Number Play
Why This Matters
Numbers are not just things you count with. They have secret habits.
Some numbers can be broken into a neat ladder of consecutive numbers, like 15 = 4 + 5 + 6. Some numbers refuse to be broken that way, no matter how hard you try. Some numbers can be checked for “is it divisible by 9?” in a few seconds, just by adding their digits — without doing any division at all.
These look like magic tricks. But they are not magic. Every one of them has a clear reason behind it. Once you see the reason, the trick becomes obvious, and you can never forget it.
This chapter is about playing with numbers and asking “why?” about everything you notice. Why can every odd number be split into two consecutive numbers? Why does adding the digits tell you about division by 9? Why do some patterns always hold?
By the end, you will not just know a few shortcuts. You will understand why they work. And understanding why is the real power of mathematics — it lets you predict patterns you have never even seen before, instead of memorising hundreds of separate rules.
The Big Idea
Numbers are built from simple parts, and those parts follow simple rules. An even number always has a 2 hidden inside it. A multiple of 9 always has a “9 + 1” hidden in every place value. Once you write a number using its hidden parts — using a bit of algebra or a picture of dots — every pattern explains itself. Instead of memorising what happens, you can prove what must happen, for every number at once, without checking them one by one.
Let’s Break It Down
We will explore four big ideas. First, which numbers are sums of consecutive numbers, and why. Then parity (whether a number is odd or even) and how it behaves. Then quick divisibility tests for 9, 3 and 11, and the reasons they work. Finally, patterns in factors, multiples and digital roots.
Before we start, let us quickly refresh a few words we will use again and again.
Sums of consecutive numbers
Anshu noticed something fun. Many numbers can be written as a sum of consecutive numbers:
7 = 3 + 4, 12 = 3 + 4 + 5, 15 = 7 + 8 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5.
Some numbers can even be split in more than one way, like 15 above. Let us see this with dots, because a picture makes the idea click.
The diagram below shows 15 split three different ways. Each time, the same 15 dots are just regrouped.
Now for the big question: can every number be written this way? And which numbers can be split in many ways?
Start with the easiest pattern. Every odd number is the sum of two consecutive numbers. For example, 13 = 6 + 7, and 25 = 12 + 13. Why is this always true?
Here is the reason, shown with dots. The picture below splits 13 into two near-equal rows.
Let us make that reasoning exact with a tiny bit of algebra.
Show, using algebra, that every odd number can be written as the sum of two consecutive numbers.
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Write any odd number in its hidden form. An odd number is always 2n + 1 for some whole number n. For example, if n = 6, then 2n + 1 = 13.
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We want two consecutive numbers that add to 2n + 1. Two consecutive numbers look like n and n + 1.
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Add them: n + (n + 1) = 2n + 1. That is exactly our odd number.
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So 2n + 1 = n + (n + 1). Every odd number 2n + 1 equals the two consecutive numbers n and n + 1. For 13, n = 6, so 13 = 6 + 7. This works for every odd number, with no exceptions.
What about other numbers? It turns out almost every number can be written as a sum of consecutive numbers — except the powers of 2: 1, 2, 4, 8, 16, 32, and so on. These are the only numbers that simply refuse. (The deep reason is studied in higher classes, but you can test it yourself: try to split 8 or 16 into consecutive numbers and you will fail every time.)
Can the number 9 be written as a sum of consecutive numbers? Find at least one way.
Yes. 9 is odd, so it is the sum of two consecutive numbers: 9 = 4 + 5. It can also be written as 9 = 2 + 3 + 4. So 9 can be split in two different ways. (9 is not a power of 2, so it is allowed to split.)
Parity — how odd and even behave
Here is a surprising pattern from the textbook. Take any 4 consecutive numbers, say 3, 4, 5, 6. Put + or − signs between them in every possible way:
3 + 4 + 5 + 6 = 18, 3 + 4 − 5 + 6 = 8, 3 − 4 − 5 − 6 = −12, and so on.
