arrangements larger than it, $7{,}641$ , $7{,}614$ and $7{,}461$ , end in $41$ , $14$ and $61$ , and none of those is a multiple of $4$ .

Its final two digits confirm divisibility.
$16=4\times4$
Dividing checks the whole number.
$7{,}416=4\times1{,}854$

Answer

$7{,}416$ .

Key idea

A condition on the last two digits can be settled before maximizing the remaining places.
Problem 6: Mark complete
Problem 7 The battery packs

A store sells batteries in sealed packs of $2$ . A clerk says that whenever the store's total number of batteries is a multiple of $4$ , the packs can be put in boxes of four packs ( $8$ batteries), with no pack opened or left out. The clerk's reason: the packs supply the $2$ , the total supplies the $4$ , and $2\times4=8$ . Is the clerk right? Explain.
1. Hint 1
  Each box takes $4$ packs, so boxing works exactly when the number of packs is a multiple of $4$ .
2. Hint 2
  Try a total that is a multiple of $4$ but not of $8$ , and count the packs it makes.
3. Hint 3
  Coming in packs of $2$ and being a multiple of $4$ are two tests whose divisors share the factor $2$ .
Answer

No; for example, $44$ batteries ( $22$ packs) cannot be boxed this way. Any multiple of $4$ that is not a multiple of $8$ , such as $12$ or $20$ , also shows it.

Full solution

Each box holds $4$ packs of $2$ .
$4\times2=8$
So the packs can be boxed with none left out exactly when the number of packs is a multiple of $4$ , which is when the total is a multiple of $8$ .

Test a total of $44$ batteries, which is a multiple of $4$ .
$44=4\times11$
It makes $22$ packs.
$44\div2=22$
Five boxes take $20$ packs and leave $2$ packs, or $4$ batteries, over.
$44=8\times5+4$
So the clerk's claim fails for $44$ batteries.

The clerk has combined two tests whose divisors share a factor above $1$ .
Coming in packs of $2$ passes the test for $2$ , and the total passes the test for $4$ .
Since $2$ and $4$ share the factor $2$ , passing both does not guarantee that the total is a multiple of $8$ .

Answer

No; for example, $44$ batteries ( $22$ packs) cannot be boxed this way. Any multiple of $4$ that is not a multiple of $8$ , such as $12$ or $20$ , also shows it.

Key idea

Passing two tests whose divisors share a factor above $1$ does not guarantee divisibility by their product.
Problem 7: Mark complete
Problem 8 The final digit

A student says $5{,}742$ is not divisible by $9$ , because its last digit, $2$ , is not a multiple of $9$ . Is the student's conclusion right? Is the reasoning sound? Explain.
1. Hint 1
  Ask which part of a number the test for $9$ actually reads, and run that test on $5{,}742$ .
2. Hint 2
  Split $5{,}742$ into $5{,}740+2$ and ask whether the $5{,}740$ part is sure to be a multiple of $9$ .
3. Hint 3
  Compare two numbers with the same last digit, such as $18$ and $28$ , under the test for $9$ .
Answer

The conclusion is wrong: $5{,}742$ is divisible by $9$ ( $5{,}742=9\times638$ ). The reasoning is not sound either.

Full solution

The test for $9$ reads the digit sum, not the last digit.
$5+7+4+2=18$
The sum is a multiple of $9$ .
$18=9\times2$
So $5{,}742$ is divisible by $9$ , and dividing confirms it.
$5{,}742=9\times638$
The student's conclusion is wrong.

The last digit settles $2$ , $5$ and $10$ because everything above it is a multiple of $10$ , and each of those divisors divides $10$ .
For $9$ that fails.
$10=9+1$
So the part $5{,}740$ is a multiple of $10$ but need not be a multiple of $9$ , and here it is not.
$5{,}740=9\times637+7$
The last digit alone cannot decide.

For example, $18$ and $28$ both end in $8$ , yet only $18$ is divisible by $9$ .
So the student's reasoning is not sound.

Answer

The conclusion is wrong: $5{,}742$ is divisible by $9$ ( $5{,}742=9\times638$ ). The reasoning is not sound either.

Key idea

The last digit settles $2$ , $5$ and $10$ because each divides $10$ ; $9$ does not divide $10$ , so the last digit cannot settle divisibility by $9$ .
Problem 8: Mark complete
Problem 9 The inserted zero

Choose any whole number written in digits, and insert a single $0$ somewhere in its written form. A student says the old and new numbers either both pass or both fail the test for $3$ , and the same is true for $9$ . Is the student correct? Explain.
1. Hint 1
  Compare the sum of the digits before and after inserting the zero.
2. Hint 2
  Each place value is one more than a multiple of $9$ , so moving a digit to another place still leaves one copy of that digit in the test.
Answer

Yes; the student is correct for both $3$ and $9$ .

Full solution

Inserting $0$ does not change the digit sum.
It contributes nothing, and all other digits still occur the same number of times.

Each place value contributes a multiple of $9$ plus $1$ .
$10=9+1$ $100=99+1$ $1{,}000=999+1$
Therefore each digit contributes a multiple of $9$ plus one copy of itself, regardless of which place it occupies.
The remaining copies add to the digit sum.

The old and new numbers consequently have the same test result for $9$ .
Each of $9$ , $99$ and $999$ is also a multiple of $3$ , so the same split shows each number is a multiple of $3$ plus its digit sum, and the result for $3$ is unchanged too.

Answer

Yes; the student is correct for both $3$ and $9$ .

Key idea

Inserting a zero changes place values but preserves the digit sum and the tests for three and nine.
Problem 9: Mark complete
Problem 10 Three hundred more

A storeroom holds a number of parts that is divisible by $9$ . Then $300$ more parts arrive. Must the new total be divisible by $3$ ? Can it be divisible by $9$ ? Explain both answers.
1. Hint 1
  Split the new total into the old total and the $300$ that arrived, and ask what each of those two amounts leaves over on division by $3$ and by $9$ .
2. Hint 2
  A multiple of $9$ is also a multiple of $3$ , since $9=3\times3$ .
3. Hint 3
  Write $300$ as a multiple of $3$ , and then as a multiple of $9$ plus a leftover smaller than $9$ .
Answer

It must be divisible by $3$ ; it can never be divisible by $9$ .

Full solution

The old total is a multiple of $9$ , and $9=3\times3$ , so it is also a multiple of $3$ .
The $300$ added is a multiple of $3$ too.
$300=3\times100$
The sum of two multiples of $3$ is a multiple of $3$ , so the new total must be divisible by $3$ .

For $9$ , split the $300$ into $297$ and $3$ .
$297=9\times33$
The old total and the $297$ are both multiples of $9$ , so together they make a multiple of $9$ .
The new total is that multiple of $9$ plus $3$ .

Because $3$ is less than $9$ , the new total leaves remainder $3$ on division by $9$ , whatever the old total was.
So it can never be divisible by $9$ .
For example, $3{,}600=9\times400$ becomes $3{,}900$ , whose digit sum $12$ is not a multiple of $9$ .

Answer

It must be divisible by $3$ ; it can never be divisible by $9$ .

Key idea

Adding an amount to a multiple of a number gives the same remainder, on division by that number, as the amount alone.
Problem 10: Mark complete

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