Figure it out

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1. The sum of four consecutive numbers is 34. What are these numbers?

Let the four consecutive numbers be:

x, x + 1, x + 2, x + 3

According to the question:

x + (x + 1) + (x + 2) + (x + 3) = 34

4x + 6 = 34

4x = 28

x = 7

Therefore, the four numbers are:

7, 8, 9 and 10

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2. Suppose p is the greatest of five consecutive numbers. Describe the other four numbers in terms of p.

Since p is the greatest number, the four numbers before it are:

• p − 1
• p − 2
• p − 3
• p − 4

Therefore, the five consecutive numbers are:

p − 4, p − 3, p − 2, p − 1, p

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3. Determine whether each statement is always true, sometimes true, or never true.

(i) The sum of two even numbers is a multiple of 3.

Let the two even numbers be 2a and 2b.

Their sum is:

2a + 2b = 2(a + b)

The sum is always even, but it is not always a multiple of 3.

Example:

2 + 4 = 6 (multiple of 3)

Non-example:

4 + 8 = 12 (multiple of 3), but 2 + 8 = 10 (not a multiple of 3)

Therefore, the statement is sometimes true.

(ii) If a number is not divisible by 18, then it is also not divisible by 9.

Consider the number 9.

9 is divisible by 9 but not divisible by 18.

So the statement fails.

Therefore, the statement is sometimes true.

(iii) If two numbers are not divisible by 6, then their sum is not divisible by 6.

Consider 1 and 5.

Neither 1 nor 5 is divisible by 6.

But:

1 + 5 = 6

which is divisible by 6.

Therefore, the statement is not always true.

Example:

1 + 5 = 6 (divisible by 6)

Non-example:

1 + 2 = 3 (not divisible by 6)

Hence, the statement is sometimes true.

(iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3.

Let the numbers be 6a and 9b.

Their sum is:

6a + 9b = 3(2a + 3b)

Since 3 is a factor, the sum is always divisible by 3.

Therefore, the statement is always true.

(v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.

Let the numbers be 6a and 3b.

Their sum is:

6a + 3b = 3(2a + b)

The sum is always divisible by 3, but not necessarily by 9.

Example:

6 + 3 = 9 (multiple of 9)

Non-example:

12 + 3 = 15 (not a multiple of 9)

Therefore, the statement is sometimes true.

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4. Find numbers that leave a remainder of 2 when divided by 3 and by 4.

Examples:

2, 14, 26, 38, 50, 62, ...

Each of these numbers gives remainder 2 when divided by 3 and also when divided by 4.

Since the LCM of 3 and 4 is 12, all such numbers can be written as:

12n + 2

where n is a whole number.

Therefore, the required algebraic expression is:

12n + 2

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5. Pebble Puzzle

Conditions:

• When grouped in 3's, remainder = 1
• When grouped in 2's, remainder = 1 (odd number)
• When grouped in 5's, remainder = 1
• When grouped in 7's, remainder = 0
• More than one hundred

Let the number be N.

Since N leaves remainder 1 when divided by 3, 5 and 2:

N − 1 is divisible by 2, 3 and 5.

LCM(2, 3, 5) = 30

Therefore:

N = 30k + 1

Also, N must be divisible by 7.

Checking values greater than 100:

30 × 4 + 1 = 121

121 ÷ 7 = 17 remainder 2

30 × 6 + 1 = 181

181 ÷ 7 = 25 remainder 6

30 × 7 + 1 = 211

211 ÷ 7 = 30 remainder 1

30 × 8 + 1 = 241

241 ÷ 7 = 34 remainder 3

30 × 9 + 1 = 271

271 ÷ 7 = 38 remainder 5

30 × 11 + 1 = 331

331 ÷ 7 = 47 remainder 2

30 × 14 + 1 = 421

421 ÷ 7 = 60 remainder 1

The smallest number greater than 100 satisfying all conditions is:

91 (but not greater than 100)

Next such number:

301

Check:

• 301 ÷ 2 leaves remainder 1
• 301 ÷ 3 leaves remainder 1
• 301 ÷ 5 leaves remainder 1
• 301 ÷ 7 = 43 exactly

Therefore, the number of pebbles is 301.

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6. Tathagat's Claim

Each number leaves a remainder of 2 when divided by 6.

So each number can be written as:

6a + 2, 6b + 2, 6c + 2

Their sum is:

(6a + 2) + (6b + 2) + (6c + 2)

= 6a + 6b + 6c + 6

= 6(a + b + c + 1)

Since 6 is a factor, the sum is always divisible by 6.

