Number System - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\left( {2464} \right)^{1793}} \times {\left( {615} \right)^{317}} \times {\left( {131} \right)^{491}} \cr & {4^1} \to 4 \to 4 \cr & {4^2} \to 16 \to 6 \cr & {4^3} \to 64 \to 4 \cr} $$
So, odd power of 4 will have 4 as unit digit and even power of 4 will have 6 as unit digit 5 and 1 have same unit digits respectively.
(2464)¹⁷⁹³ have the odd power therefore unit digit is 4
(615)³¹⁷ have unit digit 5 and
(131)⁴⁹¹ have unit digit 1.
i.e Unit digit of the product will be 4 × 5 × 1 = 20.
∴ 0 is the unit digit

Let the number be x
According to question
$$\eqalign{ & \left( {3 \times {x^2}} \right){\text{ - }}\left( {4 \times x} \right) = 50 + x.....(i) \cr & \Rightarrow 3{x^2} - 4x = 50 + x \cr & \Rightarrow 3{x^2} - 5x - 50 = 0 \cr & \Rightarrow 3{x^2} - 15x + 10x - 50 = 0 \cr & \Rightarrow 3{x^{}}(x - 5) + 10(x - 5) = 0 \cr & \Rightarrow (x - 5)(3x + 10) = 0 \cr & x = 5\,\,or\,\, - \frac{{10}}{3} \cr} $$
Since the natural number is x = 5

Shortcut method :
$$ \Rightarrow 3{x^2} - 4x = 50 + x.....(i)$$
Now put the value of x from option (b)
$$\eqalign{ & x = 5 \cr & 3 \times {(5)^2} - 4 \times 5 = 50 + 5 \cr & 75 - 20 = 55 \cr & 55 = 55 \cr} $$
LHS = RHS (it satisfies the conditions)
$${\text{so, x = 5}}$$

$$999\,\frac{1}{7} + 999\frac{2}{7} + 999\frac{3}{7} + $$     $$999\frac{4}{7} + $$   $$999\frac{5}{7} + $$   $$999\frac{6}{7}$$
$$ \Rightarrow \left( {999 \times 6} \right) + $$   $$\left( {\frac{1}{7} + \frac{2}{7} + \frac{3}{7} + \frac{4}{7} + \frac{5}{7} + \frac{6}{7}} \right)$$
$$\eqalign{ & \Rightarrow 5994 + \left( {\frac{{21}}{7}} \right) \cr & \Rightarrow 5994 + 3 \cr & \Rightarrow 5997 \cr} $$

In this case:
122 (powers of the last digit 2 )

2¹ = 2
2² = 4
2³ = 8
2⁴ = 16
2⁵ = 32
2⁶ = 64
2⁷ = 128
2⁸ = 256
2⁹ = 512
……

As you can see, the pattern repeats every 4^th power (2, 4, 8, 6).

We have to find 122¹⁷³
Hence, remainder when 173 is divided by 4 = 1
(which corresponds to the 1^st number of the pattern)
Thus, unit digit of 122¹⁷³ = 2

$$\eqalign{ & \Rightarrow {2^{16}} - 1\,\, \cr & = \left( {{2^8} - {1^8}\,\,} \right)\left( {{2^8} + {1^8}\,} \right) \cr & = \left( {{2^4} - 1\,\,} \right)\left( {{2^4} + 1\,\,} \right)\left( {{2^8} + 1\,\,} \right) \cr & = \left( {16 - 1} \right)\left( {16 + 1} \right)\left( {{2^8} + 1\,\,} \right) \cr & = 15 \times 17\left( {{2^8} + 1\,\,} \right) \cr & \therefore {2^{16}} - 1\,\,{\text{is divisible by 17}} \cr} $$

a = 4011
b = 3989
a × b = 4011 × 3989
         = 15999879

Shortcut method :
a = 4011 ⇒ (4000 + 11)
b = 3989 ⇒ (4000 - 11)
[(a - b) (a + b) = a² - b²]
According to the question,
a × b = (4000 + 11) (4000 - 11)
        = 4000² -11²
        = 16000000 - 121
        = 15999879

5349 + 3957 - 7062 = 2244
This is not divisible by 7.

Sum = 5 + 10 + 15 + 20 + .....+ 95 , It is an arithmetic progression
Number of terms (n)
= $$\frac{{{\text{Last term }} - {\text{ First term}}}}{{{\text{Differnce}}}}$$     + 1
⇒ n = $$\frac{95 - 5}{5}$$ + 1 = 19
So, sum = $$\frac{n}{2}$$ [2a + (n - 1)d]
Here, a = 5, n = 19, d = 5
Sum = $$\frac{19}{2}$$ [10 + (19 - 1)5]
        = 950

Since 13 is factor of 52. So divide its remainder directly by 13.
$$\eqalign{ & \Rightarrow \frac{{{\text{remainder}}}}{{13}} = \frac{{45}}{{13}} \cr & \Rightarrow {\text{remainder }} = 6 \cr} $$

L.C.M. of powers (3, 2, 1 and 6) = 6
$$\eqalign{ & {\left( 3 \right)^{\frac{1}{3}}} \Rightarrow {\left( {{3^2}} \right)^{\frac{1}{6}}} = {{\bf{9}}^{\frac{{\bf{1}}}{{\bf{6}}}}} \cr & {\left( 2 \right)^{\frac{1}{2}}} \Rightarrow {\left( {{2^3}} \right)^{\frac{1}{6}}} = {8^{\frac{1}{6}}} \cr & (1) \Rightarrow {\left( {{1^6}} \right)^{\frac{1}{6}}} = {1^{\frac{1}{6}}} \cr & {\left( 6 \right)^{\frac{1}{6}}} \Rightarrow {\left( {{6^1}} \right)^{\frac{1}{6}}} = {6^{\frac{1}{6}}} \cr} $$