Aptitude - Number System Online Test, Number System Questions and Answers updated

Solution: 13601 = 87 x 156 + 29
Here 29 is the remainder.

Solution: Divisor = (5 x 46) = 230
10 x Quotient = 230 = 230 = 23 10 Dividend = (Divisor x Quotient) + Remainder = (230 x 23) + 46 = 5290 + 46 = 5336.

Solution: Let x = 296q + 75 = (37 x 8q + 37 x 2) + 1 = 37 (8q + 2) + 1 Thus, when the number is divided by 37, the remainder is 1.

Solution: Given Exp. = -84 x (30 - 1) + 365 = -(84 x 30) + 84 + 365 = -2520 + 449 = -2071

Solution: Clearly, 4864 is divisible by 4. So, 9P2 must be divisible by 3. So, (9 + P + 2) must be divisible by 3. P = 1.

Solution: Let x be the number and y be the quotient.
Then, x = 357 xy + 39
= (17 x 21 xy) + (17 x 2) + 5
= 17 x (21y + 2) + 5)
Required remainder = 5.

Solution: The smallest prime number is 2.

Solution: 7429 Let 8597 - x = 3071 -4358 Then, x = 8597 - 3071 ---- = 5526 3071 ----

Solution: Required numbers are 102, 108, 114, ... , 996
This is an A.P. in which a1 = 102, d = 6 and l = an = 996
Let the number of terms be n.
an= a1+ (n – 1)d
Then, a1+ (n – 1)d = 996
=> 102 + (n - 1) x 6 = 996
=> (n - 1) x 6 = 996 - 102
=>6 x (n - 1) = 894
=>(n - 1) = 149
=> n = 150.

Solution: Required numbers are 24, 30, 36, 42, ..., 96 This is an A.P. in which a = 24, d = 6 and l = 96 Let the number of terms in it be n.
Then tn = 96 a + (n - 1)d = 96 24 + (n - 1) x 6 = 96 (n - 1) x 6 = 72 (n - 1) = 12 n = 13 Required number of numbers = 13.

Solution: 3-digit number divisible by 6 are: 102, 108, 114,... , 996 This is an A.P. in which a = 102, d = 6 and l = 996 Let the number of terms be n. Then tn = 996.
Therefore, a + (n - 1)d = 996 => 102 + (n - 1) x 6 = 996 => 6 x (n - 1) = 894 => (n - 1) = 149 n = 150 Therefore Number of terms = 150.

Solution: Let the number be x and on dividing x by 5, we get k as quotient and 3 as remainder.
x = 5k + 3
x^2 = (5k + 3)^2
= (25k^2 + 30k + 9)
= 5(5k ^2 + 6k + 1) + 4
On dividing x^2 by 5, we get 4 as remainder.

Solution: Formula: (Divisor*Quotient) + Remainder = Dividend. Soln:
(56*Q)+29 = D -------(1) D%8 = R -------------(2) From equation(2), ((56*Q)+29)%8 = R. => Assume Q = 1. => (56+29)%8 = R. => 85%8 = R => 5 = R.

Solution: (Local value of 7) - (Face value of 7) = (70000 - 7) = 69993

Solution: Sum of digits = (4 + 8 + 1 + x + 6 + 7 + 3) = (29 + x), which must be divisible by 9. x = 7

Solution: x + 3699 + 1985 - 2047 = 31111 => x + 3699 + 1985 = 31111 + 2047 => x + 5684 = 33158 => x = 33158 - 5684 = 27474.

Solution: 24 = 3 x8, where 3 and 8 co-prime. Clearly, 35718 is not divisible by 8, as 718 is not divisible by 8. Similarly,
63810 is not divisible by 8 and 537804 is not divisible by 8. Consider option (D), Sum of digits = (3 + 1 + 2 + 5 + 7 + 3 + 6) = 27,
which is divisible by 3. Also, 736 is divisible by 8. 3125736 is divisible by (3 x 8), i.e., 24.

Solution: 19657 Let x - 53651 = 9999 33994 Then, x = 9999 + 53651 = 63650 ----- 53651 -----

Solution: (4 + 5 + 2) - (1 + 6 + 3) = 1, not divisible by 11. (2 + 6 + 4) - (4 + 5 + 2) = 1, not divisible by 11. (4 + 6 + 1) - (2 + 5 + 3) = 1,
not divisible by 11. (4 + 6 + 1) - (2 + 5 + 4) = 0, So, 415624 is divisible by 11.

Solution: The smallest 3-digit number is 100, which is divisible by 2. Therefore, 100 is not a prime number. ?
101 < 11 and 101 is not divisible by any of the prime numbers 2, 3, 5, 7, 11. 101 is a prime number.
Hence 101 is the smallest 3-digit prime number.