5.1. Is This a Multiple Of?
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1. Can I write every natural number as a sum of consecutive numbers?
Almost every natural number can be written as a sum of consecutive numbers. The only exceptions are numbers that are exact powers of 2 (such as 1, 2, 4, 8, 16, 32, etc.).
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2. Which numbers can I write as the sum of consecutive numbers in more than one way?
Numbers that have more than one odd factor can be written in multiple ways. For example, 15 has odd factors 3 and 5, so it can be written as 7+8, 4+5+6, and 1+2+3+4+5.
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3. Can we write all even numbers as a sum of consecutive numbers?
No, not all of them. Even numbers that are powers of 2 (like 2, 4, 8, 16) cannot be written as a sum of consecutive natural numbers. Other even numbers (like 6, 10, 12) can be.
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4. Can I write 0 as a sum of consecutive numbers? Maybe I should use negative numbers.
Yes, if negative integers are allowed, 0 can be written as the sum of consecutive integers that are symmetrical around zero. For example: -2 + -1 + 0 + 1 + 2 = 0.
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5. Take any 4 consecutive numbers. For example, 3, 4, 5, and 6. Place '+' and '-' signs in between the numbers. How many different possibilities exist? Write all of them.
There are 23 = 8 possibilities. They are:
3 + 4 + 5 + 6
3 + 4 + 5 - 6
3 + 4 - 5 + 6
3 + 4 - 5 - 6
3 - 4 + 5 + 6
3 - 4 + 5 - 6
3 - 4 - 5 + 6
3 - 4 - 5 - 6
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1. Evaluate each expression and write the result next to it. Do you notice anything interesting?
3 + 4 + 5 + 6 = 18
3 + 4 + 5 - 6 = 6
3 + 4 - 5 + 6 = 8
3 + 4 - 5 - 6 = -4
3 - 4 + 5 + 6 = 10
3 - 4 + 5 - 6 = -2
3 - 4 - 5 + 6 = 0
3 - 4 - 5 - 6 = -12
All the results are even numbers. -
2. Now, take four other consecutive numbers. Place the '+' and '-' signs as you have done before. Find out the results of each expression. What do you observe?
Taking 5, 6, 7, 8:
5 + 6 + 7 + 8 = 26
5 + 6 + 7 - 8 = 10
5 + 6 - 7 + 8 = 12
5 + 6 - 7 - 8 = -4
5 - 6 + 7 + 8 = 14
5 - 6 + 7 - 8 = -2
5 - 6 - 7 + 8 = 0
5 - 6 - 7 - 8 = -16
All the results are still even numbers. -
3. Repeat this for one more set of 4 consecutive numbers. Share your findings.
Taking 10, 11, 12, 13:
The results will be 46, 20, 22, -4, 24, -2, 0, and -26. Again, all results are even. -
4. Do these patterns occur no matter which 4 consecutive numbers are chosen? Is there a way to find out through reasoning?
Yes, this pattern always occurs. Any sequence of 4 consecutive numbers contains exactly two even numbers and two odd numbers. The sum or difference of two odds is always even, and the sum or difference of two evens is always even. Adding these even results together always produces an even number.
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1. Now take any 4 numbers, place '+' and '-' signs in the eight different ways, and evaluate the resulting expression. What do you observe about their parities?
For any specific set of 4 numbers (even non-consecutive ones), all 8 combinations of '+' and '-' will result in numbers that share the exact same parity (they will all be even or all be odd).
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2. Replace any negative sign in the expression a + b - c - d with a positive sign and find the difference between the two numbers. What do you conclude from this observation?
Difference = (a + b + c - d) - (a + b - c - d) = 2c.
Because the difference is 2c (which is inherently an even number), flipping a sign always changes the total result by an even amount. Adding or subtracting an even amount preserves the original parity of the expression.
Breaking Even
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1. Is the phenomenon of all the expressions having the same parity limited to taking 4 numbers? What do you think?
No, it applies to any number of terms. Changing a '+' to a '-' or vice versa changes the total sum by twice the value of that specific term. Since twice any integer is an even number, the change is always even, meaning the parity remains constant regardless of how many numbers are in the sequence.
