Number Play · Lesson 1 of 4
Is This a Multiple Of?
“Explore consecutive sums, parity, multiples, mathematical claims and remainder puzzles.”
• Express consecutive integers and multiples using letters. • Explain parity patterns under sign changes and arithmetic operations. • Decide when two even numbers add to a multiple of four. • Support always, sometimes and never claims with reasoning. • Use remainders to solve grouping and number puzzles.
Imagine arranging a handful of counters into equal rows. Sometimes every counter fits; sometimes a few remain. You can answer one question by counting, but what if someone changes the number of counters? A useful rule should still tell you what happens. In this lesson, you will investigate sums, signs and equal groups, then use letters to explain why a pattern continues even when the numbers change.
Begin by trying a few small cases. Look for what stays the same, and ask whether your observation must always be true. A calculation checks one case; an explanation using groups or algebra can cover many cases at once. Unless a question says otherwise, letters in our number patterns stand for integers, and divisors are positive integers.
Sum of Consecutive Numbers
The numbers 3, 4 and 5 are consecutive integers: each is one more than the previous number. So are −2, −1, 0 and 1. If the first number is n, the next numbers are n + 1, n + 2 and so on. The letter lets us describe every starting point at once.
Anshu notices that 7 = 3 + 4, 10 = 1 + 2 + 3 + 4, and 12 = 3 + 4 + 5. The number 15 has several descriptions: 7 + 8, 4 + 5 + 6, and 1 + 2 + 3 + 4 + 5. The total stays the same although the starting number and number of terms change.
First investigate sums of at least two consecutive positive integers. This restriction matters: allowing a single term would make every positive integer an immediate answer. Later, allow zero and negative integers and see what changes.
Integers written in order with a difference of 1 between each adjacent pair. Starting at n gives n, n + 1, n + 2, … .
Problem
Write 23 as the sum of two consecutive positive integers.
- 1.Write the numbers as n and n + 1. Their sum is 2n + 1.
- 2.Set 2n + 1 = 23. Subtract 1 to obtain 2n = 22; divide by 2 to obtain n = 11.
- 3.Thus 23 = 11 + 12. In general, every odd integer has the form 2n + 1 and is a sum of two consecutive integers. For two positive terms, the odd total must be at least 3.
Problem
Write 18 as consecutive positive sums, and decide whether 8 can be written this way.
- 1.Try three terms around 6: 5 + 6 + 7 = 18. Another representation is 3 + 4 + 5 + 6 = 18. Even totals are possible.
- 2.For 8, two terms have an odd sum and cannot work. Three terms have sum 3(n + 1), so the total must be divisible by 3; 8 is not.
- 3.The smallest positive sum with four terms is 1 + 2 + 3 + 4 = 10. Five or more positive terms give an even larger sum. Therefore 8 has no such representation.
- 4.This is a complete argument for 8. It does not prove that every even number fails: 18 is a counterexample to that claim.
Try the totals from 1 to 20 using at least two consecutive positive integers. Record every representation you find. You can stop increasing the number of terms once the smallest possible sum, 1 + 2 + …, exceeds your target. Describe any pattern you notice as a conjecture until you have a reason it always works.
Problem
Write 0 and 8 as sums of consecutive integers.
- 1.Opposite integers cancel, so −2 + (−1) + 0 + 1 + 2 = 0. The shorter sum −1 + 0 + 1 also works.
- 2.For 8, use −7 + (−6) + … + (−1) + 0 + 1 + … + 7 + 8. Every pair from −7 and 7 through −1 and 1 cancels, leaving 8.
- 3.More generally, for a positive integer N, the consecutive integers from −(N − 1) to N sum to N. When N = 1, use 0 + 1 to keep at least two terms.
- 4.The earlier impossibility for 8 applied only to positive terms. Always state which terms are allowed.
Problem
Four consecutive integers have sum 34. Find them. Also describe five consecutive integers whose greatest member is p.
- 1.Write n + (n + 1) + (n + 2) + (n + 3) = 34. Collect terms: 4n + 6 = 34.
- 2.Subtract 6: 4n = 28. Divide by 4: n = 7. The numbers are 7, 8, 9, 10. Check: 7 + 8 + 9 + 10 = 34.
