CLASS 8 MATHS · PART 1 · CHAPTER 5 · STUDY NOTES
Number Play
Understand number play through short explanations, labelled models and original practice. Read one topic, explain it in your own words, then check.
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Letters describe every consecutive-number example
Write the first integer as n. The next ones are n + 1, n + 2 and so on.Neighbours balance around the middle
Value (number)
- Number-line marks · Only horizontal position represents the number.
- 118 · (18, 0)
- 219 · (19, 0)
- 320 · (20, 0)
The two outside numbers have average 19, so all three together have sum 3 × 19.
Read the plotted values
| Value | Height |
|---|---|
| 18 | 0 |
| 19 | 0 |
| 20 | 0 |
- Consecutive integers
- Integers differing by 1, in order.
- General expression
- A formula describing a whole family of cases.
Two consecutive integers add to n + (n + 1) = 2n + 1, always odd.
Three consecutive integers centred on m add to (m − 1) + m + (m + 1) = 3m. The answer depends on whether negative integers, zero and one-term sums are allowed.
Let’s look a little closer.
Go deeper: Specify the allowed numbers
With at least two positive consecutive terms, some totals cannot be made. Allowing negative integers changes which representations are possible. A claim about all sums must state these rules.
Take a moment. What do you think?
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Question
Three consecutive integers add to 57. Find them.
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Changing a plus to a minus keeps parity
Replacing +b by −b changes an integer expression by 2b, an even amount.Changing a sign changes two copies
- Even integer
- An integer that can be written 2k for an integer k.
- Odd integer
- An integer that can be written 2k + 1.
Integers of the same parity differ by an even number. Therefore changing signs in a ± b ± c ± d never changes the expression’s parity.
Four consecutive integers contain two even and two odd values. Their sum is even, so every sign choice has even parity.
Show a factor of 2
6p + 4q = 2(3p + 2q)For integer p and q, the bracket is an integer, so this expression is always even.
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Question
Why is n(n + 1) even for every integer n?
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Even numbers have two remainder types modulo 4
An even integer either fills groups of four exactly or leaves two units.Two leftovers complete one group
- Remainder
- For division by a positive integer d, the leftover r in n = dq + r, with 0 ≤ r < d.
The two forms are 4p and 4p + 2. Two type-2 leftovers combine into a full group of four.
Thus the sum is divisible by 4 when the two even integers have the same remainder type: 0 + 0 or 2 + 2. Mixed types leave remainder 2.
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Question
Do 14 and 22 add to a multiple of 4? Explain without full addition.
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A shared factor stays in a sum or difference
Write each number as a multiple of the shared divisor.Multiples stay on the same grid
Value (number)
- Number-line marks · Only horizontal position represents the number.
- 10 · (0, 0)
- 25 · (5, 0)
- 310 · (10, 0)
- 415 · (15, 0)
- 520 · (20, 0)
These are consecutive multiples of 5. Adding or subtracting whole jumps of 5 lands on another multiple.
Read the plotted values
| Value | Height |
|---|---|
| 0 | 0 |
| 5 | 0 |
| 10 | 0 |
| 15 | 0 |
| 20 | 0 |
- Divisibility
- d divides n when n = dk for some integer k, with d ≠ 0.
If M = da and N = db, then M + N = d(a + b) and M − N = d(a − b). The common factor d remains.
The reverse claim is false: 2 + 10 is divisible by 6, but neither summand is. If d divides A, it also divides every integer multiple of A.
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Question
If 5 divides each of two integers, must it divide their difference?
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Combine divisibility tests using the LCM
Two successful tests imply divisibility by the least common multiple, not always their product.Take enough of each prime factor
- Least common multiple
- The smallest positive integer divisible by both given positive integers.
Divisible by 4 and 6 means divisible by LCM(4,6) = 12. It does not guarantee divisibility by 24.
To test 24, use 3 and 8: their prime factors together give 3 × 2³, and LCM(3,8) = 24.
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Question
Why do the tests for 6 use 2 and 3 successfully?
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A remainder is a fixed offset from multiples
Numbers with the same remainder sit at equally spaced positions.Four past each multiple of seven
Value (number)
- Number-line marks · Only horizontal position represents the number.
- 14 · (4, 0)
- 211 · (11, 0)
- 318 · (18, 0)
- 425 · (25, 0)
Each marked number is a multiple of 7 plus 4. The spacing is 7.