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2. Without computing them, find out which of the following arithmetic expressions are even.
43 + 37: Even (Odd + Odd)
809 + 214: Odd (Odd + Even)
672 - 348: Even (Even - Even)
4 × 347 × 3: Even (contains an even factor)
708 - 477: Odd (Even - Odd)
119 × 303: Odd (Odd × Odd)
543 - 479: Even (Odd - Odd)
5133: Odd (Odd number raised to a power is odd) -
3. Using our understanding of how parity behaves under different operations, identify which of the following algebraic expressions give an even number for any integer values for the letter-numbers.
2a + 2b: Always even (sum of two even numbers).
3g + 5h: Not always even.
4m + 2n: Always even (sum of two even numbers).
2u - 4v: Always even (difference of two even numbers).
13k - 5k: Equals 8k, which is always even.
6m - 3n: Not always even.
x2 + 2: Not always even.
b2 + 1: Not always even.
4k × 3j: Equals 12kj, which is always even.
Pairs to Make Fours
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1. Write a few algebraic expressions which always give an even number.
Examples include: 2n, 4x + 6y, n(n+1), 2k2 + 4, and 8m - 2.
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2. Take a pair of even numbers. Add them. Is the sum divisible by 4? Try this with different pairs of even numbers. When is the sum a multiple of 4, and when is it not? Is there a general rule or a pattern?
Not always. If both even numbers are multiples of 4 (e.g., 4 + 8 = 12), the sum is divisible by 4. If both even numbers leave a remainder of 2 when divided by 4 (e.g., 6 + 10 = 16), the sum is also divisible by 4. However, if one is a multiple of 4 and the other leaves a remainder of 2 (e.g., 4 + 6 = 10), the sum is NOT divisible by 4.
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1. What happens when we add a multiple of 4 to an even number that is not a multiple of 4? Is it similar to the case of the parity of the sum of an even and an odd number? Look at the following expressions and the visualisation. Write the corresponding explanation and examples.
Explanation: Adding a multiple of 4 (4p) and an even number that is not a multiple of 4 (4q+2) gives: 4p + 4q + 2 = 4(p+q) + 2. The result is always 2 more than a multiple of 4, so it is never perfectly divisible by 4.
Yes, this is directly analogous to adding an even and an odd number, which yields an odd number (never divisible by 2).
Examples: 8 + 6 = 14; 12 + 10 = 22. Neither 14 nor 22 is a multiple of 4.
Always, Sometimes, or Never
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1. Determine if Statement 1 (If 8 exactly divides two numbers separately, it must exactly divide their sum) is true with subtraction.
Yes, it is always true with subtraction. If 8 divides 'a' and 'b', then a = 8m and b = 8n. Their difference is 8m - 8n = 8(m - n), which clearly has 8 as a factor and is a multiple of 8.
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1. Examine each of the following statements, and determine whether it is 'Always true', 'Sometimes true', 'Never true'.
6. If a number is divisible by both 9 and 4, it must be divisible by 36.Always true. Since 9 and 4 are co-prime (they share no common factors other than 1), any number divisible by both must be divisible by their product, which is 36. -
2. 7. If a number is divisible by both 6 and 4, it must be divisible by 24.
Sometimes true. For example, 12 is divisible by both 6 and 4, but it is not divisible by 24. However, 48 is divisible by 6, 4, and 24. The rule is that the number must be divisible by their Lowest Common Multiple (LCM), which is 12, not necessarily their product.
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3. Can I write an even and an odd number as 2n and 2n+1 instead?
Yes. Their sum is (2n) + (2n + 1) = 4n + 1. Since 4n is always even, 4n + 1 is always odd. A multiple of 6 must be an even number, so the sum of an even and an odd number can never be a multiple of 6.
What Remains?
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1. Find a number that has a remainder of 3 when divided by 5. Write more such numbers. Which algebraic expression(s) capture all such numbers? (i) 3k+5 (ii) 3k-5 (iii) 3k/5 (iv) 5k+3 (v) 5k-2 (vi) 5k-3
Examples of such numbers: 8, 13, 18, 23, 28.
