Numbers are not just symbols we use for counting – they follow patterns and rules that make mathematics interesting. This chapter helps us study how numbers behave when we add, subtract, multiply or divide them. We’ll look at divisibility rules, remainders, and number puzzles that develop logical thinking.
Is This a Multiple Of?
Sum of Consecutive Numbers
Anshu is working with consecutive numbers and trying to express different numbers as their sums. He writes examples like:
- 7 = 3 + 4
- 10 = 1 + 2 + 3 + 4
- 12 = 3 + 4 + 5
- 15 = 7 + 8 (also 15 = 4 + 5 + 6 and 15 = 1 + 2 + 3 + 4 + 5)
This exploration raises several interesting questions. Can every natural number be written as a sum of consecutive numbers? Which numbers can be written in more than one way? We know that all odd numbers can be written as sum of two consecutive numbers – for example 7 = 3 + 4, 9 = 4 + 5, 11 = 5 + 6. But what about even numbers?
Another interesting question is whether we can write 0 as a sum of consecutive numbers if we use negative numbers. For example, 0 = -2 + (-1) + 0 + 1 + 2. These are questions worth exploring with your classmates.
Working with Four Consecutive Numbers
Let’s take any four consecutive numbers like 3, 4, 5, 6. If we place plus and minus signs between them, how many different expressions can we make?
All Possible Expressions:
- 3 + 4 – 5 + 6 = 8
- 3 + 4 – 5 – 6 = -4
- 3 + 4 + 5 – 6 = 6
- 3 + 4 + 5 + 6 = 18
- 3 – 4 – 5 + 6 = 0
- 3 – 4 – 5 – 6 = -12
- 3 – 4 + 5 – 6 = -2
- 3 – 4 + 5 + 6 = 10
There are eight such expressions possible. Now look at the results carefully – they are all even numbers (8, -4, 6, 18, 0, -12, -2, 10). This is not a coincidence. No matter which four consecutive numbers we choose and how we arrange the signs, the result will always be even.
Why This Happens:
If we try with another set like 5, 6, 7, 8:
- 5 + 6 – 7 + 8 = 12
- 5 – 6 – 7 – 8 = -16
- (and six more expressions)
Again, all results are even. The reason is that when we have four consecutive numbers, their behaviour with addition and subtraction always produces even results. This can be explained using algebra.
Let’s say the four numbers are a, b, c, d. When we change one sign in an expression (like changing +b to -b), the value changes by 2b, which is an even number. Since all expressions are related by adding or subtracting even amounts, they all have the same parity (all even or all odd). And since at least one combination gives an even result, all must be even.
Breaking Even
Understanding when expressions always give even results is imp for number theory. Let’s study some algebraic expressions:
| Expression | Always Even? | Explanation |
|---|---|---|
| 2a + 2b | Yes | Both terms have factor 2, so sum is even |
| 4m + 2n | Yes | Can be written as 2(2m + n), contains factor 2 |
| 2u – 4v | Yes | Can be written as 2(u – 2v), contains factor 2 |
| x² + 2 | Sometimes | Depends on whether x is even or odd |
For the last expression x² + 2:
- If x = 6 (even), then x² + 2 = 36 + 2 = 38 (even)
- If x = 3 (odd), then x² + 2 = 9 + 2 = 11 (odd)
So it’s not always even.
Pairs to Make Fours
When we add two even numbers, their sum is always even. But when does the sum become a multiple of 4? This requires understanding two types of even numbers:
Type 1: Multiples of 4
Numbers like 4, 8, 12, 16, 20, 24… These leave remainder 0 when divided by 4.
Type 2: Even but not multiples of 4
Numbers like 2, 6, 10, 14, 18, 22… These leave remainder 2 when divided by 4.
Now let’s see what happens when we add different combinations:
Case 1: Both numbers are multiples of 4
Example: 12 + 16 = 28
Using algebra, if both numbers are 4p and 4q, their sum is 4p + 4q = 4(p + q), which is definitely a multiple of 4.