- 3.If p is greatest, count backwards by one each time. In increasing order the five numbers are p − 4, p − 3, p − 2, p − 1, p.
Now keep four consecutive integers in their original order and place either + or − in each of the three gaps. The first term stays positive as written. Each gap offers two choices, so there are 2 × 2 × 2 = 8 sign patterns. Work systematically by fixing the first sign, then the second, then changing the third.
| Sign pattern | For 3, 4, 5, 6 | For n, n + 1, n + 2, n + 3 |
|---|---|---|
| + + + | 18 | 4n + 6 |
| + + − | 6 | 2n |
| + − + | 8 | 2n + 2 |
| + − − | −4 | −4 |
| − + + | 10 | 2n + 4 |
| − + − | −2 | −2 |
| − − + | 0 | 0 |
| − − − | −12 | −2n − 6 |
Every result in the last column has a factor of 2. The results −4, −2 and 0 appear for every choice of starting integer. For example, n + (n + 1) − (n + 2) − (n + 3) simplifies to −4 because all four n terms cancel.
There is also a shorter explanation of the parity pattern. Four consecutive integers contain two even and two odd numbers, so their all-plus sum is even. Changing a term from +b to −b changes the total by −2b, an even amount. Changing −b to +b changes it by +2b. Adding or subtracting an even number never changes parity.
Whether an integer is even or odd. An even integer has the form 2k; an odd integer has the form 2k + 1, where k is an integer. Zero and negative multiples of 2 are even.
This reasoning works for any fixed list of integers, not only four consecutive ones. For 1, 2, 4 and 6 the all-plus total is 13, which is odd; all eight sign patterns therefore give odd results. The common parity depends on the chosen numbers. The result is not always even.
A token model gives the same explanation: replacing b positive tokens by b negative tokens changes their value by two for each token. There are b such changes if b is positive, so the overall change is −2b. Algebra extends the argument to negative and zero values too. The number of tokens may change in a model, but the value changes by an even integer.
Breaking Even
You often do not need an exact answer to know its parity. The sum 43 + 37 is even because two odd quantities can pair their leftover ones. The difference 708 − 477 is odd because removing an odd quantity from an even one leaves odd parity.
For multiplication, one even factor supplies a factor of 2 to the whole product. A product of odd factors stays odd. A positive whole-number power repeats the same factor, so it has the same parity as its base. We do not include exponent zero in this statement: for example, 2⁰ = 1.
| Operation | Result |
|---|---|
| even ± even | even |
| odd ± odd | even |
| even ± odd or odd ± even | odd |
| even × any integer | even |
| odd × odd | odd |
| positive integer power of an even / odd base | even / odd respectively |
Problem
Determine the parity of 672 − 348, 4 × 347 × 3, 809 + 214, 119 × 303, 543 − 479 and 513³.
- 1.672 and 348 are even, so their difference is even. The product with factor 4 is even.
- 2.809 is odd and 214 is even, so their sum is odd. Both 119 and 303 are odd, so their product is odd.
- 3.543 and 479 are odd, so their difference is even. The cube 513³ multiplies three odd factors and is odd.
To decide whether an algebraic expression is always even, look for a factor of 2 that remains for every integer input. Trying several even answers is useful for guessing, but it does not establish an always claim. To disprove an always claim, one valid odd result is enough.
For two independently chosen numbers, use separate letters, such as 2m and 2n + 1. Writing 2n and 2n + 1 with the same n describes neighbouring integers only; it does not describe every possible even–odd pair.
Problem
Explain why 4m + 2q is even for all integers m and q.
- 1.Factor out 2: 4m + 2q = 2(2m + q).
- 2.The expression 2m + q is an integer because sums and products of integers are integers. Therefore the whole expression is twice an integer.
- 3.If m = 4 and q = −9, the result is 16 − 18 = −2. Negative even answers still satisfy the rule.
Problem
Is x² + 2 always even? What about b² + 1?
- 1.If x = 6, then x² + 2 = 38, which is even. But x = 3 gives 11, which is odd. Therefore it is only sometimes even.
- 2.Squaring preserves the parity of x. Adding 2 also preserves parity. Thus x² + 2 is even exactly when x is even.
