Number System - Aptitude MCQ Questions and Solutions with Explanations

I. We have (195 - 195) = 0
∴ 195195 is divisible by 7

II. We have (181 - 181) = 0
∴ 181181 is divisible by 7

III. We have (120 - 120) = 0
∴ 120120 is divisible by 7

IV. We have (891 - 891) = 0
∴ 891891 is divisible by 7

Hence, all are divisible by 7

Let ,   $$\frac{1x5y4}{148}$$ = 78
Then,
10000 + 1000x + 500 + 10y + 4 = 148 × 78
⇒ 10000 + 1000x + 500 + 10y + 4 = 11544
⇒ 10000 + 1000x + 500 + 10y + 4 = 10000 + 1000 + 500 + 40 + 4
⇒ 1000x = 1000
⇒ x = 1

The given series is such that the sum of first hundred terms is zero, and 101^st term is 2. So, the sum of 101 terms is 2.

We have : (2¹²ⁿ - 6⁴ⁿ)
= (2¹²ⁿ - 2⁴ⁿ × 3⁴ⁿ)
= 2⁴ⁿ (2⁸ⁿ - 3⁴ⁿ)
Putting n = 1, we get the number 2⁴ (2⁸ - 3⁴)
= 16 (256 - 81)
= (16 × 175)
= 2800
Hence, the number formed by last two digits is 00

Let the thousand's, hundred's, ten's, and one's digits be x, y, z, w respectively.
Then, the number is
1000x + 100y + 10z + w
= 10³x + 10²y + 10z + w

Unit digit of 7⁴ is 1.
So, the unit digit of (7⁴)²³ is 1
∴ Unit digit of 7⁹⁵
= Unit digit of (7⁹² × 7³) = (1 × 3) = 3
Unit digit of 3⁴ is 1
So, the unit digit of (3⁴)¹⁴ is 1
∴ Unit digit of 3⁵⁸
= unit digit of (3⁵⁶ × 3² ) = (1 × 9) = 9
Hence, the unit digit of
(7⁹⁵ - 3) = (13 - 9) = 4

Given ,
a + b + c = 6
ab + bc + ca = 10
∴ (a + b + c)² = 36
⇒ a² + b² + c² + 2ab + 2bc + 2ca = 36
⇒ a² + b² + c² + 2(ab + bc + ca) = 36
⇒ a² + b² + c² + 2 × 10 = 36
⇒ a² + b² + c² = 16
As we know :
$$ \Rightarrow \frac{{{a^3} + {b^3} + {c^3} - 3abc}}{{{a^2} + {b^2} + {c^2} - ab - bc - ca}}$$       $$ = \left( {a + b + c} \right)$$
$$\eqalign{ & \Rightarrow \frac{{{a^3} + {b^3} + {c^3} - 3abc}}{{16 - \left( {ab + bc + ca} \right)}} = 6 \cr & \Rightarrow \frac{{{a^3} + {b^3} + {c^3} - 3abc}}{{16 - 10}} = 6 \cr & \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 6 \times 6 \cr & \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 36 \cr} $$

Required number = 30005

We keep on dividing 33333... by 7 till we get 0 as remainder.

∴ Required number = 47619

If a and b are two rational numbers, then $$\frac{a + b}{2}$$   is a rational number lying between a and b.