Case 2: Both numbers are Type 2 (remainder 2)
Example: 6 + 10 = 16
Using algebra, if both numbers leave remainder 2, they can be written as (4p + 2) and (4q + 2). Their sum is 4p + 2 + 4q + 2 = 4(p + q + 1), which is also a multiple of 4.
Case 3: One from each type
Example: 8 + 6 = 14
Using algebra: 4p + (4q + 2) = 4(p + q) + 2, which leaves remainder 2 when divided by 4, so not a multiple of 4.
Summary Table:
| First Number Type | Second Number Type | Sum Multiple of 4? |
|---|---|---|
| Multiple of 4 | Multiple of 4 | Always Yes |
| Remainder 2 | Remainder 2 | Always Yes |
| Multiple of 4 | Remainder 2 | Never |
Always, Sometimes, or Never
Now we study different statements about numbers and determine whether they are always true, sometimes true, or never true.
Statement 1: If 8 divides two numbers, then it divides their sum
This is ALWAYS TRUE. If two numbers are multiples of 8, we can write them as 8a and 8b. Their sum is 8a + 8b = 8(a + b), which clearly contains 8 as a factor.
Statement 2: If a number is divisible by 8, then any two numbers that add up to it must each be divisible by 8
This is SOMETIMES TRUE. For example, 16 is divisible by 8. We can write 16 = 8 + 8 (both divisible by 8), but also 16 = 10 + 6 (neither divisible by 8). So it depends on which pair we choose.
Statement 3: If a number is divisible by 7, then all its multiples are divisible by 7
This is ALWAYS TRUE. If n = 7k, then any multiple of n is n × m = 7k × m = 7(km), which is divisible by 7.
Statement 4: If divisible by 12, then divisible by all factors of 12
This is ALWAYS TRUE. The factors of 12 are 1, 2, 3, 4, 6, and 12. If a number is divisible by 12, it must be divisible by all these factors because they are contained in 12.
Statement 5: If divisible by 7, then divisible by all multiples of 7
This is SOMETIMES TRUE. For example, 14 is divisible by 7, but 14 is not divisible by 21 (which is 7 × 3). However, 21 is divisible by both 7 and 21. So it depends on the specific numbers.
Statement 6: If divisible by both 9 and 4, then divisible by 36
This is ALWAYS TRUE. If a number is divisible by both 9 and 4, it must be divisible by their LCM. Since LCM(9, 4) = 36, the number must be divisible by 36.
Statement 7: If we add an odd number and an even number, the result is a multiple of 6
This is NEVER TRUE. When we add odd and even, we always get odd. But multiples of 6 are always even. So the sum can never be a multiple of 6.
What Remains?
Understanding remainders is imp for many problems. When we divide a number by another number, the remainder follows a pattern.
Example: Numbers leaving remainder 3 when divided by 5
These numbers are: 3, 8, 13, 18, 23, 28, 33…
We can write these as 5k + 3, where k = 0, 1, 2, 3…
Another way to look at these same numbers – they are 2 less than the next multiple of 5:
- 3 = 5 – 2
- 8 = 10 – 2
- 13 = 15 – 2
- 18 = 20 – 2
So we can also write them as 5k – 2, where k = 1, 2, 3, 4…
Both expressions 5k + 3 and 5k – 2 describe the same set of numbers, just written differently.
Checking Divisibility Quickly
Instead of actually dividing, we can use shortcuts to check if a number is divisible by another number. These shortcuts are called divisibility rules.
Divisibility by 10, 5, and 2
For 10:
A number is divisible by 10 if and only if its last digit is 0.
Examples: 50, 120, 3450 are divisible by 10, but 45, 123, 3459 are not.
For 5:
A number is divisible by 5 if its last digit is either 0 or 5.
Examples: 45, 120, 3455 are divisible by 5, but 43, 127 are not.
For 2:
A number is divisible by 2 if its last digit is even (0, 2, 4, 6, or 8).
Examples: 44, 126, 3450 are divisible by 2, but 45, 123, 3457 are not.
