Let's Investigate!
Look at the following statements about factors and multiples. Decide whether each statement is:
- Always True – true in every case.
- Sometimes True – true only for some cases.
- Never True – not true in any case.
If \(a\) divides \(M\) and \(a\) divides \(N\), then a divides \(M + N\) and \(a\) divides \(M – N\).
| Understanding the Idea | Using Algebra | Worked Examples |
|---|---|---|
| Suppose \(7\) divides two numbers exactly. These numbers can be written as \(7a\) and \(7b\). | First number \(= 7a\) Second number \(= 7b\) |
\(21\) and \(49\)
\(35\) and \(56\)
\(63\) and \(91\)
|
| Since both numbers have \(7\) as a common factor, adding or subtracting them keeps \(7\) as a common factor. |
Addition: (\(7a + 7b\) \(= 7(a+b)\))
Subtraction: (\(7a - 7b\) \(= 7(a-b)\))
|
Addition: \(21 + 49\) \(= 70 = 7 × 10\)
Subtraction: \(49 − 21\) \(= 28 = 7 × 4\)
|
If \(A\) is divisible by \(k\), then all multiples of \(A\) are divisible by \(k\).
| Understanding the Idea | Using Algebra | Worked Examples |
|---|---|---|
|
Suppose \(18\) is divisible by \(\
6\). Any multiple of \(18\) is formed by multiplying \(18\) by a whole number. Therefore, every multiple of \(18\) is also divisible by \(6\).
|
Let \(A = 6a\), where \(A\) is divisible by \(6\). Any multiple of \(A\) can be written as \(nA\). (\(nA = n(6a)=6(na)\)) Since \(6\) is still a factor, every multiple of \(A\) is divisible by \(6\). |
\(18 × 2 = 36\);
\(36 ÷ 6 = 6\)
\(18 × 5 = 90\);
\(90 ÷ 6 = 15\)
|
If \(A\) is divisible by \(k\), then \(A\) is divisible by all factors of \(k\).
| Understanding the Idea | Algebraic Reasoning | Worked Example |
|---|---|---|
| If a number is divisible by \(k\), it can be divided exactly into \(k\) equal groups. Since every factor of \(k\) divides \(k\) exactly, the same number can also be divided into groups equal to each factor of \(k\) without any remainder. |
If \(A\) is divisible by \(k\), then \(A = k × m\) for some whole number \(m\). Let \(f\) be a factor of \(k\). Then \(k = f × t\) for some whole number t.
Therefore, \(A = (f × t) × m = f × (tm)\). Since \(tm\) is a whole number, \(A\) is divisible by \(f\). Hence, \(A\) is divisible by every factor of \(k\).
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\(24\) is divisible by \(12\) because \(24 = 12 × 2\). The factors of \(12\) are \(1, 2, 3, 4, 6, 12\).
\(24 ÷ 1 = 24\)
\(24 ÷ 2 = 12\)
\(24 ÷ 3 = 8\)
\(24 ÷ 4 = 6\)
\(24 ÷ 6 = 4\)
\(24 ÷ 12 = 2\) Since \(24\) is divisible by \(12\), it is also divisible by every factor of \(12\).
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If \(A\) is divisible by \(k\) and \(A\) is also divisible by \(m\), then \(A\) is divisible by the LCM of \(k\) and \(m\).
| Understanding the Idea | Algebraic Reasoning | Worked Example |
|---|---|---|
|
If a number is divisible by both \(k\) and \(m\), it has all the factors of both numbers. However, it will be divisible by their LCM only when it contains all the prime factors of both numbers in the required highest powers. Therefore, the statement is not always true.
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If (\(A=kx=my\)), then (\(A\)) is a common multiple of (\(k\)) and (\(m\)). |
\(60\) is divisible by \(4\) and by \(6\).
\(LCM(4,6) = 12\).
\(60 ÷ 12 = 5\)
\(42\) is divisible by \(7\) and by \(14\).
\(LCM(7,14) = 14\).
\(42 ÷ 14 = 3\)
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What Remains? - Remainders:
When we divide a number by another number, two things are possible:
- The number is divided exactly, leaving no remainder.
- The number is not divided exactly, so a remainder is left.
Numbers that leave the same remainder when divided by a given number follow a fixed pattern.
| Number | Quotient | Remainder | Notice the fact |
|---|---|---|---|
| \(3\) | \(0\) | \(3\) | \(3 = 5 × 0 + 3\) |
| \(8\) | \(1\) | \(3\) | \(8 = 5 × 1 + 3\) |
| \(13\) | \(2\) | \(3\) | \(13 = 5 × 2 + 3\) |
| \(18\) | \(3\) | \(3\) | \(18 = 5 × 3 + 3\) |
| \(23\) | \(4\) | \(3\) | \(23 = 5 × 4 + 3\)
|
So, all numbers that leave a remainder 3 when divided by 5 follow the same pattern.
Algebraic Representation of Number:
If \(k\) is any whole number, then \(\boxed{5k+3}\)represents all numbers that leave a remainder of \(3\) when divided by \(5\).
Here,
- \(5k\) represents any multiple of \(5\).
- \(+3\) means we move \(3\) more than that multiple.
Another way to represent the expression:
Instead of expressing \(3\) more than a multiple of \(5\), we can write \(2\) less than the next multiple of \(5\). So, the same numbers can also be written as \(\boxed{5k-2}\) and \(k\geq1\).
Important!
If a number leaves a remainder \(r\) when divided by \(d\), then all such numbers can be written as \(dk+r\) with \(k \geq 0\) or \(dk-(d-r)\) with \(k \geq 1\).
Examples:
Find a few numbers that leave a remainder of \(1\) when divided by \(2\) and a remainder of \(1\) when divided by \(3\). Write an algebraic expression to describe all such numbers.
The numbers that leave a remainder of \(1\) when divided by \(2\) must be of the form \(2m+1\) with \(m \geq 0\).
The numbers that leave a remainder of \(1\) when divided by \(3\) must be of the form \(3n+1\) with \(n \geq 0\).
If a number leaves the same remainder \(r\) when divided by \(a\) and \(b\), then
\(\text{Number} = \text{LCM}(a,b)×k+r\)
\(\text{LCM}(2,3) = 6\)
Number \(= 6 \times k + 1\), \(k \geq 0\)
Numbers are \(1, 7, 13, 19,...\)