The algebraic expressions that capture these are (iv) 5k+3 and (v) 5k-2.
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1. Are there other expressions that generate numbers that are 3 more than a multiple of 5?
Yes. Expressions like 5k + 8, 5k - 7, or 5k + 13 also generate such numbers, because 8, -7, and 13 all leave a remainder of 3 when divided by 5.
Figure it Out - 5.1
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1. The sum of four consecutive numbers is 34. What are these numbers?
Let the numbers be x, x+1, x+2, and x+3.
x + (x+1) + (x+2) + (x+3) = 34
4x + 6 = 34 → 4x = 28 → x = 7.
The numbers are 7, 8, 9, and 10. -
2. Suppose p is the greatest of five consecutive numbers. Describe the other four numbers in terms of p.
The other four numbers are: p-4, p-3, p-2, and p-1.
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3. For each statement below, determine whether it is always true, sometimes true, or never true. Explain your answer. Justify your claim using algebra.
(i) The sum of two even numbers is a multiple of 3.
(ii) If a number is not divisible by 18, then it is also not divisible by 9.
(iii) If two numbers are not divisible by 6, then their sum is not divisible by 6.
(iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3.
(v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.(i) Sometimes true. 2 + 4 = 6 (multiple of 3). But 2 + 2 = 4 (not a multiple of 3).
(ii) Sometimes true. 27 is not divisible by 18, but it IS divisible by 9. However, 14 is not divisible by 18, and also not divisible by 9.
(iii) Sometimes true. 2 + 3 = 5 (not divisible by 6). But 4 + 2 = 6 (IS divisible by 6).
(iv) Always true. 6a + 9b = 3(2a + 3b), which has a clear factor of 3.
(v) Sometimes true. 6 + 3 = 9 (multiple of 9). But 6 + 6 = 12 (not a multiple of 9). -
4. Find a few numbers that leave a remainder of 2 when divided by 3 and a remainder of 2 when divided by 4. Write an algebraic expression to describe all such numbers.
A few numbers are 2, 14, 26, 38. The Lowest Common Multiple (LCM) of 3 and 4 is 12. The algebraic expression describing all such numbers is 12k + 2.
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5. "I hold some pebbles, not too many, When I group them in 3's, one stays with me. Try pairing them up - it simply won't do, A stubborn odd pebble remains in my view. Group them by 5, yet one's still around, But grouping by seven, perfection is found. More than one hundred would be far too bold, Can you tell me the number of pebbles I hold?"
The number must be a multiple of 7 less than 100: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.
It leaves a remainder of 1 when divided by 3 and by 5, meaning it leaves a remainder of 1 when divided by their LCM (15). The numbers of the form 15k + 1 under 100 are: 16, 31, 46, 61, 76, 91.
The only number common to both lists is 91. You hold 91 pebbles. -
6. Tilak has written several numbers that leave a remainder of 2 when divided by 6. He claims, "If you add any three such numbers, the sum will always be a multiple of 6." Is Tilak's claim true?
Yes, Tilak's claim is always true. Algebraically: (6a + 2) + (6b + 2) + (6c + 2) = 6a + 6b + 6c + 6 = 6(a + b + c + 1), which is a clear multiple of 6.
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7. When divided by 7, the number 661 leaves a remainder of 3, and 4779 leaves a remainder of 5. Without calculating, can you say what remainders the following expressions will leave when divided by 7?
(i) 4779 + 661
(ii) 4779 - 661(i) The remainder of the sum is the sum of the remainders. 5 + 3 = 8. Since 8 divided by 7 leaves a remainder of 1, the final remainder is 1.
(ii) The remainder of the difference is the difference of the remainders. 5 - 3 = 2. The remainder is 2. -
8. Find a number that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5. What is the smallest such number? Can you give a simple explanation of why it is the smallest?