Read the plotted values
| Value | Height |
|---|---|
| 4 | 0 |
| 11 | 0 |
| 18 | 0 |
| 25 | 0 |
- Remainder form
- n = dq + r, with integer q and 0 ≤ r < d for positive d.
Numbers leaving remainder 4 on division by 7 have the form 7q + 4. For nonnegative numbers, q ≥ 0 here.
When adding remainders, regroup if their sum reaches the divisor. If remainders are 5 and 4 modulo 7, their sum leaves 2 because 9 = 7 + 2.
Convert a negative offset to a remainder
−4 = (−1) × 7 + 3A remainder on division by 7 must be between 0 and 6, so −4 itself is not the remainder.
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Question
A and B leave remainders 2 and 6 on division by 7. What remainder does A − B leave?
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Last-digit tests come from powers of ten
Separate the last one, two or three digits from the earlier part.Split at the relevant place
- Decimal expansion
- Writing an integer as the sum of its digits times powers of 10.
| Divisor | Test |
|---|---|
| 2 | Units digit even |
| 5 | Units digit 0 or 5 |
| 10 | Units digit 0 |
| 4 | Last two digits divisible by 4 |
| 8 | Last three digits divisible by 8 |
For 2, 5 and 10, the earlier places are already divisible by the divisor. Only the last digit matters.
For 4, earlier places are multiples of 100, so test the last two digits. For 8, they are multiples of 1000, so test the last three.
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Question
Why is 12,316 divisible by 4?
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Digit sums work for 9 and 3 because 10 leaves 1
Each power of ten is one more than a multiple of 9, and also one more than a multiple of 3.Discard complete groups of nine
- Digit sum
- The sum of a number’s decimal digits.
Use 10 = 9 + 1 and 100 = 99 + 1. In 253, all the large multiple-of-9 parts can be removed, leaving 2 + 5 + 3 = 10.
A number is divisible by 9 exactly when its digit sum is. The same method tests 3. Reduce the digit sum again to obtain a remainder within the allowed range.
Explain the shortcut
253 = 2 × 99 + 5 × 9 + (2 + 5 + 3)The first two terms are divisible by 9. Only the final digit-sum part affects the remainder.
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Question
Check 7,146 for divisibility by 3 and by 9.
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For 11, neighbouring place values have opposite offsets
From the units place, use plus, minus, plus, minus in order.Alternate from the units place
- Alternating digit sum
- A sum of decimal digits with alternating plus and minus signs.
The offsets are 1 = 0 × 11 + 1, 10 = 1 × 11 − 1, and 100 = 9 × 11 + 1. They continue alternating.
A number is divisible by 11 exactly when the alternating sum is a multiple of 11, including zero. A negative alternating sum must be adjusted to obtain a nonnegative remainder.
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Question
Check 5,203 for divisibility by 11.
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Digital root 9 means remainder zero, not remainder nine
Keep adding digits until one digit remains. Notice the special case for multiples of 9.Do not confuse the two outputs
- Digital root
- The single digit obtained by repeatedly summing the digits of a nonnegative integer.
For a positive multiple of 9 the digital root is 9, but its remainder on division by 9 is 0. For other positive numbers the digital root equals that remainder. Zero has digital root 0.
Digital roots can expose some arithmetic errors, but passing a digital-root check does not prove an answer correct.
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Question
Find the digital root of 6,237 and its remainder on division by 9.
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Use place value and carries to solve hidden digits
One letter keeps one digit throughout, different letters have different digits, and a leading digit is not zero.A carry links the two columns
- Cryptarithm
- An arithmetic puzzle whose digits are represented by letters.
- Carry
- A group exchanged into the next place value during calculation.
Try the original puzzle A7 + 3B = 8A. Work from the units column and include a possible carry.
The tens equation is A + 3 + carry = 8. If the carry were 0, A would be 5 and the units equation would be 7 + B = 5, impossible. So carry = 1.
- 1
With carry 1, the tens column gives A + 3 + 1 = 8, so A = 4.
- 2
The units column gives 7 + B = 10 + 4, so B = 7.
- 3
Check the whole puzzle: 47 + 37 = 84.
- 4
A and B represent different digits, and no leading digit is zero.
Go deeper: Bounds help multiplication puzzles
For a two-digit product, bound the whole number first. A two-digit number multiplied by 7 must be at most 14 to stay below 100. Such bounds reduce the possible digits before you test columns.
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Question
Why must a carry in addition of two decimal digits be either 0 or 1?
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