Divisibility Questions with Solutions | Set 1 (Q1 to Q10)

This set contains Divisibility Questions with Solutions — Set 1 (Q1 to Q10) covering divisibility rules, divisibility tests and mixed problems — from basic to advanced — exactly as asked in real competitive exams.

Solutions are written in a simple, step-by-step notebook style for easy self-study and quick understanding. Each solution is broken down step by step so even the toughest question feels easy. These questions are hand-picked for students preparing for SSC CGL, SSC CHSL, CAT, Bank PO, Bank Clerk, UPSC CSAT, Railway RRB, AMCAT, eLitmus, TCS NQT and all campus placement aptitude tests. International students preparing for GRE, GMAT, SAT, ACT, MAT and all Numerical Reasoning Tests will find these equally useful.

✏️ Attempt each question on your own first — then check the solution below.

1. If number 4696×24 is divisible by 9 then find the value of x.

2. When 235 is added to 6×9, the result is 8y4. If 8y4 is divisible by 3, what is the largest possible value of x ?

3. If 6823x is divisible by 11, find the value of x.

4. Both the end digit of a 99 digit number N are 3. If N is divisible by 11 then which one of the following can be middle digit.
(i) 3      (ii) 4         (iii) 5          (iv) 6

5. Both the end digits of a 100 digit number N are 3. N is divisible by 11. Then which one of the following can be any digit.
(i) 3         (ii) 4          (iii) Any digit       (iv) 5

6. (2⁸¹ + 2⁸² + 2⁸³ + 2⁸⁴) is divisible by
(i) 13         (ii) 9         (iii) 10         (iv) 11

7. A 4-digit number is formed by repeating a 2-digit number such as 2828, 5252 etc. Any number of this form is exactly divisible by:
(i) 9 only             (ii) 11 only            (iii) 13 only       (iv) smallest prime number of 3 digit

8. A 8-digit number is formed by repeating a four digit number; for example 25892589, 87618761 etc. Any number of t his form is always exactly divisible by
(i) 7 only       (ii) 11 only        (iii) 13 only            (iv) 10001

9. Which of the following number will always divide a six digit number ababab when 1 ≤ a ≤ 9 & 1 ≤ b ≤ 9
(i) 101         (ii) 10101         (iii) 1001         (iv) 11010

10. The divisor is 28 times the quotient and 4 times the remainder. If the quotient is 12, Find the dividend.

1. If number 4696×24 is divisible by 9 then find the value of x.
Sol:
(4 + 9 + 6 + 6 + x+ 2 + 4) = (31 + x) is also divisible by 9.
Hence x = 5     Answer

2. When 235 is added to 6×9, the result is 8y4. If 8y4 is divisible by 3, what is the largest possible value of x ?
Sol:
8y4 is divisible by 3
∴ 8 + y + 4 = (12 + y) is divisible by 3
∴ possible values of y = 0, 3, 6, 9
now 235 + 6×9 = 8y4
⟹ 6×9 = 8y4 – 235
only for y = 9 we get 894 – 235 = 659 which is of for 6×9.
So max value of x = 5    Answer

3. If 6823x is divisible by 11, find the value of x.
Sol:

∴ which makes number divisible by 11
∴ x = 3      Answer

4. Both the end digit of a 99 digit number N are 3. If N is divisible by 11 then which one of the following can be middle digit.
(i) 3      (ii) 4         (iii) 5          (iv) 6
Sol:
Out of total 99 digits \(\frac{{99 + 1}}{2}\) = 50 digits will be at odd place & 49 digits will be at even places.
Out fo 50 digits which are at odd places 1ˢᵗ place and last place digits are 3.
∴ 3 + 3 + 48x = 49x
x = 6     Answer

5. Both the end digits of a 100 digit number N are 3. N is divisible by 11. Then which one of the following can be any digit.
(i) 3         (ii) 4          (iii) Any digit       (iv) 5
Sol:
Total digits are 100 ⟹ there will be 50 odd places and 50 even places
1ˢᵗ & 100ᵗʰ place digit is 3

6. (2⁸¹ + 2⁸² + 2⁸³ + 2⁸⁴) is divisible by
(i) 13         (ii) 9         (iii) 10         (iv) 11
Sol:
2⁸¹(1 + 2 + 4 + 8) = 2⁸¹ × 15 = 2⁸¹ ×3 × 5 = 2⁸⁰ × 3 × 10
∴ Divisible by (iii) 10 ✅      Answer

7. A 4-digit number is formed by repeating a 2-digit number such as 2828, 5252 etc. Any number of this form is exactly divisible by:
(i) 9 only             (ii) 11 only            (iii) 13 only       (iv) smallest prime number of 3 digit
Sol:
Let digit at unit place = x
& digit at 10’s place = y
∴ number = 1000x + 100y + 10x + y
 = 1010x + 101y
= 101(10x + y) ⟵ divisible by 101
(iv) divisible by smallest 3-digit prime number     Answer

8. A 8-digit number is formed by repeating a four digit number; for example 25892589, 87618761 etc. Any number of t his form is always exactly divisible by
(i) 7 only       (ii) 11 only        (iii) 13 only            (iv) 10001
Sol:
abcdabcd
= 10⁷a + 10⁶b + 10⁵c + 10⁴d + 10³a + 10² b + 10c + d
= 10001 × 10³a + 10001 × 10²b + 1001 × 10c + 10001d
= 10001(10³a + 10²b + 10c + d) ⟵ divisible by 10001
(i) 10001 ✅    Answer

9. Which of the following number will always divide a six digit number ababab when 1 ≤ a ≤ 9 & 1 ≤ b ≤ 9
(i) 101         (ii) 10101         (iii) 1001         (iv) 11010
Sol:
ababab
= ab × 10000 + ab × 100 + ab × 1
= 10101ab
⟹ (ii)10101 ✅      Answer

10. The divisor is 28 times the quotient and 4 times the remainder. If the quotient is 12, Find the dividend.
Sol:
divisor = 28Q = 4R
Q = 12 ⟹ 28 × 12 = 4R ⟹ R = 28 × 3
∴ dividend = divisor × Q = R
= 28 × 12 × 12 + 28 × 3
= 4116    Answer

✅ Well done on completing Set 1!

Continue practising with Divisibility Questions 11 to 20 → Set 2 or revisit the Divisibility Concept Page to strengthen your formulas and tricks before moving ahead.

Consistent practice is the key to mastering divisibility for SSC CGL, SSC CHSL, CAT, Bank PO, Bank Clerk, UPSC CSAT, Railway RRB, AMCAT, eLitmus, TCS NQT and international exams including GRE, GMAT, SAT, ACT, MAT and all Numerical Reasoning Tests. Want to understand the concept better? Read about Divisibility Rules on Wikipedia before attempting the next set.

This page is part of our complete series of divisibility questions with solutions for competitive exams — covering every question type from basic to advanced so you can build speed, accuracy and confidence. Practising these questions regularly will also strengthen your core divisibility concept before your exam day.