8.02a–8.02d - Bases, divisibility and the division algorithm | OCR A-Level Further Mathematics A

Read and calculate with numbers in different bases, use familiar divisibility tests, and build a test for an unfamiliar prime. You will also express division precisely using a quotient and a remainder, so that a calculation becomes a statement you can prove.

Place value in any base

Why is the palindrome $1221$ divisible by $11$ , and does the same pattern survive if you change the number base? The digits look innocent; their place values contain the explanation.

In base ten, $234 = 2\times100+3\times10+4$ . In base five, the same string of digits means $234_5=2\times25+3\times5+4=69_{10}$ . The subscript tells you how large a place becomes when you move one step left.

You already know

An integer power gives repeated multiplication, and $n^0=1$ . Expanding and factorising polynomials lets you work with a base that is not yet known.

Each digit must fit inside a single place: a digit equal to the base must be carried into the next place.

Base notation

In an integer base $n\ge2$ , the digits are $0,1,\ldots,n-1$ . A non-zero number has a non-zero first digit. For example,

2013_n=2n^3+n+3 \qquad (n>3)

The letters $\mathrm A,\mathrm B,\mathrm C,\mathrm D,\mathrm E,\mathrm F$ represent $10,11,12,13,14,15$ , respectively, when the base is large enough. Thus $\mathrm A_{11}=10_{10}$ , but $\mathrm B_{11}$ is not a valid digit.

The value of a digit string is its place-value expansion. To go the other way, divide repeatedly by the new base: the first remainder is the units digit, the next is the next digit, and so on. This place-value rule and conversion method are not in the formula booklet: learn them.

[DIAGRAM: asset_name: Place values in base five and base twelve; asset_slug: h245_8_02_place_values; caption: Changing the base changes the place values, not the meaning of a digit; description: Two rows of three pale boxes show 234 in base five above place values 25, 5, 1 and 2A4 in base twelve above place values 144, 12, 1. Arrows connect each digit to its contribution. The totals 69 and 412 are shown in base ten.]
Changing the base changes the place values, not the meaning of a digit

The same conversion can be checked in reverse, which is useful when several remainders look alike.

Worked example 1: converting both ways

Question. Express $234_5$ in base ten and $97_{10}$ in base five.

Plan. Expand the first number in powers of five; divide the second repeatedly by five and read its remainders upwards.

\begin{aligned} 234_5&=2\times5^2+3\times5+4 &&\text{place values}\\ &=50+15+4=69_{10} &&\text{add contributions}\\ 97&=5\times19+2 &&\text{units digit }2\\ 19&=5\times3+4 &&\text{next digit }4\\ 3&=5\times0+3 &&\text{first digit }3 \end{aligned}

Why this step? Dividing by five removes the units place and shifts every other digit one place right, so the remainders emerge from right to left.

Answer. $234_5=69_{10}$ and $97_{10}=342_5$ .

Check. $342_5=3\times25+4\times5+2=97$ , as required.

The next example changes the base to twelve. The method is identical; a remainder of ten now needs a letter.

Worked example 2: a letter is a digit

Question. Express $2\mathrm A4_{12}$ in base ten and $391_{10}$ in base twelve.

Plan. Use $\mathrm A=10$ , then apply the same place-value and repeated-division methods with twelve.

\begin{aligned} 2\mathrm A4_{12}&=2\times12^2+10\times12+4 &&\mathrm A=10\\ &=288+120+4=412_{10} &&\text{add contributions}\\ 391&=12\times32+7 &&\text{units digit }7\\ 32&=12\times2+8 &&\text{next digit }8\\ 2&=12\times0+2 &&\text{first digit }2 \end{aligned}

Why this step? A letter in a numeral records one digit, not an algebraic variable or two separate digits.

Answer. $2\mathrm A4_{12}=412_{10}$ and $391_{10}=287_{12}$ .

Check. $287_{12}=2\times144+8\times12+7=391$ . All three digits are less than twelve.

Read the subscript before attempting any arithmetic.

Watch out

Wrong: $12_5=12_{10}$ . Right: $12_5=1\times5+2=7_{10}$ . Also, $57_7$ is invalid: the digit $7$ cannot occur in base seven.

Try the two directions with a fresh pair of numbers.

Algebra with a base

When the base is unknown, a numeral is a polynomial waiting to be expanded.

For example, $23_n=2n+3$ , so $23_n=17_{10}$ gives $n=7$ . You must also check $n>3$ , because the largest digit is three. Solving the equation does not by itself make a base valid.

The same expansion reveals patterns that hold for every permitted base. If $a,b$ are digits and $a\ne0$ , the four-digit palindrome $abba_n$ has value

\begin{aligned} an^3+bn^2+bn+a &=a(n^3+1)+bn(n+1)\\ &=(n+1)\bigl(a(n^2-n+1)+bn\bigr) \end{aligned}

The bracket is an integer, so the numeral is divisible by $n+1$ . This identity is not in the formula booklet: learn how to obtain it by expansion and factorisation. In particular, $1221_{10}=11(91+20)=11\times111$ : the opening pattern is already taking shape.