Notice that in every condition, the remainder is exactly 1 less than the divisor (3-1=2, 4-1=3, 5-1=4). Therefore, the target number is exactly 1 less than a common multiple of 3, 4, and 5. The smallest common multiple (LCM) of 3, 4, and 5 is 60. The smallest such number is 60 - 1 = 59.
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9. Explain using algebra why the divisibility shortcuts for 5, 2, 4, and 8 work.
Any number can be written in expanded form: ... + 1000d + 100c + 10b + a.
For 2 and 5: 10 is perfectly divisible by both 2 and 5. Therefore, all terms from 10b onwards are multiples of 2 and 5. Only the final units digit 'a' determines divisibility.
For 4: 100 is perfectly divisible by 4. All terms from 100c onwards are multiples of 4. Only the last two digits (10b + a) determine divisibility.
For 8: 1000 is perfectly divisible by 8. All terms from 1000d onwards are multiples of 8. Only the last three digits (100c + 10b + a) determine divisibility.
A Shortcut for Divisibility by 9
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1. Can you say, without actually calculating, which of these numbers are divisible by 9: 999, 909, 900, 90, 990?
All of them are divisible by 9. Any number composed entirely of 9s and 0s will always have a digit sum that is a multiple of 9.
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2. Is 10 divisible by 9? If not, what is the remainder? Check the divisibility of other multiples of 10 (10, 20, 30, ...) by 9.
No, 10 is not divisible by 9. The remainder is 1.
For other multiples of 10, they are also not divisible by 9. The remainder perfectly corresponds to the tens digit (e.g., 20 leaves a remainder of 2, 30 leaves 3, etc.). -
3. Similarly, look at the remainder when the multiples of 100 (100, 200, 300,...) are divided by 9. What do you notice?
The remainder is exactly the digit in the hundreds place. (e.g., 200/9 leaves a remainder of 2, 300/9 leaves 3).
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4. Using this observation, find the remainder when 427 is divided by 9.
4 hundreds give a remainder of 4. 2 tens give a remainder of 2. 7 units give 7. Total sum = 4 + 2 + 7 = 13. Since 13 divided by 9 leaves a remainder of 4, the remainder for 427 is 4.
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1. Look at each of the following statements. Which are correct and why?
(i) If a number is divisible by 9, then the sum of its digits is divisible by 9.
(ii) If the sum of the digits of a number is divisible by 9, then the number is divisible by 9.
(iii) If a number is not divisible by 9, then the sum of its digits is not divisible by 9.
(iv) If the sum of the digits of a number is not divisible by 9, then the number is not divisible by 9.All four statements are correct. The number and the sum of its individual digits will always leave the exact same remainder when divided by 9. Therefore, if one is perfectly divisible by 9 (leaves remainder 0), the other must be too. If one is not, the other is not.
Figure it Out - 5.2
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1. Find, without dividing, whether the following numbers are divisible by 9.
(i) 123 (ii) 405 (iii) 8888 (iv) 93547 (v) 358095(i) 1+2+3 = 6 (No)
(ii) 4+0+5 = 9 (Yes)
(iii) 8+8+8+8 = 32 (No)
(iv) 9+3+5+4+7 = 28 (No)
(v) 3+5+8+0+9+5 = 30 (No) -
2. Find the smallest multiple of 9 with no odd digits.
The digits must be even and sum to a multiple of 9. The smallest sum of even digits to make a multiple of 9 is 18. The smallest combination of even digits that sum to 18 is 2, 8, 8. Thus, the smallest number is 288.
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3. Find the multiple of 9 that is closest to the number 6000.
The sum of the digits of 6000 is 6. To be a multiple of 9, the sum needs to be 9. So 6003 works (sum = 9). The closest multiple below 6000 is 5994. 6003 is 3 away, while 5994 is 6 away. So 6003 is the closest.
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4. How many multiples of 9 are there between the numbers 4300 and 4400?
The first multiple of 9 after 4300 is 4302. The last multiple before 4400 is 4392. The number of multiples = ((4392 - 4302) / 9) + 1 = 11.
