Number System - Aptitude MCQ Questions and Solutions with Explanations

The two digit number which gives a remainder of 3 when divided by 7 are:
10, 17, 24 ..... 94.
Now, these number are in AP series with
1st Term, a = 10;
Number of Terms, n = 13;
Last term, L = 94 and
Common Difference, d = 7.
Sum,
$$\eqalign{ & = \left\{ {{\text{n}} \times \frac{{{\text{a}} + {\text{L}}}}{2}} \right\} \cr & = 13 \times 52 \cr & = 676 \cr} $$

NOTE: aⁿ + bⁿ is always divisible by (a + b) when n is odd.

So, (15²³ + 23²³) always divisible by 38.
And 38 is a multiple of 19, So, the number which is divisible by 38, is divisible by 19 too.
(15²³ + 23²³) is divisible by 19.
So, $$\frac{{{{15}^{23}} + {{23}^{23}}}}{{19}}$$   == Remainder ⇒ $$\frac{{15 + 23}}{{19}}$$  == Remainder ⇒ $$\frac{{38}}{{19}}$$ == Remainder ⇒ 0

In this type of problem we find the difference of divisors and their remainders.
Here difference,
32 - 10 = 22
40 - 18 = 22
72 - 50 = 22
Here, in each case difference is same i.e. 22
Then required number of pebbles is given by
[(LCM of 32, 40, 72) -22]

32 = 2 × 2 × 2 × 2 × 2
40 = 2 × 2 × 2 × 5
72 = 2 × 2 × 2 × 3 × 3
Hence,
LCM = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 1440
Thus,
Required number of pebbles,
= 1440 - 22
= 1418

Since, HCF of two numbers = 57
Hence, numbers are multiples of 57
Let, the numbers be 57x and 57y, where x and y are prime to each other.
According to the sum,
57x + 57y = 684
Or, x + y = 12
Hence, required possible pair of values of x and y which are prime to each other are (1, 11) and (5, 7).
Thus, required pairs of numbers are,
{57 × 1 = 57 and 57 × 11 = 627}

{57 × 5 = 285 and 57 × 7 = 399}

Let, these two numbers be 3x and 4x then their LCM = 12x
Now, according to question,
12x = 240
Or, x = 20
Thus, the numbers are (3x = 3 × 20) = 60 and (4x = 4 × 20) = 80
Then smaller in this two is 60

Let the numbers be 17x and 17y where x and y are co-prime.
LCM = 17xy
Now, 17xy = 714
Or, xy = 42 = 6 × 7
→x = 6 and y = 7
Or, x = 7 and y = 6
1^st number = 17 × 6 = 102
2^nd number = 17 × 7 = 119
Sum = 102 + 119 = 221

LCM of 20, 28, 32, 35 will be the greatest number which is divisible by these numbers.
Firstly, we find LCM of 20, 28, 32, 35
20 = 2 × 2 × 5
28 = 2 × 2 × 7
32 = 2 × 2 × 2 × 2 × 2
35 = 5 × 7
LCM = 2 × 2 × 2 × 2 × 2 × 5 × 7 = 1120
Required greatest number which subtract from 5834 are divide by 20, 28, 32 and 35
= 5834 - 1120
= 4714

LCM in power 2, 3, 4, 6, 12 is 12
We multiplied the LCM to the power of the numbers.
$${2^{\frac{{1 \times 12}}{2}}},{3^{\frac{{1 \times 12}}{3}}},{4^{\frac{{1 \times 12}}{4}}},{6^{\frac{{1 \times 12}}{6}}}{\kern 1pt} {\text{and}}\,{12^{\frac{{1 \times 12}}{{12}}}}$$
We get,
= 2⁶, 3⁴, 4³, 6², 12
= 64, 84, 64, 36, 12
Hence, greatest number would be $${3^{\frac{1}{3}}}$$

Let two numbers be N₁ and N₂.
Now, we have
HCF of the numbers × LCM of the numbers = Multiplication of the numbers.
936 × 4 = 72 × N₁
Or, N₁ = $$\frac{{936 \times 4}}{{72}}$$
Or, N₁ = 52

(30)²⁷²⁰, we can write it as[(30)⁴]⁶⁸⁰
Or,[(10 × 3)⁴]⁶⁸⁰
The right most non-zero digit depends on the unit digit of [(3)⁴]⁶⁸⁰
Unit digit of [(3)⁴]⁶⁸⁰,
Or, (81)⁶⁸⁰
The unit digit of 81 is 1 so any power of 81 will always give its unit digit as 1
Thus, required unit digit is 1