Class 6 | Exercise 3.3

1. Using divisibility tests, determine which numbers are divisible by 2, 3, 4, 5, 6, 8, 9, 10, 11.

Quick Divisibility Rules Reminder:

Divisibility by 2: Last digit even (0,2,4,6,8)
Divisibility by 3: Sum of digits divisible by 3
Divisibility by 4: Last 2 digits form a number divisible by 4
Divisibility by 5: Last digit 0 or 5
Divisibility by 6: Divisible by both 2 and 3
Divisibility by 8: Last 3 digits divisible by 8
Divisibility by 9: Sum of digits divisible by 9
Divisibility by 10: Last digit 0
Divisibility by 11: (Sum of digits at odd places) − (Sum at even places) = 0 or divisible by 11

Number Divisible	By 2	By 3	By 4	By 5	By 6	By 8	By 9	By 10	By 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	No

2. Using divisibility tests, determine which of the following numbers are divisible by 4; by 8:

Rule:
• Divisible by 4 → last 2 digits divisible by 4.
• Divisible by 8 → last 3 digits divisible by 8.

(a) 572
Last 2 digits = 72 → \(72 \div 4 = 18\) → divisible by 4 → Yes
Last 3 digits = 572 → \(572 \div 8 = 71.5\) → not divisible by 8 → No

(b) 726352
Last 2 digits = 52 → \(52 \div 4 = 13\) → Yes
Last 3 digits = 352 → \(352 \div 8 = 44\) → Yes

(c) 5500
Last 2 digits = 00 → \(0 \div 4 = 0\) → Yes
Last 3 digits = 500 → \(500 \div 8 = 62.5\) → No

(d) 6000
Last 2 digits = 00 → Yes
Last 3 digits = 000 → \(0 \div 8 = 0\) → Yes

(e) 12159
Last 2 digits = 59 → \(59 \div 4 = 14.75\) → No
Last 3 digits = 159 → \(159 \div 8 = 19.875\) → No

(f) 14560
Last 2 digits = 60 → \(60 \div 4 = 15\) → Yes
Last 3 digits = 560 → \(560 \div 8 = 70\) → Yes

(g) 21084
Last 2 digits = 84 → \(84 \div 4 = 21\) → Yes
Last 3 digits = 084 = 84 → \(84 \div 8 = 10.5\) → No

(h) 31795072
Last 2 digits = 72 → \(72 \div 4 = 18\) → Yes
Last 3 digits = 072 = 72 → \(72 \div 8 = 9\) → Yes

(i) 1700
Last 2 digits = 00 → Yes
Last 3 digits = 700 → \(700 \div 8 = 87.5\) → No

(j) 2150
Last 2 digits = 50 → \(50 \div 4 = 12.5\) → No
Last 3 digits = 150 → \(150 \div 8 = 18.75\) → No

3. Using divisibility tests, determine which of the following numbers are divisible by 6:

Rule: A number is divisible by 6 if it is divisible by both 2 and 3.
→ Check: (i) last digit even? (ii) sum of digits divisible by 3?

(a) 297144
Even? Yes (ends with 4)
Sum = \(2+9+7+1+4+4 = 27\) → \(27 \div 3 = 9\) → divisible by 3 → Yes

(b) 1258
Even? Yes
Sum = \(1+2+5+8 = 16\) → \(16 \div 3 \approx 5.33\) → not divisible by 3 → No

(c) 4335
Even? No (ends with 5) → No

(d) 61233
Even? No (ends with 3) → No

(e) 901352
Even? Yes
Sum = \(9+0+1+3+5+2 = 20\) → \(20 \div 3 \approx 6.67\) → No

(f) 438750
Even? Yes
Sum = \(4+3+8+7+5+0 = 27\) → divisible by 3 → Yes

(g) 1790184
Even? Yes
Sum = \(1+7+9+0+1+8+4 = 30\) → \(30 \div 3 = 10\) → Yes

(h) 12583
Even? No → No

(i) 639210
Even? Yes
Sum = \(6+3+9+2+1+0 = 21\) → divisible by 3 → Yes

(j) 17852
Even? Yes
Sum = \(1+7+8+5+2 = 23\) → \(23 \div 3 \approx 7.67\) → No

4. Using divisibility tests, determine which of the following numbers are divisible by 11:

Rule: Find (sum of digits at odd places) − (sum at even places).
Count places from the right (units = place 1).
If result is 0 or divisible by 11, then the number is divisible by 11.