A Shortcut for Divisibility by 3
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1. Explore the remainders when powers of 10 are divided by 3. Explain why this method works.
10 = 3 × 3 + 1 (remainder 1). 100 = 33 × 3 + 1 (remainder 1). Every power of 10 leaves a remainder of exactly 1 when divided by 3. Therefore, the value of any digit multiplied by its place value will leave a remainder equal to the digit itself. Summing these digits gives the total remainder for the whole number.
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1. If this difference is 11 or a multiple of 11, what does that say about the remainder obtained when the number is divisible by 11?
It means the remainder is 0. The number is exactly divisible by 11.
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2. Using this shortcut, find out whether the following numbers are divisible by 11. Further, find the remainder if the number is not divisible by 11.
(i) 158 (ii) 841 (iii) 481 (iv) 5529 (v) 90904 (vi) 857076(i) 158: 8 - 5 + 1 = 4. (No, remainder 4).
(ii) 841: 1 - 4 + 8 = 5. (No, remainder 5).
(iii) 481: 1 - 8 + 4 = -3. Remainder = 11 - 3 = 8. (No, remainder 8).
(iv) 5529: 9 - 2 + 5 - 5 = 7. (No, remainder 7).
(v) 90904: 4 - 0 + 9 - 0 + 9 = 22. (Yes, divisible by 11).
(vi) 857076: 6 - 7 + 0 - 7 + 5 - 8 = -11. (Yes, divisible by 11).
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1. Is this method similar to or different from the method we saw just before?
It is essentially the exact same method mathematically. Placing alternating '+' and '-' signs starting from the units digit achieves the exact same result as summing alternate digit positions and taking their difference.
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2. Fill in the following table.
Number 2 3 4 5 6 8 9 10 11 128 Yes No Yes No No Yes No No No 990 Yes Yes No Yes Yes No Yes Yes Yes 1586 Yes No No No No No No No No 275 No No No Yes No No No No Yes 6686 Yes No No No No No No No No 639210 Yes Yes No Yes Yes No No Yes Yes 429714 Yes Yes No No Yes No Yes No No 2856 Yes Yes Yes No Yes Yes No No No 3060 Yes Yes Yes Yes Yes No Yes Yes No 406839 No Yes No No No No No No Yes
More on Divisibility Shortcuts
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1. How can we find out if a number is divisible by 6? Will checking its divisibility by its factors 2 and 3 work? Use the shortcuts for 2 and 3 on these numbers and divide each number by 6 to verify - 38, 225, 186, 64.
Yes, a number is divisible by 6 if it is divisible by both 2 (even) and 3 (sum of digits is a multiple of 3).
38: Divisible by 2, not 3. (Not div by 6).
225: Not div by 2. (Not div by 6).
186: Div by 2 (ends in 6), Div by 3 (1+8+6=15). (Yes, divisible by 6).
64: Div by 2, not 3. (Not div by 6). -
2. How about checking divisibility by 24? Will checking the divisibility by its factors, 4 and 6, work? Why or why not? Explain using prime factorisation why checking divisibility by 3 and 8 works for checking divisibility by 24, but checking divisibility by 4 and 6 is not sufficient for checking divisibility by 24.
Checking 4 and 6 does not work because they are not co-prime (they share a common factor of 2). For instance, 12 is divisible by both 4 and 6, but not 24.
3 and 8 work perfectly because they are co-prime; their prime factorisations share no common factors (3 = 31, 8 = 23). Thus, if a number contains all the prime factors of 3 and 8, it must naturally contain all the factors of their product, 24. -
3. What property do you think this digital root will have?
The digital root of a number is equal to the exact remainder obtained when the number is divided by 9. If the digital root is 9, the number is exactly divisible by 9.
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4. Between the numbers 600 and 700, which numbers have the digital root: (i) 5, (ii) 7, (iii) 3?
(i) 5: 608, 617, 626, 635, 644, 653, 662, 671, 680, 689, 698.
(ii) 7: 601, 610, 619, 628, 637, 646, 655, 664, 673, 682, 691.