There are 8 such expressions. If you work them all out, you notice something: the answer is always even. Never odd. And this happens no matter which 4 numbers you start with. Why?
The secret is parity. Let us first nail down how odd and even behave when you add or subtract.
odd ± odd = even
even ± even = even
odd ± even = odd
Notice the most important fact here: changing a + into a − never changes the parity. For example, 7 + 3 = 10 (even) and 7 − 3 = 4 (also even). The parity stays the same.
Here is the clean reason, straight from the textbook. Take an expression like a + b − c − d. Now flip one sign, say change + b to − b, to get a − b − c − d. How much did the value change?
(a + b − c − d) − (a − b − c − d) = 2b
The change is 2b, which is always even. So flipping a sign moves the value by an even amount. And if two numbers differ by an even amount, they must have the same parity. So flipping signs never changes whether the answer is odd or even.
Starting from one expression, you can reach all 8 by flipping signs. So all 8 expressions have the same parity. For 3, 4, 5, 6 the plain sum 3 + 4 + 5 + 6 = 18 is even, so all 8 must be even too.
Let us use this idea to judge an expression without computing it.
Without calculating, decide whether 4m + 2q is even for every whole-number value of m and q.
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Look for a hidden common factor. Both terms share a 2: 4m = 2 × 2m, and 2q = 2 × q.
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Factor the 2 out: 4m + 2q = 2(2m + q).
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The whole thing is 2 times something. Any number of the form 2 × (a whole number) is even, no matter what 2m + q turns out to be.
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So 4m + 2q is always even. For example, if m = 4 and q = −9, then 4m + 2q = 16 − 18 = −2, which is even. The “2 is a factor” reason works for every m and q at once.
Be careful though — not every expression is always even. Take x² + 2. If x is even, x² is even, so x² + 2 is even. But if x is odd, x² is odd, so x² + 2 is odd. For example, x = 3 gives 9 + 2 = 11, which is odd. So x² + 2 is only sometimes even.
When do two even numbers add to a multiple of 4?
Take two even numbers and add them. Is the sum always a multiple of 4? Try a few:
2 + 6 = 8 (yes, multiple of 4), but 2 + 4 = 6 (no, not a multiple of 4).
So sometimes yes, sometimes no. When exactly? The trick is to sort even numbers into two types, by their remainder when divided by 4.
Type A — multiples of 4: numbers like 4, 8, 12, 16. They leave remainder 0 when divided by 4. In algebra: 4p.
Type B — even but not multiples of 4: numbers like 2, 6, 10, 14. They leave remainder 2 when divided by 4. In algebra: 4q + 2.
Now there are three cases. Two Type-A numbers, two Type-B numbers, or one of each. Let us picture the trickiest one: two Type-B numbers, like 6 + 10. The diagram below arranges every number in rows of 4 and shows what happens to the leftovers.
In algebra this is just as clean: (4p + 2) + (4q + 2) = 4p + 4q + 4 = 4(p + q + 1), which is a multiple of 4. The two 2s combine into a 4.
Here are all three cases together. Notice the pattern is exactly like the parity of odd and even.
| What you add | Algebra | Is the sum a multiple of 4? |
|---|---|---|
| multiple of 4 + multiple of 4 | 4p + 4q = 4(p + q) | Yes — always |
| remainder-2 + remainder-2 | (4p + 2) + (4q + 2) = 4(p + q + 1) | Yes — the two 2s make a 4 |
| multiple of 4 + remainder-2 | 4p + (4q + 2) = 4(p + q) + 2 | No — remainder 2 is left over |
So two even numbers add to a multiple of 4 exactly when they are both Type A, or both Type B. A mismatch leaves a remainder of 2.
Is 14 + 22 a multiple of 4? Decide using the remainder rule, not by adding.
Both 14 and 22 leave remainder 2 when divided by 4 (14 = 4 × 3 + 2, and 22 = 4 × 5 + 2). They are both Type B. Two Type-B numbers always add to a multiple of 4, because the two leftover 2s join into a 4. So yes, 14 + 22 is a multiple of 4. (Check: 14 + 22 = 36 = 4 × 9.)