An equation between two bases needs a separate expansion on each side.

Worked example 3: finding an unknown base

Question. Given that $132_n=72_{10}$ , determine $n$ .

Plan. Expand in powers of $n$ , solve the resulting quadratic, and keep only an integer base that permits every digit.

\begin{aligned} n^2+3n+2&=72 &&\text{expand }132_n\\ n^2+3n-70&=0 &&\text{collect terms}\\ (n+10)(n-7)&=0 &&\text{factorise}\\ n&=7 &&n\text{ is positive and }n>3 \end{aligned}

Why this step? The negative algebraic root is a root of the polynomial, but it cannot be a base in this setting.

Answer. $n=7$ .

Check. $132_7=49+21+2=72_{10}$ .

The second example changes one feature: the value on the right is itself written in a base involving $n$ . Expand that side too.

Worked example 4: two related bases

Question. Given that $132_n=112_{n+1}$ , determine $n$ .

Plan. Expand each numeral using its own base, then check the resulting base restrictions.

\begin{aligned} n^2+3n+2&=(n+1)^2+(n+1)+2 &&\text{expand both sides}\\ n^2+3n+2&=n^2+3n+4 &&\text{simplify} \end{aligned}

The equality would require $2=4$ , which is impossible.

Why this step? The same digit string can have different values in different bases; the base must be substituted into every power, not added to the final value.

Answer. There is no such base $n$ .

Check. At $n=4$ , the two values are $30$ and $32$ ; their difference remains two for every $n$ .

An impossible equation is a complete answer when the algebra proves that no allowed base works.

Watch out

Wrong: $132_{n+1}=132_n+1$ . Each place value changes. Right: $132_{n+1}=(n+1)^2+3(n+1)+2$ .

In a proof about a digit pattern, state that the factor left after factorising is an integer: that is what turns algebra into divisibility.

Standard divisibility tests

A divisibility test keeps just enough information to decide whether a number is an exact multiple.

Divisibility

For integers $a\ne0$ and $b$ , $a\mid b$ means that $b=ak$ for some integer $k$ . For example, $7\mid42$ , $7\mid(-14)$ and $7\mid0$ , but $7\nmid20$ .

These standard tests apply to base-ten numerals. You may use them without proof; the tests are not in the formula booklet: learn them.

Divisor	Test
$2$	The last digit is even.
$3$	The sum of the digits is divisible by $3$ .
$4$	The number formed by the last two digits is divisible by $4$ .
$5$	The last digit is $0$ or $5$ .
$8$	The number formed by the last three digits is divisible by $8$ .
$9$	The sum of the digits is divisible by $9$ .
$11$	The alternating sum of the digits is divisible by $11$ , including zero.

For $11$ , either sign convention works: $462$ has alternating sum $4-6+2=0$ , while $2728$ has $2-7+2-8=-11$ . Both are divisible by eleven. For a short numeral, the “last three digits” test simply uses the whole numeral.

To test a composite divisor, combine tests for coprime factors, or divide by one factor and test the quotient for the next. For example, divisibility by $72=8\times9$ needs both tests, since $8$ and $9$ have no common factor greater than one. Checking divisibility by $4$ and $6$ only establishes divisibility by $12$ , not necessarily by $24$ .

Here are two tests with the number held fixed, so you can see exactly what changes with the divisor.

Worked example 5: combining coprime tests

Question. Use standard tests to determine whether $5832$ is divisible by $72$ .

Plan. Test $8$ and $9$ , whose product is $72$ and which are coprime.

\begin{aligned} 832&=8\times104 &&\text{last three digits}\\ 5+8+3+2&=18=9\times2 &&\text{digit sum} \end{aligned}

Why this step? The two tests account for the complete factors $8$ and $9$ ; neither repeats a prime factor from the other.

Answer. Both tests pass, so $72\mid5832$ .

Check. $5832=72\times81$ .

The next example changes only the divisor. It now contains one more factor of two, so the previous two tests do not finish the job.

Worked example 6: repeated tests on a quotient

Question. Use standard tests to determine whether $5832$ is divisible by $144$ .

Plan. Since $144=8\times18$ , divide by eight and test the quotient for divisibility by eighteen.

\begin{aligned} 5832&=8\times729 &&\text{first exact division}\\ 7+2+9&=18 &&9\mid729\\ 729&\text{ is odd} &&2\nmid729 \end{aligned}

Why this step? Testing the quotient makes repeated prime factors count again; testing the original number for eight and eighteen would only establish divisibility by their least common multiple, seventy-two.

Answer. $18\nmid729$ , so $144\nmid5832$ .

Check. $5832=144\times40+72$ , so the division is not exact.

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8.02a–8.02d - Bases, divisibility and the division algorithm

Place value in any base

Algebra with a base

Standard divisibility tests

Continue this lesson.

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