- 3.For b² + 1, adding 1 reverses the parity of b². It is even when b is odd and odd when b is even: 3² + 1 = 10, whereas 2² + 1 = 5.
| Expression | When is it even? | Reason |
|---|---|---|
| 2a + 2b | Always | 2(a + b) |
| 3g + 5h | When g and h have the same parity | Odd coefficients preserve each input’s parity |
| 4m + 2n | Always | 2(2m + n) |
| 2u − 4v | Always | 2(u − 2v) |
| 13k − 5k | Always | 8k = 2(4k) |
| 6m − 3n | Exactly when n is even | 6m is even; 3n has the parity of n |
| x² + 2 | Exactly when x is even | Squaring and adding 2 preserve parity |
| b² + 1 | Exactly when b is odd | Adding 1 reverses parity |
| 4k × 3j | Always | 12kj = 2(6kj) |
Pairs to Make Fours
Two even numbers always have an even sum, but is that enough to make a multiple of 4? Compare 6 + 10 = 16 with 6 + 8 = 14. Both pairs contain even numbers, yet only the first sum is divisible by 4.
Arrange an even number into groups of four. Its remainder cannot be 1 or 3, because a multiple of four is even and an odd remainder would make the total odd. The only possibilities are 0 and 2. Thus an even number is either 4p or 4p + 2.
Adding two numbers now means combining their complete groups and their leftovers. Complete groups cause no difficulty. Whether an extra group can be made depends on the two remainders.
Problem
Explain why 12 + 16 and 6 + 10 are multiples of 4.
- 1.12 = 4 × 3 and 16 = 4 × 4. Combining them gives seven groups: 12 + 16 = 4(3 + 4) = 28.
- 2.6 = 4 × 1 + 2 and 10 = 4 × 2 + 2. The two leftover pairs make a new group of four.
- 3.Hence 6 + 10 = 4(1 + 2 + 1) = 16. Both remainder types match: either 0 with 0, or 2 with 2.
Problem
Will 20 + 14 be divisible by 4?
- 1.20 = 4 × 5 has no leftover. 14 = 4 × 3 + 2 has two leftovers.
- 2.The sum is 4(5 + 3) + 2 = 34. It still leaves remainder 2.
- 3.Therefore it is even but not a multiple of 4. A multiple of 4 plus an even nonmultiple of 4 always behaves this way.
Always, Sometimes, or Never
An always claim needs a reason that covers every allowed case. A sometimes claim needs at least one case where it works and one where it fails. A never claim needs a reason that excludes every case. These are different jobs: a long list of successful trials is not a proof of an always claim.
For a nonzero integer d, saying that d divides N means N = dk for some integer k. We also say N is a multiple of d and d is a factor of N. Grouping pictures and factored expressions both make this relationship visible.
Problem
If 8 divides M and N, must it divide their sum and difference?
- 1.Write M = 8a and N = 8b, where a and b are integers.
- 2.Then M + N = 8(a + b), and M − N = 8(a − b). Both bracketed expressions are integers.
- 3.Both results are multiples of 8. For example, 56 + 16 = 72 and 56 − 16 = 40. Reversing the subtraction gives −40, also a multiple of 8.
- 4.The same argument works for any positive divisor d: da ± db = d(a ± b).
Problem
If a sum is divisible by 8, must both addends be divisible by 8?
- 1.72 = 48 + 24 is a successful case: both addends are divisible by 8.
- 2.But 72 = 50 + 22 is another decomposition, and neither addend is divisible by 8.
- 3.The claim is sometimes true for a chosen pair. Divisibility of the sum does not force divisibility of each part.
A multiple of a multiple keeps the original factor. If A = 7j, then mA = 7(jm), so every integer multiple of A is divisible by 7. Also, a multiple of 12 is divisible by every factor of 12: 1, 2, 3, 4, 6 and 12. For instance, 12m = 3(4m) shows the factor 3.
Do not reverse that reasoning. Being divisible by 7 does not force divisibility by every multiple of 7. The number 42 is divisible by 14 but not by 28. More precisely, 7k is divisible by 7m exactly when m divides k, for nonzero m.