Always, sometimes, or never — divisibility facts
The same kind of reasoning settles many divisibility statements. The big tools are these four facts, all provable with the “factor out” trick:
If a divides M and a divides N, then a divides M + N and M − N.
If A is divisible by k, then every multiple of A is divisible by k.
If A is divisible by k, then A is divisible by every factor of k.
If A is divisible by k and by m, then A is divisible by the LCM of k and m.
The first one is easy to see. If 8 divides two numbers, say 8a and 8b, then their sum is 8a + 8b = 8(a + b), which is still a multiple of 8. The shared factor of 8 never disappears when you add.
Let us use these to judge a statement.
Always, sometimes, or never true: 'When you add an odd number to an even number you get a multiple of 6.'
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Check the parity first. A multiple of 6 is always even, because 6 itself is even (6 = 2 × 3).
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Now look at the left side: odd + even. From our parity rules, odd + even = odd. So the left side is always odd.
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But the right side (a multiple of 6) is always even. An odd number can never equal an even number.
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So the statement is never true. No odd-plus-even sum can ever be a multiple of 6, because one side is forced to be odd and the other even.
A quick test for divisibility by 9 (and why it works)
Here is a famous shortcut: a number is divisible by 9 if the sum of its digits is divisible by 9. For example, 5 + 4 = 9 tells you 54 is divisible by 9. No division needed.
But why does just adding digits work? This is the heart of the chapter. The reason is beautiful, and the picture below shows it using 427.
So the rule is really about remainders. Every place value (10, 100, 1000, …) is one more than a multiple of 9. The multiple-of-9 parts contribute nothing to the remainder. What is left over is exactly the digit sum. If the digit sum is a multiple of 9, the whole number is too.
Let us use it.
Is 93,547 divisible by 9? Use the digit-sum test.
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Add all the digits: 9 + 3 + 5 + 4 + 7 = 28.
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Is 28 a multiple of 9? The multiples of 9 are 9, 18, 27, 36. 28 is not among them. To be sure, add the digits of 28: 2 + 8 = 10, then 1 + 0 = 1. The single digit is 1, not 9.
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Since the digit sum is not a multiple of 9, the number is not divisible by 9. In fact the remainder is 1.
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So 93,547 is not divisible by 9; it leaves remainder 1. We answered without doing any long division at all.
The test for 3 works exactly the same way, for the same reason: every place value is also one more than a multiple of 3 (10 = 9 + 1, 100 = 99 + 1, and 9 and 99 are multiples of 3 too). So a number is divisible by 3 if the sum of its digits is divisible by 3.
A quick test for divisibility by 11 (and why it works)
The test for 11 has a twist: you alternately add and subtract the digits. Why the alternating signs? Because the place values do not all behave the same way — they flip between one more and one less than a multiple of 11. The picture below shows this.
So the rule is: start at the units digit, alternately add and subtract the digits, and if the result is 0 or a multiple of 11, the number is divisible by 11. Let us try it.
Is 5,529 divisible by 11? If not, what is the remainder?
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Write the digits and attach alternating signs, starting from the units digit with a plus: the units 9 is +, the tens 2 is −, the hundreds 5 is +, the thousands 5 is −.
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Compute the alternating sum: +9 − 2 + 5 − 5 = 7.
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Is 7 a multiple of 11? No. The nearest multiples of 11 are 0 and 11, and 7 is neither.
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So 5,529 is not divisible by 11. The alternating sum is 7, which tells us it is 7 more than a multiple of 11 — so the remainder is 7.
Factors, multiples and digital roots
Many number patterns come from how factors nest inside one another. A clear example: the multiples of 4, 8 and 32. Every multiple of 32 is also a multiple of 8, and every multiple of 8 is also a multiple of 4. The picture below shows this as circles inside circles.