Therefore, Tathagat's claim is true.

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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, find the remainders.

Given:

661 = 7a + 3

4779 = 7b + 5

where a and b are whole numbers.

(i) 4779 + 661

Algebraic Method:

(7b + 5) + (7a + 3)

= 7a + 7b + 8

= 7(a + b + 1) + 1

Therefore, the remainder when divided by 7 is:

1

Visual Method:

661 leaves 3 extra objects after making groups of 7.

4779 leaves 5 extra objects after making groups of 7.

Adding the remainders:

3 + 5 = 8

From these 8 objects, one complete group of 7 can be formed, leaving:

8 − 7 = 1

Hence the remainder is 1.

(ii) 4779 − 661

Algebraic Method:

(7b + 5) − (7a + 3)

= 7b − 7a + 2

= 7(b − a) + 2

Therefore, the remainder when divided by 7 is:

2

Visual Method:

The remainder of 4779 is 5 and the remainder of 661 is 3.

Subtracting the remainders:

5 − 3 = 2

Therefore, after removing equal groups of 7, the remainder left is:

2

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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?

We need a number N such that:

• N leaves remainder 2 when divided by 3.
• N leaves remainder 3 when divided by 4.
• N leaves remainder 4 when divided by 5.

Notice that in each case the remainder is one less than the divisor.

Therefore:

N + 1 is divisible by 3, 4, and 5.

So N + 1 must be a common multiple of 3, 4, and 5.

The least common multiple (LCM) of 3, 4, and 5 is:

LCM(3, 4, 5) = 60

Hence:

N + 1 = 60

N = 59

Verification:

• 59 ÷ 3 = 19 remainder 2 ✓
• 59 ÷ 4 = 14 remainder 3 ✓
• 59 ÷ 5 = 11 remainder 4 ✓

Therefore, the smallest such number is:

59

Simple Explanation:

Since the remainders are 2, 3, and 4, the number is exactly one less than a multiple of 3, 4, and 5. The smallest number that is divisible by all three is 60. Therefore, one less than 60, which is 59, is the smallest number satisfying all the conditions.

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Figure it out

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1. Find, without dividing, whether the following numbers are divisible by 9.

Rule: A number is divisible by 9 if the sum of its digits is divisible by 9.

(i) 123

Sum of digits = 1 + 2 + 3 = 6

Since 6 is not divisible by 9, 123 is not divisible by 9.

(ii) 405

Sum of digits = 4 + 0 + 5 = 9

Since 9 is divisible by 9, 405 is divisible by 9.

(iii) 8888

Sum of digits = 8 + 8 + 8 + 8 = 32

Since 32 is not divisible by 9, 8888 is not divisible by 9.

(iv) 93547

Sum of digits = 9 + 3 + 5 + 4 + 7 = 28

Since 28 is not divisible by 9, 93547 is not divisible by 9.

(v) 358095

Sum of digits = 3 + 5 + 8 + 0 + 9 + 5 = 30

Since 30 is not divisible by 9, 358095 is not divisible by 9.

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2. Find the smallest multiple of 9 with no odd digits.

We need the smallest multiple of 9 whose digits are all even.

Checking multiples of 9:

• 9 → contains odd digit 9
• 18 → contains odd digit 1
• 27 → contains odd digit 7
• 36 → contains odd digit 3
• 45 → contains odd digit 5
• 54 → contains odd digit 5
• 63 → contains odd digit 3
• 72 → contains odd digit 7
• 81 → contains odd digit 1
• 90 → contains odd digit 9
• 108 → contains odd digit 1
• 126 → contains odd digit 1
• 144 → contains odd digit 1
• 162 → contains odd digit 1
• 180 → contains odd digit 1
• 198 → contains odd digits 1 and 9
• 216 → all digits are even ✓

Therefore, the smallest multiple of 9 with no odd digits is:

216

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3. Find the multiple of 9 that is closest to the number 6000.

Divide 6000 by 9 mentally:

9 × 666 = 5994

9 × 667 = 6003

The distances from 6000 are:

• 6000 − 5994 = 6
• 6003 − 6000 = 3

Since 3 is smaller than 6, the closest multiple of 9 is:

6003

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4. How many multiples of 9 are there between the numbers 4300 and 4400?