(iii) 3: 606, 615, 624, 633, 642, 651, 660, 669, 678, 687, 696. -
5. Write the digital roots of any 12 consecutive numbers. What do you observe?
The digital roots follow a rigid, repeating sequential pattern: 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3...
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1. Now, find the digital roots of some consecutive multiples of (i) 3, (ii) 4, and (iii) 6.
(i) Multiples of 3: 3, 6, 9, 3, 6, 9...
(ii) Multiples of 4: 4, 8, 3, 7, 2, 6, 1, 5, 9...
(iii) Multiples of 6: 6, 3, 9, 6, 3, 9... -
2. What are the digital roots of numbers that are 1 more than a multiple of 6? What do you notice? Try to explain the patterns noticed.
The numbers are 7, 13, 19, 25... The digital root pattern is 7, 4, 1, 7, 4, 1...
This happens because the digital roots of multiples of 6 follow the strict pattern 6, 3, 9. Adding 1 to each of these roots gives 7, 4, 10(which reduces to 1). -
3. The largest odd single-digit I proudly claim. What's my number? What's my name?
The largest odd single digit is 9. The digital root is 9.
Figure it Out - 5.3
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1. The digital root of an 8-digit number is 5. What will be the digital root of 10 more than that number?
Adding 10 is equivalent to adding a digital root of 1 (since 1+0=1). So, 5 + 1 = 6. The new digital root is 6.
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2. Write any number. Generate a sequence of numbers by repeatedly adding 11. What would be the digital roots of this sequence of numbers? Share your observations.
Adding 11 adds 2 to the digital root (1+1=2). The sequence of digital roots will go up by 2 each time, looping after 9. (e.g., if starting at 1, the roots will be: 1, 3, 5, 7, 9, 2, 4, 6, 8...).
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3. What will be the digital root of the number 9a + 36b + 13?
9a and 36b are multiples of 9, so their digital roots are 9 (which acts like a 0 in digital root addition). The overall digital root is solely determined by the remaining term, 13. The digital root of 13 is 1+3 = 4.
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4. Make conjectures by examining if there are any patterns or relations between
(i) the parity of a number and its digital root.
(ii) the digital root of a number and the remainder obtained when the number is divided by 3 or 9.(i) There is no relation between parity and digital root. An even number can have any digital root, and an odd number can have any digital root.
(ii) The digital root IS the exact remainder when divided by 9 (if the root is 9, the remainder is 0). If you take the digital root modulo 3, you get the exact remainder when divided by 3.
5.3 Digits in Disguise
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1. Solve the cryptarithms given below.
(i) A1 + 1B = B0
(ii) AB + 37 = 6A
(iii) ON + ON = PO
(iv) GR + QR = PRR(i) Units: 1 + B ends in 0 → B = 9 (carry 1). Tens: A + 1 + 1 = 9 → A = 7. (71 + 19 = 90).
(ii) Units: B + 7 ends in A. Tens: A + 3 (+carry) = 6. If A=2, then B+7=12 (B=5, carry 1). Tens check: 2+3+1 = 6. (25 + 37 = 62). So A=2, B=5.
(iii) O cannot be 0. O must be even (N+N). Let O=2. Units: N+N=12 → N=6 (carry 1). Tens: 2+2+1 = 5 = P. (26 + 26 = 52). O=2, N=6, P=5.
(iv) Units: R+R=R → R=0. Tens: G+Q=10 (with carry 1 to make P=1). So R=0, P=1, and G & Q are any distinct numbers adding to 10. -
2. (vi) Try this now: GH × H = 9K. Pick the solution to this question from the options given below.
From the options, the only one that matches the pattern (H × H ends in K) and reaches the 90s is 24 × 4 = 96. So, G=2, H=4, K=6.
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3. (vii) BYE × 6 = RAY. What can you say about 'Y'? What digits are possible/not possible?
Y must be an even number because multiplying any number by 6 always results in an even product. So Y can be 0, 2, 4, 6, 8. As stated, Y cannot be 7 or more.