This is just the third tool from before: if a number is divisible by 32, it is divisible by every factor of 32, including 8 and 4.
Now a fun pattern called the digital root. Take a number, add its digits, then keep adding the digits of the result until a single digit is left. That single digit is the digital root.
For example, for 489710: 4 + 8 + 9 + 7 + 1 + 0 = 29, then 2 + 9 = 11, then 1 + 1 = 2. So the digital root of 489710 is 2.
This is the very same process as the divisibility test for 9. So the digital root tells you the remainder when the number is divided by 9. (If the digital root is 9, the number is an exact multiple of 9.)
What is the digital root of 7,309, and what does it tell you about dividing 7,309 by 9?
Add the digits: 7 + 3 + 0 + 9 = 19. Add again: 1 + 9 = 10. Add again: 1 + 0 = 1. So the digital root is 1. Because the digital root equals the remainder on division by 9, this tells us 7,309 leaves remainder 1 when divided by 9.
Common Mistakes
These are the slip-ups students make most often in this chapter. Spot them now to keep your reasoning clean.
The digit-sum test for 9 is just a memorised trick with no real reason; you simply add digits because the teacher said so.
The rule is taught as a quick shortcut to pass exams, and it works so reliably that students never feel the need to ask where it comes from.
It has a clear reason. Every place value (10, 100, 1000, ...) is one more than a multiple of 9. The multiple-of-9 parts leave no remainder, so the leftover is exactly the digit sum. That is why the digit sum decides divisibility by 9.
To check divisibility by 24, just check divisibility by its factors 4 and 6, since 4 × 6 = 24.
It feels natural that if a number passes both factor tests it must be divisible by their product, and 4 times 6 does equal 24, so the logic looks airtight.
It fails because 4 and 6 share the factor 2, so they overlap. The number 12 is divisible by both 4 and 6 but not by 24. To test for 24 correctly, check divisibility by 3 and 8, whose only common factor is 1.
Changing a plus sign to a minus sign in an expression can change whether the answer is odd or even.
A plus and a minus look like opposite operations, and they certainly change the value, so it seems they should change the parity too.
Flipping a sign changes the value by an even amount (by 2 times one of the numbers). Two values that differ by an even amount have the same parity. So a + b and a − b are always both odd or both even — the parity never changes.
Quick Check
Try these quick questions. The explanation appears after you answer, so read it either way.
Which of these numbers CANNOT be written as a sum of two or more consecutive numbers?
16 is a power of 2 (16 = 2⁴), and powers of 2 are the only numbers that cannot be split into a sum of consecutive numbers. The others can: 15 = 7 + 8, 9 = 4 + 5, and 21 = 10 + 11. So the answer is 16.
Without dividing, decide: is 405 divisible by 9?
Add the digits: 4 + 0 + 5 = 9, which is divisible by 9. So 405 is divisible by 9 (405 = 9 × 45). The ending digit and oddness have nothing to do with the test for 9 — only the digit sum matters.
Use the alternating-sum test: is 5,929 divisible by 11?
Starting from the units digit and alternating: +9 − 2 + 9 − 5 = 11. Since 11 is a multiple of 11, the number is divisible by 11 (5,929 = 11 × 539). The digits being odd is irrelevant — it is the alternating sum that decides.
Two even numbers are added. When is the sum guaranteed to be a multiple of 4?
Even numbers split into two types by their remainder on dividing by 4: remainder 0 (multiples of 4) or remainder 2. Two of the same type always work — two multiples of 4 obviously, and two remainder-2 numbers because their leftover 2s join into a 4. A mismatch leaves remainder 2. So the sum is a multiple of 4 exactly when both have the same remainder.
Practice Problems
Try each one yourself first. Then tap to check your full solution.
Easy
Write 21 as a sum of two consecutive numbers, and also as a sum of consecutive numbers in one other way.
21 is odd, so it is the sum of two consecutive numbers. Halve it: 21 ÷ 2 is between 10 and 11, so the two numbers are 10 and 11. Check: 10 + 11 = 21.