Note: Odd numbers are highlighed in yellow colour and Even numbers in green color

(a) 5445

Digits (right to left): 5(1), 4(2), 4(3), 5(4)
Odd places (1,3): \(5 + 4 = 9\)
Even places (2,4): \(4 + 5 = 9\)
Difference = \(9 - 9 = 0\) → Yes

(b) 10824

Digits: 4(1), 2(2), 8(3), 0(4), 1(5)
Odd: \(4 + 8 + 1 = 13\)
Even: \(2 + 0 = 2\)
Difference = \(13 - 2 = 11\) → divisible by 11 → Yes

(c) 7138965

Digits: 5(1), 6(2), 9(3), 8(4), 3(5), 1(6), 7(7)
Odd: \(5 + 9 + 3 + 7 = 24\)
Even: \(6 + 8 + 1 = 15\)
Difference = \(24 - 15 = 9\) → not divisible by 11 → No

(d) 70169308

Digits: 8(1), 0(2), 3(3), 9(4), 6(5), 1(6), 0(7), 7(8)
Odd: \(8 + 3 + 6 + 0 = 17\)
Even: \(0 + 9 + 1 + 7 = 17\)
Difference = \(17 - 17 = 0\) → Yes

(e) 10000001

Digits: 1(1), 0(2), 0(3), 0(4), 0(5), 0(6), 0(7), 1(8)
Odd: \(1 + 0 + 0 + 0 = 1\)
Even: \(0 + 0 + 0 + 1 = 1\)
Difference = \(1 - 1 = 0\) → Yes

(f) 901153

Digits: 3(1), 5(2), 1(3), 1(4), 0(5), 9(6)
Odd: \(3 + 1 + 0 = 4\)
Even: \(5 + 1 + 9 = 15\)
Difference = \(4 - 15 = -11\) → |−11| = 11 → divisible by 11 → Yes

5. Write the smallest and greatest digit in the blank so the number is divisible by 3.

Rule: We know, a Number is divisible by 3 if sum of all digits is divisible by 3.

(a) __ 6724
Sum of given digits: __+ \(6 + 7 + 2 + 4 = 19\)
We need \(19 + x\) divisible by 3.
We will check all numbers from 1 to 9 for \(x\):
\(19 + 1 = 20\) (No), \(19 + 2 = 21\) (Yes), \(19 + 3 = 22\) (No), \(19 + 4 = 23\) (No), \(19 + 5 = 24\) (Yes), \(19 + 6 = 25\) (No), \(19 + 7 = 26\) (No), \(19 + 8 = 27\) (Yes), \(19 + 9 = 28\) (No)
So Possible numbers are: 2, 5 and 8

Smallest = 2, Greatest = 8

(b) 4765 __ 2
Sum of given digits: \(4 + 7 + 6 + 5 + 2 = 24\)
Let the missing digit be \(x\). Total sum = \(24 + x\)
We need \(24 + x\) divisible by 3.
Since 24 is already divisible by 3, \(x\) must also be divisible by 3.
Possible digits for \(x\) are: 0, 3, 6, 9
Check:
\(24 + 0 = 24\) (Yes), \(24 + 3 = 27\) (Yes), \(24 + 6 = 30\) (Yes), \(24 + 9 = 33\) (Yes)

Therefore, Smallest digit = 0 & Greatest digit = 9

6. Write a digit in the blank so the number is divisible by 11.

Rule for 11: (Sum of digits at odd places) − (Sum at even places) = 0 or multiple of 11.

(a) 92 __ 389

Label positions from right (1 = units):
Digits: 9 (6), 2 (5), x (4), 3 (3), 8 (2), 9 (1)
Odd places (1,3,5): \(9 + 3 + 2 = 14\)
Even places (2,4,6): \(8 + x + 9 = 17 + x\)
Difference = \(|14 - (17 + x)| = | -3 - x | = x + 3\)
Set \(x + 3 = 0\) or 11 → \(x = 8\) (since \(8+3=11\))

So the missing digit = 8

(b) 8 __ 9484

Digits: 8 (6), x (5), 9 (4), 4 (3), 8 (2), 4 (1)
Odd places: \(4 + 4 + x = 8 + x\)
Even places: \(8 + 9 + 8 = 25\)
Difference = \(|(8 + x) - 25| = |x - 17|\)
Set = 0 or 11 → \(x = 6\) (since \(|6 - 17| = 11\))

So the missing digit = 6

You’ve mastered divisibility! These patterns make big numbers easy to handle. Keep going! 🌟