First multiple of 9 greater than 4300:

9 × 478 = 4302

Last multiple of 9 less than 4400:

9 × 488 = 4392

Therefore, the multiples are:

4302, 4311, 4320, ..., 4392

This is an arithmetic sequence with:

• First term = 4302
• Last term = 4392
• Common difference = 9

Number of terms:

= (4392 − 4302) ÷ 9 + 1

= 90 ÷ 9 + 1

= 10 + 1

= 11

Therefore, there are 11 multiples of 9 between 4300 and 4400.

Figure it out

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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?

The digital root of a number is the single digit obtained by repeatedly adding its digits.

A digital root is closely related to the remainder when the number is divided by 9.

Given:

Digital root of the number = 5

This means the number leaves a remainder of 5 when divided by 9.

When 10 is added:

10 ≡ 1 (mod 9)

So the new remainder becomes:

5 + 1 = 6

Therefore, the digital root of the new number is:

6

Example:

14 has digital root 5.

14 + 10 = 24

Digital root of 24 = 2 + 4 = 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? Share your observations.

Let us start with the number 25.

Sequence:

25, 36, 47, 58, 69, 80, 91, 102, ...

Their digital roots are:

Observation:

Adding 11 is equivalent to adding 2 modulo 9 because:

11 ≡ 2 (mod 9)

Therefore, each new digital root is obtained by adding 2 to the previous digital root (cycling through 1 to 9).

The digital roots follow a repeating pattern:

7, 9, 2, 4, 6, 8, 1, 3, 5, 7, ...

Thus, the digital roots repeat in a cycle.

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3. What will be the digital root of the number 9a + 36b + 13?

Consider:

9a + 36b + 13

Since 9a is a multiple of 9 and 36b is also a multiple of 9:

9a + 36b leaves remainder 0 when divided by 9.

Therefore, the digital root depends only on 13.

Digital root of 13:

1 + 3 = 4

Hence, the digital root of:

9a + 36b + 13

is always:

4

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4. Make conjectures by examining patterns or relations.

(i) Relation between the parity of a number and its digital root

Parity means whether a number is even or odd.

After examining many examples, we find that there is no fixed relationship between the parity of a number and the parity of its digital root.

Examples:

Conjecture:

There is no definite relationship between the parity of a number and the parity of its digital root.

(ii) Relation between the digital root and the remainder when a number is divided by 3 or 9

The digital root is directly related to divisibility by 9.

A number and the sum of its digits leave the same remainder when divided by 9.

Examples:

Conjecture:

• If the digital root is 9, the number is divisible by 9.
• If the digital root is 3, 6, or 9, the number is divisible by 3.
• The digital root gives the remainder when dividing by 9, except that a digital root of 9 corresponds to remainder 0.

Thus, digital roots provide a quick way to check divisibility by 3 and 9.

Figure it out

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1. If 31z5 is a multiple of 9, where z is a digit, what is the value of z?

A number is divisible by 9 if the sum of its digits is divisible by 9.

For the number 31z5:

Sum of digits = 3 + 1 + z + 5 = 9 + z

Since 9 + z must be a multiple of 9, possible values are:

9 + z = 9 or 18

If 9 + z = 9, then z = 0

If 9 + z = 18, then z = 9

Therefore, the possible values of z are:

z = 0 or z = 9

Verification:

• 3105 → 3 + 1 + 0 + 5 = 9 ✓
• 3195 → 3 + 1 + 9 + 5 = 18 ✓

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2. Snehal's Claim

One number leaves a remainder of 8 when divided by 12.

Let it be:

12a + 8

Another number is 4 short of a multiple of 12.

Let it be:

12b − 4

Their sum is:

(12a + 8) + (12b − 4)

= 12a + 12b + 4

= 12(a + b) + 4

= 4[3(a + b) + 1]

For the sum to be a multiple of 8, the bracket must be even.

But 3(a + b) + 1 is not always even.

Example:

a = 1, b = 1

Sum = 20 + 8 = 28

28 is not divisible by 8.

Therefore, Snehal's claim is not always true.

It is sometimes true.

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3. When is the sum of two multiples of 3 a multiple of 6?

Let the multiples of 3 be:

3a and 3b

Their sum is:

3a + 3b = 3(a + b)

For this sum to be divisible by 6, (a + b) must be even.

Case 1: Both multiples are even

Example:

6 + 12 = 18

18 is divisible by 6.

Case 2: Both multiples are odd

Example:

3 + 9 = 12

12 is divisible by 6.

Case 3: One even multiple and one odd multiple

Example:

6 + 9 = 15

15 is not divisible by 6.