Why These Rules Work:
Any number can be written in expanded form. For example:
- 1234 = 1000 + 200 + 30 + 4 = (1 × 10³) + (2 × 10²) + (3 × 10) + 4
When we divide by 10:
- 1000 ÷ 10 = 100 (exact division)
- 200 ÷ 10 = 20 (exact division)
- 30 ÷ 10 = 3 (exact division)
- 4 ÷ 10 leaves remainder 4
So only the last digit determines whether division by 10 is exact. The same logic applies for 5 and 2.
Shortcut for Divisibility by 9
A number is divisible by 9 if and only if the sum of its digits is divisible by 9.
Example 1: Is 7309 divisible by 9?
Sum of digits = 7 + 3 + 0 + 9 = 19
19 is not divisible by 9, so 7309 is not divisible by 9.
Example 2: Is 6318 divisible by 9?
Sum of digits = 6 + 3 + 1 + 8 = 18
18 is divisible by 9 (18 = 2 × 9), so 6318 is divisible by 9.
Why This Works:
Let’s understand with 7309:
7309 = 7000 + 300 + 0 + 9
= 7(1000) + 3(100) + 0(10) + 9
= 7(999 + 1) + 3(99 + 1) + 0(9 + 1) + 9
= 7(999) + 7 + 3(99) + 3 + 0(9) + 9
= [7(999) + 3(99) + 0(9)] + [7 + 3 + 0 + 9]
The first bracket contains multiples of 9 (since 999, 99, and 9 are all divisible by 9). The second bracket is just the sum of digits. So when we divide by 9, only the sum of digits determines the remainder.
Finding Remainder:
To find remainder when 7309 is divided by 9:
Sum of digits = 19
19 = 2(9) + 1
So remainder is 1.
We can verify: 7309 = 812(9) + 1
Digital Root:
If we keep adding digits until we get single digit, that’s called digital root.
For 7309: 7 + 3 + 0 + 9 = 19 → 1 + 9 = 10 → 1 + 0 = 1
Digital root is 1, which is the remainder when divided by 9.
Shortcut for Divisibility by 3
A number is divisible by 3 if and only if the sum of its digits is divisible by 3.
This works exactly like the rule for 9, because 10 leaves remainder 1 when divided by 3, just like it does when divided by 9.
Example: Is 8532 divisible by 3?
Sum of digits = 8 + 5 + 3 + 2 = 18
18 is divisible by 3, so 8532 is divisible by 3.
Shortcut for Divisibility by 11
For divisibility by 11, we use an alternating sum of digits.
Steps:
- Start from the rightmost digit (units place)
- Add the digit
- For the next digit, subtract it
- For the next, add it
- Continue alternating between adding and subtracting
- If final result is 0 or divisible by 11, the number is divisible by 11
Example 1: Is 7293 divisible by 11?
Starting from right: 3 – 9 + 2 – 7 = -11
Since -11 is divisible by 11, the number 7293 is divisible by 11.
Example 2: Is 328105 divisible by 11?
Starting from right: 5 – 0 + 1 – 8 + 2 – 3 = -3
Since -3 is not divisible by 11, the number 328105 is not divisible by 11.
Why This Works:
Consider 7293 = 7000 + 200 + 90 + 3
We need to know remainders when powers of 10 are divided by 11:
- 10 ÷ 11 leaves remainder -1 (or 10)
- 100 ÷ 11 leaves remainder 1
- 1000 ÷ 11 leaves remainder -1 (or 10)
- 10000 ÷ 11 leaves remainder 1
The pattern alternates between 1 and -1.
So: 7293 = 7(1000) + 2(100) + 9(10) + 3
When divided by 11, remainder = 7(-1) + 2(1) + 9(-1) + 3 = -7 + 2 – 9 + 3 = -11
Since -11 is divisible by 11, the original number is divisible by 11.
Divisibility by 6
A number is divisible by 6 if it is divisible by both 2 and 3.
Since 6 = 2 × 3, and 2 and 3 are coprime (have no common factors), a number must be divisible by both to be divisible by 6.
Example: Is 2346 divisible by 6?