When two divisibility conditions hold, the guaranteed divisor is their least common multiple, or LCM. This is the smallest positive number containing all prime factors needed by both divisors, with each factor included often enough. Multiplying the divisors can count shared factors twice.
Problem
Classify: divisible by 9 and 4 implies divisible by 36; divisible by 6 and 4 implies divisible by 24.
- 1.9 = 3² and 4 = 2². Together they require 2² × 3² = 36, so the first statement is always true.
- 2.6 = 2 × 3 and 4 = 2². Together they require only 2² × 3 = 12. The factor 2 in 6 is already included among the two factors in 4.
- 3.12 is divisible by 6 and 4 but not by 24, whereas 24 satisfies all three conditions. The second statement is sometimes true.
Problem
Can the sum of an even integer and an odd integer be a multiple of 6?
- 1.Write the sum as 2m + (2n + 1) = 2(m + n) + 1. It is odd.
- 2.Every multiple of 6 has form 6k = 2(3k) and is even.
- 3.An integer cannot be both even and odd, so the sum is never a multiple of 6.
| Claim | Classification | Example or reason |
|---|---|---|
| Sum of two even integers is a multiple of 3 | Sometimes | 2 + 4 = 6; 2 + 6 = 8 |
| Not divisible by 18 implies not divisible by 9 | Sometimes | 30 satisfies both nondivisibility conditions; 27 is divisible by 9 but not 18 |
| Two nonmultiples of 6 have a nonmultiple sum | Sometimes | 9 + 11 = 20; 8 + 10 = 18 |
| A multiple of 6 plus a multiple of 9 is a multiple of 3 | Always | 6a + 9b = 3(2a + 3b) |
| A multiple of 6 plus a multiple of 3 is a multiple of 9 | Sometimes | 18 + 9 = 27; 12 + 9 = 21 |
The table’s sometimes statements can also be analysed algebraically. Two even numbers sum to 2(a + b), which is divisible by 3 precisely when a + b is. Two nonmultiples of 6 can leave remainders that add to 6, such as 2 and 4, or fail to do so. Finally, 6a + 3b = 3(2a + b) is divisible by 9 only when 2a + b is divisible by 3. These conditions explain the examples rather than merely listing them.
What Remains?
Suppose you put 18 counters into groups of 5. You make three complete groups and have three counters left. The equation 18 = 5 × 3 + 3 records both parts. The remainder must be smaller than the group size, otherwise you could make another complete group.
Numbers with remainder 3 on division by 5 are 3, 8, 13, 18, 23, … . They are all three more than a multiple of 5, so write them as 5k + 3 for k = 0, 1, 2, … . The same list is two less than a positive multiple of 5: 5k − 2 for k = 1, 2, 3, … . The parameter starts at a different value in the two descriptions.
In N = dq + r, q counts the complete groups of size d and r is the leftover. For a nonnegative integer N and a positive integer d, q and r are nonnegative integers and r is less than d.
Problem
Which of 3k + 5, 3k − 5, 3k/5, 5k + 3, 5k − 2 and 5k − 3 describe all nonnegative integers with remainder 3 on division by 5?
- 1.5k + 3 works for k = 0, 1, 2, … . For 5k − 2, rewrite it as 5(k − 1) + 3; it works for k = 1, 2, 3, … .
- 2.5k − 3 = 5(k − 1) + 2 gives remainder 2, not 3.
- 3.The expressions 3k + 5 and 3k − 5 do not keep a fixed remainder of 3 as k varies. For example, k = 5 gives 20 and 10, both remainder 0.
- 4.3k/5 need not even be an integer: k = 1 gives 3/5. Thus only the two specified families work with their stated parameter ranges.
Problem
661 leaves remainder 3 and 4779 leaves remainder 5 when divided by 7. Find the remainders of their sum and of 4779 − 661.
- 1.Write 661 = 7q + 3 and 4779 = 7p + 5.
- 2.Their sum is 7(p + q) + 8 = 7(p + q + 1) + 1. The eight leftover counters make another group of seven and leave one.
- 3.Their difference is 7(p − q) + (5 − 3) = 7(p − q) + 2. Removing three of the five leftovers leaves two.
- 4.The remainders are 1 and 2. Always reduce a leftover total if it reaches the divisor.