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4. Solve the following:
(i) UT × 3 = PUT
(ii) AB × 5 = BC
(iii) L2N × 2 = 2NP
(iv) XY × 4 = ZX
(v) PP × QQ = PRP
(vi) JK × 6 = KKK(i) P=1, U=5, T=0 (50 × 3 = 150).
(ii) A=1, B=2, C=0 (12 × 5 = 60) or A=2, B=5, C=0 (25 × 5 = 125, wait A=2, B=5 no). 10 × 5 = 50. Let's use 10. (10 × 5 = 50).
(iii) L=1, N=4, P=8 (124 × 2 = 248).
(iv) X=8, Y=2, Z=3 (82 × 4 = 328).
(v) P=2, Q=1, R=4 (22 × 11 = 242).
(vi) J=7, K=4 (74 × 6 = 444).
Figure it Out - 5.4
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1. If 31z5 is a multiple of 9, where z is a digit, what is the value of z? Explain why there are two answers to this problem.
The sum of the digits is 3+1+z+5 = 9+z. For this to be a multiple of 9, z can be 0 (sum=9) or 9 (sum=18). There are two answers because the sum of the known digits is exactly 9, leaving space for the unknown digit to be 0 or 9 without breaking the rule.
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2. "I take a number that leaves a remainder of 8 when divided by 12. I take another number which is 4 short of a multiple of 12. Their sum will always be a multiple of 8", claims Sonu. Examine his claim and justify your conclusion.
Sonu's claim is Never True. First number: 12a + 8. Second number: 12b - 4. Their sum is 12a + 8 + 12b - 4 = 12(a+b) + 4. This sum leaves a remainder of 4 when divided by 12. It will not always be a multiple of 8 (e.g. 20 + 8 = 28, not div by 8).
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3. When is the sum of two multiples of 3, a multiple of 6 and when is it not? Explain the different possible cases, and generalise the pattern.
Multiples of 3 can be odd (e.g., 3, 9) or even (e.g., 6, 12). The sum of two even multiples is even (a multiple of 6). The sum of two odd multiples is even (a multiple of 6). However, the sum of an odd multiple and an even multiple is odd, so it is NOT a multiple of 6.
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4. Sreelatha says, "I have a number that is divisible by 9. If I reverse its digits, it will still be divisible by 9".
(i) Examine if her conjecture is true for any multiple of 9.
(ii) Are any other digit shuffles possible such that the number formed is still a multiple of 9?(i) Yes, always true. The divisibility rule for 9 depends only on the total sum of the digits. Reversing the digits does not alter their sum.
(ii) Yes, any permutation or random shuffle of the digits will result in a number that is still a multiple of 9. -
5. If 48a23b is a multiple of 18, list all possible pairs of values for a and b.
It must be divisible by 2 (so b is even: 0, 2, 4, 6, 8) and 9 (sum = 17+a+b is a multiple of 9).
If b=0, a=1.
If b=2, a=8.
If b=4, a=6.
If b=6, a=4.
If b=8, a=2. -
6. If 3p7q8 is divisible by 44, list all possible pairs of values for p and q.
It must be divisible by 4 (q8 must be div by 4 → q ∈ {0, 2, 4, 6, 8}) and by 11 ((8+7+3) - (q+p) = 18 - (p+q) must be a multiple of 11 → p+q = 7).
Pairs: (p=7, q=0), (p=5, q=2), (p=3, q=4), (p=1, q=6). -
7. Find three consecutive numbers such that the first number is a multiple of 2, the second number is a multiple of 3, and the third number is a multiple of 4. Are there more such numbers? How often do they occur?
Example: 2, 3, 4. Or 14, 15, 16. Yes, there are infinitely many. They occur exactly once every 12 numbers.
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8. Write five multiples of 36 between 45,000 and 47,000. Share your approach with the class.
Approach: 45000 / 36 = 1250. So 1250 × 36 = 45000 exactly. Multiples are 45036, 45072, 45108, 45144, 45180.
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9. The middle number in the sequence of 5 consecutive even numbers is 5p. Express the other four numbers in sequence in terms of p.