- Last digit is 6 (even), so divisible by 2 ✓
- Sum of digits = 2 + 3 + 4 + 6 = 15, divisible by 3 ✓
- Therefore, 2346 is divisible by 6 ✓
Divisibility by 24
Since 24 = 8 × 3, and these are coprime, we need to check divisibility by both 8 and 3.
Common Mistake: Checking divisibility by 4 and 6 instead.
Why this doesn’t work: 12 is divisible by both 4 and 6, but not by 24. This is because 4 and 6 are not coprime (they share factor 2). The correct factors to check are 8 and 3.
Example: Is 6312 divisible by 24?
- Last three digits 312: 312 ÷ 8 = 39 (exact), so divisible by 8 ✓
- Sum of digits = 6 + 3 + 1 + 2 = 12, divisible by 3 ✓
- Therefore, 6312 is divisible by 24 ✓
Digital Roots
A digital root is obtained by repeatedly adding digits until we get a single digit.
Example: Find digital root of 489710
- 4 + 8 + 9 + 7 + 1 + 0 = 29
- 2 + 9 = 11
- 1 + 1 = 2
Digital root is 2.
Properties of Digital Roots:
| Type of Number | Digital Root |
|---|---|
| Multiples of 9 | 9 |
| 1 more than multiple of 9 | 1 |
| 2 more than multiple of 9 | 2 |
| … | … |
| 8 more than multiple of 9 | 8 |
The digital root cycles through 1, 2, 3, 4, 5, 6, 7, 8, 9 as we go through consecutive numbers.
Connection with Divisibility by 9:
Digital root equals the remainder when number is divided by 9 (except when remainder is 0, then digital root is 9).
Digits in Disguise
These are puzzles where digits are replaced by letters, and we need to find which digit each letter represents. These are called cryptarithms.
Example 1: PQ × 8 = RS
We need to find two-digit numbers PQ and RS such that when PQ is multiplied by 8, we get RS.
Let’s think systematically:
- If P = 1, then PQ is between 10-19, and PQ × 8 is between 80-152 (three digits for most values)
- If P = 2, we need to check each possibility
Actually, since RS is also two digits, PQ × 8 must be less than 100.
This means PQ must be less than 100 ÷ 8 = 12.5
So PQ can only be 10, 11, or 12.
- 10 × 8 = 80 (works! P=1, Q=0, R=8, S=0 but Q=S, may not be intended)
- 11 × 8 = 88 (works! P=1, Q=1, R=8, S=8 but Q=S again)
- 12 × 8 = 96 (works! P=1, Q=2, R=9, S=6 and all different) ✓
Answer: PQ = 12, RS = 96
Example 2: GH × H = 9K
Here, a two-digit number GH is multiplied by its units digit H to give 9K (a two-digit number starting with 9).
Let’s try different values of H:
- H = 1: GH × 1 = GH (won’t give 90s)
- H = 2: Need GH × 2 to be in 90s, so GH around 45-49
- 46 × 2 = 92 ✓ (G=4, H=6… wait, but we said H=2, contradiction)
Let me reconsider. H appears both in GH and on its own.
Let’s try:
- H = 2: Then G2 × 2 must equal 9K
- Try G = 4: 42 × 2 = 84 (not 90s)
- Try G = 4: 46 × 2 = 92 ✓ (so G=4, H=6, K=2… but we assumed H=2)
Actually, if H=6: Then G6 × 6 must equal 9K
- 16 × 6 = 96 ✓ (G=1, H=6, K=6 but H=K)
Best answer: GH = 16, H = 6, 9K = 96 (but then H = K which might not be intended)
Alternative: 46 × 2 = 92, so G=4, H=6… wait, this doesn’t match H=2.
The puzzle might have multiple solutions depending on whether letters can repeat.
Example 3: BYE × 6 = RAY
This is more complex. We need three-digit number BYE times 6 equals three-digit number RAY.