For another way, try a longer run. 6 + 7 + 8 = 21. (Also 1 + 2 + 3 + 4 + 5 + 6 = 21.)
So 21 = 10 + 11 = 6 + 7 + 8.
Without dividing, find whether 8,888 is divisible by 9.
Add the digits: 8 + 8 + 8 + 8 = 32.
Is 32 a multiple of 9? Add again: 3 + 2 = 5. The single digit is 5, not 9.
So 8,888 is not divisible by 9. (It leaves remainder 5.)
The sum of four consecutive numbers is 34. What are these numbers?
Call the four consecutive numbers n, n + 1, n + 2, n + 3.
Add them: n + (n + 1) + (n + 2) + (n + 3) = 4n + 6.
Set equal to 34: 4n + 6 = 34, so 4n = 28, so n = 7.
The numbers are 7, 8, 9 and 10. Check: 7 + 8 + 9 + 10 = 34. ✓
Medium
If 31z5 is a multiple of 9, where z is a single digit, find all possible values of z. Why are there two answers?
For divisibility by 9, the digit sum must be a multiple of 9.
Digit sum = 3 + 1 + z + 5 = 9 + z.
We need 9 + z to be a multiple of 9. Since z is a single digit (0 to 9), 9 + z ranges from 9 to 18.
- If 9 + z = 9, then z = 0.
- If 9 + z = 18, then z = 9.
Both 9 and 18 are multiples of 9, so z = 0 or z = 9. There are two answers because the digit sum can land on two different multiples of 9 within the allowed range.
Tathagat writes numbers that each leave remainder 2 when divided by 6. He claims that adding any three such numbers always gives a multiple of 6. Is he right?
Write three such numbers in their hidden form: 6a + 2, 6b + 2, 6c + 2.
Add them: (6a + 2) + (6b + 2) + (6c + 2) = 6a + 6b + 6c + 6.
Factor out the 6: = 6(a + b + c + 1).
This is 6 times a whole number, so it is always a multiple of 6.
So Tathagat is right. The three leftover 2s add to 6, which makes one more complete group of 6.
When divided by 7, the number 661 leaves remainder 3 and 4779 leaves remainder 5. Without calculating the numbers, find the remainder when 4779 + 661 is divided by 7.
Write each number in its hidden form using its remainder.
661 = 7q + 3 (remainder 3), and 4779 = 7p + 5 (remainder 5).
Add: 4779 + 661 = (7p + 5) + (7q + 3) = 7p + 7q + 8.
Now 8 is bigger than 7, so pull out one more 7: 8 = 7 + 1.
= 7p + 7q + 7 + 1 = 7(p + q + 1) + 1.
The 7(…) part is a multiple of 7, so the remainder is 1. The two remainders 3 and 5 add to 8, which is one full 7 plus 1.
Challenge
A child has fewer than 100 pebbles. Grouped in 3s, one is left over. Grouped in 2s, one is left over. Grouped in 5s, one is left over. Grouped in 7s, none is left over. How many pebbles?
Read the clues carefully. Leftover 1 when grouped by 2, by 3, and by 5. None left over when grouped by 7.
“Remainder 1 for 2, 3 and 5 together” means the number is 1 more than a common multiple of 2, 3 and 5.
The smallest common multiple (LCM) of 2, 3, 5 is 30. So the number is of the form 30k + 1: that is 31, 61, 91, … (we stop under 100).
Now apply the last clue: it must be divisible by 7.
- 31 ÷ 7 leaves a remainder. ✗
- 61 ÷ 7 leaves a remainder. ✗
- 91 ÷ 7 = 13 exactly. ✓
So the child has 91 pebbles. Check: 91 = 30 × 3 + 1 (remainder 1 for 2, 3, 5), and 91 = 7 × 13 (no remainder for 7). ✓
Is the product of three consecutive integers always a multiple of 6? Explain why.
Take any three consecutive integers, such as n, n + 1, n + 2.