Generalisation:

The sum of two multiples of 3 is divisible by 6 when both are even multiples or both are odd multiples.

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4. Sreelatha's Conjecture

(i) If a multiple of 9 is reversed, will it still be divisible by 9?

Yes.

Reversing the digits does not change the sum of the digits.

Since divisibility by 9 depends only on the sum of the digits, the reversed number is also divisible by 9.

Example:

234 = 2 + 3 + 4 = 9

Reverse:

432 = 4 + 3 + 2 = 9

Both are divisible by 9.

Therefore, the conjecture is true.

(ii) Are other digit shuffles possible?

Yes.

Any rearrangement of the digits keeps the digit sum unchanged.

Hence every digit shuffle remains divisible by 9.

Example:

729

Digit sum = 18

Possible shuffles:

• 792
• 279
• 297
• 927
• 972

All have digit sum 18 and are divisible by 9.

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5. If 48a23b is a multiple of 18, list all possible pairs of values for a and b.

A number is divisible by 18 if it is divisible by both 2 and 9.

Divisible by 2 ⇒ b must be even.

Sum of digits:

4 + 8 + a + 2 + 3 + b = 17 + a + b

This must be divisible by 9.

Therefore:

17 + a + b = 18 or 27

So:

a + b = 1 or 10

Since b must be even:

Possible pairs are:

Since b must be even, (0,1) is rejected.

For a + b = 10:

Therefore, the possible pairs are:

(1,0), (2,8), (4,6), (6,4), (8,2)

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6. If 3p7q8 is divisible by 44, list all possible pairs of values for p and q.

A number is divisible by 44 if it is divisible by both 4 and 11.

Step 1: Divisibility by 4

The last two digits are q8.

Possible values making q8 divisible by 4:

• 08
• 28
• 48
• 68
• 88

Therefore:

q = 0, 2, 4, 6, or 8

Step 2: Divisibility by 11

(3 + 7 + 8) − (p + q) must be a multiple of 11.

18 − (p + q)

Possible values are 0 or 11.

So:

p + q = 18 or 7

Checking possible values of q:

For p + q = 18:

Therefore, the possible pairs are:

(7,0), (5,2), (3,4), (1,6)

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7. Three consecutive numbers: first is a multiple of 2, second a multiple of 3, third a multiple of 4.

Let the numbers be:

n, n + 1, n + 2

Conditions:

• n is divisible by 2
• n + 1 is divisible by 3
• n + 2 is divisible by 4

Checking:

2, 3, 4 ✓

14, 15, 16 ✓

26, 27, 28 ✓

38, 39, 40 ✓

The pattern increases by 12 each time.

General form:

12k + 2, 12k + 3, 12k + 4

Therefore such numbers exist and occur every 12 numbers.

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8. Write five multiples of 36 between 45,000 and 47,000.

36 × 1250 = 45,000

Therefore:

• 45,036 = 36 × 1251
• 45,072 = 36 × 1252
• 45,108 = 36 × 1253
• 45,144 = 36 × 1254
• 45,180 = 36 × 1255

These are five multiples of 36 between 45,000 and 47,000.

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9. The middle number in the sequence of 5 consecutive even numbers is 5p. Express the other four numbers.

Consecutive even numbers differ by 2.

If the middle number is 5p, then:

Two numbers before it:

• 5p − 4
• 5p − 2

Two numbers after it:

• 5p + 2
• 5p + 4

Therefore the five numbers are:

5p − 4, 5p − 2, 5p, 5p + 2, 5p + 4

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10. Write a 6-digit number divisible by 15 such that when digits are reversed, it is divisible by 6.

Example:

123450

Check divisibility by 15:

• Ends in 0 ✓
• Digit sum = 1+2+3+4+5+0 = 15 ✓

Therefore 123450 is divisible by 15.

Reverse:

054321 = 54321

Not divisible by 6.

Let us choose:

124560

Digit sum = 18 and ends in 0.

So it is divisible by 15.

Reverse:

065421 = 65421

Ends in an odd digit, so not divisible by 6.

A correct example:

246810

Digit sum = 21 and ends in 0, so divisible by 15.

Reverse:

018642 = 18642

Ends in 2 and digit sum is 21.

Therefore 18642 is divisible by 6.

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11. Deepak's Conjecture about Multiples of 11

Let a multiple of 11 be:

11n

Doubling gives:

2 × 11n = 22n

= 11(2n)

Thus the result is still a multiple of 11.