If subtracting two remainders gives a negative result, regroup one complete group. For example, (7a + 2) − (7b + 5) = 7(a − b) − 3 = 7(a − b − 1) + 4. The standard remainder is 4, not −3. A signed leftover describes a position near a multiple; a standard remainder always lies from 0 to d − 1.
Problem
Find all nonnegative integers leaving remainder 2 when divided by either 3 or 4.
- 1.If N leaves remainder 2 in both divisions, N − 2 is divisible by 3 and by 4.
- 2.Hence N − 2 is a multiple of LCM(3, 4) = 12. Write N = 12k + 2.
- 3.For k = 0, 1, 2, 3, …, this gives 2, 14, 26, 38, … . Include 2: a quotient of zero is allowed.
- 4.Conversely, every 12k + 2 leaves remainder 2 in both divisions, so the description is complete.
Problem
A positive number below 100 leaves one pebble when grouped in 2s, 3s or 5s, but no remainder in groups of 7. Find it.
- 1.Subtract the leftover one. The remaining count is divisible by 2, 3 and 5, hence by their LCM 30.
- 2.The count has form 30k + 1. Below 100 the candidates are 1, 31, 61 and 91.
- 3.Only 91 is divisible by 7: 91 = 7 × 13. It also equals 2 × 45 + 1, 3 × 30 + 1 and 5 × 18 + 1.
- 4.Therefore there are 91 pebbles; all four clues have been checked.
Problem
Three integers each leave remainder 2 when divided by 6. Must their sum be divisible by 6?
- 1.Write the numbers as 6a + 2, 6b + 2 and 6c + 2.
- 2.Their sum is 6(a + b + c) + 6 = 6(a + b + c + 1).
- 3.The three leftover pairs form a complete group of six, so the claim is always true. For example, 8 + 14 + 20 = 42.
Problem
Find the smallest nonnegative integer leaving remainders 2, 3 and 4 when divided by 3, 4 and 5 respectively.
- 1.Each remainder is one less than its divisor. Add 1 to the number; the result must be divisible by 3, 4 and 5.
- 2.Their LCM is 60. The smallest positive common multiple is 60, so the candidate is 60 − 1 = 59.
- 3.Check: 59 = 3 × 19 + 2 = 4 × 14 + 3 = 5 × 11 + 4.
- 4.Any smaller nonnegative answer would give a positive common multiple below 60 after adding 1. That contradicts the definition of LCM, so 59 is smallest.
Quiz
What is the sum of n, n + 1, n + 2 and n + 3?
Why does replacing +b with −b preserve the parity of an integer expression?
Which pair of even-number forms always has a sum divisible by 4?
If N is divisible by both 4 and 6, which divisor is guaranteed?
Which expression gives all nonnegative integers with remainder 4 when divided by 7?
Answer: 4n + 6. There are four copies of n, and the constants add to 0 + 1 + 2 + 3 = 6.
Practice Problems
- Find four consecutive integers with sum 50.
- Is 5a + 7b always even? State exactly when it is even.
- Two even integers leave remainders 0 and 2 on division by 4. Prove that their sum is not divisible by 4.
- N leaves remainder 4 when divided by both 5 and 6. Give the general nonnegative form and its three smallest values.
- A positive number below 100 leaves remainder 2 in groups of 3 and 4, and is divisible by 7. Find every possibility.
Write n + (n + 1) + (n + 2) + (n + 3) = 50, so 4n + 6 = 50. Then 4n = 44 and n = 11. The integers are 11, 12, 13 and 14; their sum is 50.
Key Takeaways
• Consecutive integers differ by one; the allowed terms determine which sum representations are possible. • Changing a plus sign to a minus sign changes an integer total by an even amount, so it preserves parity. • Two even numbers sum to a multiple of four exactly when their remainders on division by four match. • Examples suggest patterns; general reasoning proves always claims, while one counterexample can disprove them. • A common divisor divides a sum and a difference; simultaneous divisibility guarantees the LCM. • Write N = dq + r with 0 ≤ r < d, and regroup leftovers before stating a remainder.
Previous
Start of chapter
Next · Lesson 2
Checking Divisibility Quickly