Number System - Aptitude MCQ Questions and Solutions with Explanations

Let the number be 476ab0

476ab0 is divisible by 3
⇒ 4 + 7 + 6 + a + b + 0 is divisible by 3
⇒ 17 + a + b is divisible by 3 - - - - - - - (i)

476ab0 is divisible by 11
[(4 + 6 + b) - (7 + a + 0)] is 0 or divisible by 11
⇒ [3 + (b - a)] is 0 or divisible by 11 - - - - - - - -(ii)

Substitute the values of a and b with the values given in the choices and select the values which satisfies both Equation (i) and Equation (ii).

if a = 6 and b = 2,
17 + a + b = 17 + 6 + 2 = 25 which is not divisible by 3 --- Does not meet equation(i).Hence this is not the answer

if a = 8 and b = 2,
17 + a + b = 17 + 8 + 2 = 27 which is divisible by 3 --- Meet equation(i)
[3 + (b - a)] = [3 + (2 - 8)] = -3 which is neither 0 nor divisible by 11 --- Does not meet equation(ii).Hence this is not the answer

if a = 6 and b = 5,
17 + a + b = 17 + 6 + 5 = 28 which is not divisible by 3 --- Does not meet equation (i) .Hence this is not the answer

if a = 8 and b = 5, 17 + a + b = 17 + 8 + 5 = 30 which is divisible by 3 --- Meet equation 1 [3 + (b - a)] = [3 + (5 - 8)] = 0 ---Meet equation 2 Since these values satisfies both equation 1 and equation 2, this is the answer

$$\eqalign{ & x = 13p + 11\,{\text{and}}\,x = 17q + 9 \cr & \therefore 13p + 11 = 17q + 9 \cr & \Rightarrow 17q - 13p = 2 \cr & \Rightarrow q = \frac{{2 + 13p}}{{17}} \cr & {\text{Thr}}\,{\text{least}}\,{\text{value}}\,{\text{of}}\,p\,{\text{for}}\,{\text{which}} \cr & q = \frac{{2 + 13p}}{{17}}\,{\text{is}}\,{\text{a}}\,{\text{whole}}\,{\text{number}}\,{\text{is}}\,p = 26 \cr & \therefore x = \left( {13 \times 26 + 11} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = \left( {338 + 11} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = 349 \cr} $$

Given Exp.
= (397)² + (104)² + 2 × 397 × 104
= (397 + 104)²
= (501)² = (500 + 1)²
= (500²) + (1)² + (2 × 500 × 1)
= 250000 + 1 + 1000
= 251001

Let the two consecutive even integers be 2n and (2n + 2). Then,
(2n + 2)² - (2n)² = (2n + 2 + 2n) (2n + 2 - 2n)
= 2(4n + 2)
= 4(2n + 1), which is divisible by 4

437 > 22
All prime numbers less than 22 are : 2, 3, 5, 7, 11, 13, 17, 19
161 is divisible by 7, and 221 is divisible by 13
373 is not divisible by any of the above prime numbers.
Therefore 373 is prime

Let S_n = (2 + 4 + 6 + ..... + 30). This is an A.P. in which a = 2, d = 2 and l = 30
Let the number of terms be n Then,
a + (n - 1)d = 30
⇒ 2 + (n - 1) x 2 = 30
⇒ n = 15
$$\eqalign{ & \therefore {S_n} = \frac{n}{2}\left( {a + I} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{15}}{2} \times \left( {2 + 30} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 15 \times 16 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 240 \cr} $$

$$\eqalign{ & 20 \times x = \left( {64 + 36} \right)\left( {64 - 36} \right) \cr & 20 \times x = 100 \times 28 \cr & \therefore {\kern 1pt} x = \frac{{100 \times 28}}{{20}} \cr & \,\,\,\,\,\,\,\,\,\,\, = 140 \cr} $$

Unit digit in 3⁴ = 1 ⇒ Unit digit in (3⁴)¹⁶ = 1
Therefore Unit digit in 3⁶⁵ = Unit digit in [ (3⁴)¹⁶ × 3 ] = (1 × 3) = 3
Unit digit in 6⁵⁹ = 6
Unit digit in 7⁴ ⇒ Unit digit in (7⁴)¹⁷ is 1
Unit digit in 7⁷¹ = Unit digit in [(7⁴)¹⁷ × 7³] = (1 × 3) = 3
Therefore Required digit = Unit digit in (3 × 6 × 3) = 4

4 | x             y = (5 × 1 + 4) = 9
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5 | y - 1         x = (4 × y + 1) = (4 × 9 + 1) = 37
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   | 1 - 4

Now, 37 when divided successively by 5 and 4, we get

5 | 37
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4 | 7 - 2
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   | 1 - 3
Respective remainders are 2 and 3

$$\eqalign{ & {\bf{Series :}} \cr & 103 + 108\, + ........ + \,998 \cr & a{\text{ }} = {\text{ }}103 \cr & d{\text{ }} = {\text{ }}5 \cr & {\text{Last term = 998}} \cr & {\text{Number of term }} \cr & = \frac{{998 - 103}}{5} + 1 \cr & = \frac{{895}}{5} + 1 \cr & = 180 \cr & {\text{Sum of n terms}} \cr & = \frac{n}{2}\left[ {2a{\text{ }} + \left( {n - 1} \right)d} \right] \cr & = \frac{{180}}{2}\left[ {2 \times 103 + \left( {180 - 1} \right)5} \right] \cr & = 99090 \cr} $$