A multiple of 2 is always present. Among any two consecutive integers, one must be even. So among three consecutive integers there is certainly at least one even number. The product therefore has a factor of 2.
A multiple of 3 is always present. Among any three consecutive integers, exactly one is a multiple of 3 (the multiples of 3 come every third number). So the product has a factor of 3.
Since the product always has a factor of 2 and a factor of 3, it always has the factor 2 × 3 = 6.
So yes, the product of three consecutive integers is always a multiple of 6. For example, 4 × 5 × 6 = 120 = 6 × 20. (By the same reasoning, the product of four consecutive integers is always a multiple of 2 × 3 × 4 = 24.)
Summary
- A run of consecutive numbers is a stack of growing rows. Every odd number is the sum of two consecutive numbers (2n + 1 = n + (n + 1)). Every number except a power of 2 (1, 2, 4, 8, 16, …) can be written as a sum of consecutive numbers.
- Parity (odd/even) follows simple rules: odd ± odd = even, even ± even = even, odd ± even = odd. Flipping a + to a − changes the value by an even amount, so it never changes the parity.
- An expression is always even if you can factor out a 2, like 4m + 2q = 2(2m + q).
- Two even numbers add to a multiple of 4 exactly when both are multiples of 4, or both leave remainder 2. A mismatch leaves remainder 2.
- A number is divisible by 9 (or 3) if its digit sum is, because every place value is one more than a multiple of 9 (and of 3).
- A number is divisible by 11 if the alternating sum of its digits is 0 or a multiple of 11, because place values flip between one more and one less than a multiple of 11.
- If a number is divisible by k, it is divisible by every factor of k; every multiple of it keeps the same divisors; and if divisible by k and m, it is divisible by their LCM.
- A digital root (repeatedly adding digits to one digit) equals the remainder on dividing by 9.
What’s Next
You have just seen the real power of mathematics: not memorising rules, but proving why they must be true — with a picture of dots or a line of algebra.
Next, in Chapter 6 — We Distribute, Yet Things Multiply, you will meet the distributive law and see how breaking numbers apart and putting them back together unlocks fast multiplication and powerful algebra. The “factor it out” trick you used all through this chapter is exactly the idea that comes next. Onward!
Frequently Asked Questions
Which numbers can be written as a sum of consecutive numbers and which cannot?
Every counting number except the powers of 2 (1, 2, 4, 8, 16, 32 and so on) can be written as a sum of two or more consecutive numbers. Every odd number is the sum of two consecutive numbers, and most other numbers can be split in one or more ways. Powers of 2 are the only ones that cannot be split this way.
Why does the digit-sum test for divisibility by 9 work?
Every place value is one more than a multiple of 9. For example 10 is 9 plus 1, 100 is 99 plus 1, 1000 is 999 plus 1. The multiple-of-9 parts never change the remainder, so the remainder of the whole number equals the remainder of the sum of its digits. If the digit sum is divisible by 9, the number is divisible by 9.
How do you check divisibility by 11 quickly?
Place values alternate between one more and one less than a multiple of 11. So starting from the units digit, alternately add and subtract the digits. If the final result is 0 or a multiple of 11, the number is divisible by 11. For example for 462 you compute 2 minus 6 plus 4, which equals 0, so 462 is divisible by 11.
When do two even numbers add up to a multiple of 4?
Split each even number by its remainder when divided by 4. Two multiples of 4 always add to a multiple of 4. Two numbers that each leave remainder 2 also add to a multiple of 4, because the two remainders of 2 join to make one more 4. But a multiple of 4 plus a remainder-2 number leaves remainder 2, so it is not a multiple of 4.
What is a digital root and how is it found?
A digital root is the single digit you reach by adding all the digits of a number, then adding the digits of that result, and repeating until one digit is left. For example, the digital root of 489710 is 2. The digital root of a number is the same as the remainder when the number is divided by 9, except that a digital root of 9 means the number is an exact multiple of 9.