5p - 4, 5p - 2, 5p + 2, 5p + 4.
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10. Write a 6-digit number that it is divisible by 15, such that when the digits are reversed, it is divisible by 6.
Example: 400005. It ends in 5 and the sum is 9 (divisible by 15). Reversed, it is 500004, which is even and the sum is 9 (divisible by 6).
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11. Deepak claims, "There are some multiples of 11 which, when doubled, are still multiples of 11. But other multiples of 11 don't remain multiples of 11 when doubled". Examine if his conjecture is true; explain your conclusion.
Deepak's claim is False. 11 is prime. Doubling a multiple of 11 merely multiplies it by 2. The prime factor of 11 remains perfectly intact. Thus, EVERY multiple of 11 remains a multiple of 11 when doubled.
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12. Determine whether the statements below are 'Always True', 'Sometimes True', or 'Never True'. Explain your reasoning.
(i) The product of a multiple of 6 and a multiple of 3 is a multiple of 9.
(ii) The sum of three consecutive even numbers will be divisible by 6.
(iii) If abcdef is a multiple of 6, then badcef will be a multiple of 6.
(iv) 8(7b-3) - 4(11b+1) is a multiple of 12.(i) Always True. (6a) × (3b) = 18ab = 9(2ab).
(ii) Always True. 2n + (2n+2) + (2n+4) = 6n+6 = 6(n+1).
(iii) Always True. Reversing the first two digits doesn't change the sum of the digits (so it remains div by 3). The units digit is still f. Since it was divisible by 6 initially, f is even. It remains even and div by 3, so it is div by 6.
(iv) Never True. 56b - 24 - 44b - 4 = 12b - 28. Since 28 is not a multiple of 12, the expression always leaves a remainder. -
13. Choose any 3 numbers. When is their sum divisible by 3? Explore all possible cases and generalise.
Their sum is divisible by 3 when all three numbers have the exact same remainder when divided by 3 (e.g., 1+4+7=12), or when they have all three different remainders (0, 1, and 2) (e.g., 3+4+5=12).
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14. Is the product of two consecutive integers always multiple of 2? Why? What about the product of these consecutive integers? Is it always a multiple of 6? Why or why not? What can you say about the product of 4 consecutive integers? What about the product of five consecutive integers?
Yes, 2 consecutive integers always include one even number, making the product a multiple of 2.
The product of 3 consecutive integers is always a multiple of 6 (it must contain at least one even number and exactly one multiple of 3).
The product of 4 consecutive integers is always divisible by 24 (4!).
The product of 5 consecutive integers is always divisible by 120 (5!). -
15. Solve the cryptarithms -
(i) EF × E = GGG
(ii) WOW × 5 = MEOW(i) E=3, F=7, G=1 (37 × 3 = 111).
(ii) M=2, E=8, O=7, W=5 (575 × 5 = 2875). -
16. Which of the following Venn diagrams captures the relationship between the multiples of 4, 8, and 32?
Figure (i). 32 is a multiple of 8, which is a multiple of 4. So the set of multiples of 32 is completely nested inside multiples of 8, which is completely nested inside multiples of 4 (concentric circles).
Think and Disscuss
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1. The Palindrome Secret - 4-Digit Number
A four-digit "palindrome" reads the same forward and backward, like 5225 or 7117. Check if 5225 is divisible by 11. Check if 7117 is divisible by 11. Is every four-digit palindrome a multiple of 11?5225 / 11 = 475. 7117 / 11 = 647. Yes, every 4-digit palindrome is a multiple of 11. Let the number be 'abba'. The sum of the odd places = a+b. The sum of the even places = b+a. The difference is exactly 0, so it is always perfectly divisible by 11. -
2. The Repeating Triple
Write down a six-digit number where the first three digits repeat exactly, such as 123,123 or 456,456. Divide your number by 11. Is there a remainder?No, there is no remainder. Example: 123123 / 11 = 11193. This works because any number of the form 'abcabc' can be factored as abc × 1001. And 1001 is exactly divisible by 11 (1001 = 7 × 11 × 13).
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