number systems Model Questions & Answers, Practice Test for ibps so prelims

Answer: (c)

Here $(24)^2 < 589$

So, prime numbers less than 24 are 2, 3, 5, 7, 11, 13, 17, 19, 23.

Since, 589 is divisible by 19, then 589 is a composite number.

Answer: (c)

$6^23 × 75^9 × (105)^2$

= $2^23 × 3^23 × 3^9 × 5^9 × 5^9 × 3^2 × 7^2 × 5^2$

= $2^23 × 5^20 × 3^34 × 7^2$

= $2^23 × 5^20$

10 is divided by 2 and 5

Minimum of 20 and 23 is 20

∴ n = 20

Answer: (a)

Here N = $90 × 42 × 3^{2}4 × 55$

Now 90 = 3 × 3 × 10 = $3^2$ × 10

42 = 14 × 3 = 14 × $3^1$

$3^{2}4 = 3 × 3 × 3 × 3 × 4 = 3^4 × 4$

$5^5 = 11 × 5$

N = $3^2 × 3^1 × 3^4 × 10 × 14 × 4 × 11 × 5$

N = $3^7 × 10 × 14 × 4 × 11 × 5$

Maximum value of m = 7

∴ Option (b) is correct.

Answer: (c)

$(23)_5 = (2 × 5^1 × 3 × 5^0)_10 = (13)_10 = (1 × 8^1 + 5 × 8^0)_8 = (15)_8$

also, $(47)_9 = (4 × 9^1 + 7 × 9^0)_10 = (43)_10$

= $(5 × 8^1 + 3 × 8^0 ) = (53)_8$

sum = $(13)_10 + (43)_10 = (56)_10 → (70)_8$

Answer: (b)

⇒ 231228 = 7006 × 33 + 30 ... (1)

Now, when the number divides by 33 its remainder is 30.

Therefore, 3 must be added to 231228 to make it exactly divisible by 33.