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