Examples:

• 11 × 2 = 22
• 22 × 2 = 44
• 55 × 2 = 110

Therefore every multiple of 11 remains a multiple of 11 when doubled.

Deepak's second statement is incorrect.

The conjecture is true.

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12. Determine whether the statements are Always True, Sometimes True, or Never True.

(i) Product of a multiple of 6 and a multiple of 3 is a multiple of 9.

Example:

6 × 3 = 18 ✓

But:

6 × 6 = 36 ✓

18 × 6 = 108 ✓

However:

12 × 3 = 36 ✓

Not every product necessarily contains two factors of 3.

Example:

6 × 12 = 72 (not divisible by 9)

Therefore:

Sometimes True

(ii) Sum of three consecutive even numbers is divisible by 6.

Let numbers be:

2n, 2n+2, 2n+4

Sum:

6n + 6 = 6(n+1)

Therefore:

Always True

(iii) If abcdef is a multiple of 6, then badcef is a multiple of 6.

Rearranging digits may change divisibility by 3.

Sometimes it remains divisible by 6 and sometimes not.

Therefore:

Sometimes True

(iv) 8(7b − 3) − 4(11b + 1) is a multiple of 12.

Simplifying:

56b − 24 − 44b − 4

= 12b − 28

= 4(3b − 7)

For b = 1:

12 − 28 = −16

Not divisible by 12.

For b = 2:

24 − 28 = −4

Not divisible by 12.

Therefore:

Sometimes True

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13. Choose any 3 numbers. When is their sum divisible by 3?

A sum is divisible by 3 when the sum of the remainders obtained on division by 3 is itself divisible by 3.

Possible cases:

• 0 + 0 + 0
• 1 + 1 + 1
• 2 + 2 + 2
• 0 + 1 + 2

In all these cases, the total remainder is a multiple of 3.

Therefore, the sum is divisible by 3.

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14. Products of consecutive integers

(i) Two consecutive integers

One of them must be even.

Therefore the product is always divisible by 2.

Example:

5 × 6 = 30

(ii) Three consecutive integers

Among three consecutive integers:

• One is divisible by 3
• At least one is even

Therefore the product is always divisible by:

2 × 3 = 6

(iii) Four consecutive integers

Among four consecutive integers:

• One is divisible by 4
• Another is even
• One is divisible by 3

Therefore the product is always divisible by:

4 × 2 × 3 = 24

(iv) Five consecutive integers

Among five consecutive integers:

• One is divisible by 5
• One is divisible by 4
• One is divisible by 3
• At least two contribute factors of 2

Hence the product is always divisible by:

5 × 4 × 3 × 2 = 120

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15. Solve the Cryptarithms

(i) EF × E = GGG

Since GGG = 111G,

Trying digit values:

37 × 3 = 111

Therefore:

E = 3, F = 7, G = 1

Check:

37 × 3 = 111 ✓

(ii) WOW × 5 = MEOW

One valid solution is:

WOW = 186

186 × 5 = 930

This does not fit the pattern.

Trying systematically:

217 × 5 = 1085

Pattern:

WOW = 217

MEOW = 1085

Thus:

• W = 2
• O = 1
• M = 1
• E = 8

(Students are generally expected to explore and find a valid digit arrangement through trial and verification.)

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16. Which of the following Venn diagrams captures the relationship between the multiples of 4, 8, and 32?

Let us understand the relationship between these sets.

Every multiple of 32 is also a multiple of 8 because:

32 = 8 × 4

Examples:

• 32
• 64
• 96
• 128

All of these are multiples of 8.

Similarly, every multiple of 8 is also a multiple of 4 because:

8 = 4 × 2

Examples:

• 8
• 16
• 24
• 32

All of these are multiples of 4.

Therefore:

Multiples of 32 ⊂ Multiples of 8 ⊂ Multiples of 4

This means:

• The set of multiples of 32 lies completely inside the set of multiples of 8.
• The set of multiples of 8 lies completely inside the set of multiples of 4.
• The set of multiples of 4 is the largest set.

So the correct Venn diagram must have:

• Largest circle → Multiples of 4
• Middle circle → Multiples of 8
• Smallest circle → Multiples of 32

Among the given diagrams, this is shown in Diagram (iv).

Answer: (iv)

Reason:

Multiples of 32 are contained within multiples of 8, and multiples of 8 are contained within multiples of 4. Hence the sets are nested one inside another as shown in diagram (iv).