Strategy:
- Since RAY is three digits and BYE × 6 = RAY, BYE must be less than 167 (because 167 × 6 = 1002)
- So B = 1
- Now 1YE × 6 = RAY
- Since result starts with R, and we multiply by 6, try different values
This requires systematic trial:
- If Y = 0: 10E × 6 = 6 × 10E must start with R
- If Y = 1: 11E × 6 = 60 + 6E, for E=0 gives 660 (R=6)
- 110 × 6 = 660, but then we’d have R=6, A=6
After trying: 139 × 6 = 834
So BYE = 139, RAY = 834
B=1, Y=3, E=9, R=8, A=3, Y already used for A… there’s an issue.
Let me try again: 154 × 6 = 924
BYE = 154, RAY = 924
B=1, Y=5, E=4, R=9, A=2, Y=5 (no conflict with A)
But this gives A=2, and we need to verify all constraints.
These puzzles require patient trial and logical elimination.
Questions and Answers
Can every natural number be written as a sum of consecutive numbers?
- Not every natural number can be written as a sum of consecutive positive integers – specifically, powers of 2 (like 1, 2, 4, 8, 16, 32…) cannot be written as sum of consecutive numbers except trivially as themselves
- However, all odd numbers can definitely be written as sum of two consecutive numbers (like 7 = 3 + 4, 9 = 4 + 5, etc.)
- Most other numbers can be written in at least one way as sum of consecutive numbers
- For example, 15 can be written in multiple ways: 15 = 7 + 8, or 15 = 4 + 5 + 6, or 15 = 1 + 2 + 3 + 4 + 5
Why does adding/subtracting four consecutive numbers with any combination of signs always give even result?
- This happens because of the algebraic structure – when you change one sign in an expression with four consecutive numbers, the value changes by an even amount (specifically by 2 times that number)
- Since all eight possible expressions are related to each other by adding or subtracting even numbers, they all must have the same parity (all even or all odd)
- At least one of the combinations produces an even result, which means all must be even
- For example with 3, 4, 5, 6: the sum 3+4+5+6=18 is even, so all other combinations must also be even
How do we know which two even numbers will add up to a multiple of 4?
- Even numbers fall into two categories: multiples of 4 (like 4, 8, 12, 16…) and non-multiples that leave remainder 2 (like 2, 6, 10, 14…)
- If both numbers are multiples of 4, their sum is definitely a multiple of 4 (like 8 + 12 = 20)
- If both numbers leave remainder 2 when divided by 4, their sum is also a multiple of 4 (like 6 + 10 = 16)
- But if one is a multiple of 4 and the other leaves remainder 2, their sum will leave remainder 2 and won’t be divisible by 4 (like 8 + 6 = 14)
Why does the divisibility rule for 9 work by adding digits?
- This works because every power of 10 (like 10, 100, 1000, etc.) leaves remainder 1 when divided by 9
- When we write a number in expanded form like 7309 = 7(1000) + 3(100) + 0(10) + 9, and divide by 9, the remainders from 1000, 100, and 10 are all 1
- So the total remainder depends only on 7(1) + 3(1) + 0(1) + 9 = 7 + 3 + 0 + 9, which is just the sum of the digits
- This is why checking divisibility by 9 is the same as checking if sum of digits is divisible by 9
How is divisibility by 11 different from divisibility by 9?
- For 11, the remainders when powers of 10 are divided alternate between 1 and -1 (or 10), unlike the consistent remainder of 1 for division by 9
- This is why we use alternating addition and subtraction of digits for divisibility by 11
- The pattern is: units digit gets added, tens digit gets subtracted, hundreds digit gets added, thousands digit gets subtracted, and so on
- If this alternating sum is 0 or divisible by 11, then the original number is divisible by 11
Why can’t we use divisibility by 4 and 6 to check divisibility by 24?
- This doesn’t work because 4 and 6 are not coprime – they share the common factor 2
- When two numbers share a common factor, being divisible by both doesn’t guarantee divisibility by their product
- For example, 12 is divisible by both 4 and 6, but 12 is not divisible by 24
- The correct way is to use 8 and 3 (which are coprime), or use 24 directly by checking last three digits for divisibility by 8 and sum of digits for divisibility by 3
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