Number System — Part 1 MCQs with Explanations

Remainder, $$\frac{{73 \times 75 \times 78 \times 57 \times 197 \times 37}}{{34}}$$ $$ = \frac{{5 \times 7 \times 10 \times 23 \times 27 \times 3}}{{34}}$$ [We have taken individual remainder, which means if 73 is divided by 34 individually, it will give remainder 5, 75 divided 34 gives remainder 7 and so on.] $$\eqalign{ & \frac{{5 \times 7 \times 10 \times 23 \times 27 \times 3}}{{34}} \cr & = \frac{{35 \times 30 \times 23 \times 27}}{{34}} \cr & = \frac{{1 \times - 4 \times - 11 \times - 7}}{{34}} \cr} $$ [We have taken here negative as well as positive remainder at the same time. When 30 divided by 34 it will give either positive remainder 30 or negative remainder -4. We can use any one of negative or positive remainder at any time.] $$\eqalign{ & \frac{{1 \times - 4 \times - 11 \times - 7}}{{34}} \cr & = \frac{{28 \times - 11}}{{34}} \cr & = \frac{{ - 6 \times - 11}}{{34}} \cr & = \frac{{66}}{{34}} \cr & {\text{R}} = 32 \cr} $$ Required remainder = 32

$$\eqalign{ & {\text{Remainder of}}\frac{{{{67}^{99}}}}{7} \cr & {\text{or, }}R = \frac{{{{\left( {63 + 4} \right)}^{99}}}}{7} \cr} $$ 63 is divisible by 7 for any power, so required remainder will depend on the power of 4 Required remainder $$\eqalign{ & \frac{{{4^{99}}}}{7} = = R = = \frac{{{4^{\left( {96 + 3} \right)}}}}{7} \cr & \frac{{{4^3}}}{7} \Rightarrow \frac{{64}}{7} \Rightarrow \frac{{\left( {63 + 1} \right)}}{7} = = R \Rightarrow 1 \cr & \cr & {\text{Note}}: \cr & \frac{4}{7}{\text{remainder}} = 4 \cr & \frac{{\left( {4 \times 4} \right)}}{7} = \frac{{16}}{7}{\text{remainder}} = 2 \cr & \frac{{\left( {4 \times 4 \times 4} \right)}}{7} = \frac{{64}}{7} = 1 \cr & \frac{{\left( {4 \times 4 \times 4 \times 4} \right)}}{7} = \frac{{256}}{7}{\text{remainder}} = 4 \cr & \frac{{\left( {4 \times 4 \times 4 \times 4 \times 4} \right)}}{7} = 2 \cr} $$ If we check for more power we will find that the remainder start repeating themselves as 4, 2, 1, 4, 2, 1 and so on. So when we get A number having greater power and to be divided by the other number B, we will break power in (4n + x) and the final remainder will depend on x i.e. $$\frac{{{{\text{A}}^{\text{x}}}}}{{\text{B}}}$$

Remainder, $$\eqalign{ & \frac{{1421 \times 1423 \times 1425}}{{12}} = R \cr & R \Rightarrow \frac{{5 \times 7 \times 9}}{{12}} \cr} $$ [Here, we have taken individual remainder such as 1421 divided by 12 gives remainder 5, 1423 and 1425 gives the remainder as 7 and 9 on dividing by 12.] Now, the sum is reduced to, $$\frac{{5 \times 7 \times 9}}{{12}} = \frac{{35 \times 9}}{{12}}$$ $$\frac{{35 \times 9}}{{12}}$$ = Remainder ⇒ -1 × -3 = 3 [Here, we have taken negative remainder.] So, required remainder will be 3. Note: When, $$\frac{9}{{12}}$$ it gives positive remainder as 9 and it also give a negative remainder -3. As per our convenience,we can take any time positive or negative remainder.

Since, the numbers are given in the form of ratio that means their common factors have been cancelled. Each one's common factor is HCF. And here HCF = 12, hence, the numbers are 12, 24 and 36. Alternatively, Let the numbers be x, 2x and 3x. The HCF in x, 2x and 3x is x because 1, 2, 3 are prime. Hence, x = 12; then the other numbers are 24 and 36.

In this type of question, We need to find out the LCM of the given numbers. LCM of 12, 15, 18 and 20; $$\eqalign{ & 12 = 2 \times 2 \times 3; \cr & 15 = 3 \times 5; \cr & 18 = 2 \times 3 \times 3; \cr & 20 = 2 \times 2 \times 5; \cr} $$ Hence, LCM = $$2 \times 2 \times 3 \times 5 \times 3$$ Since, the soldiers are in the form of a solid square. Hence, LCM must be a perfect square. To make the LCM a perfect square, We have to multiply it by 5, hence, The required number of soldiers = $$2 \times 2 \times 3 \times 3 \times 5 \times 5$$ = 900

We have to find the Least number, therefore we find out the LCM of 8, 12, 16 and 20. 8 = 2 × 2 × 2; 12 = 2 × 2 × 3; 16 = 2 × 2 × 2 × 2; 20 = 2 × 2 × 5; LCM = 2 × 2 × 2 × 2 × 3 × 5 = 240; This is the least number which is exactly divisible by 8, 12, 16 and 20 Thus, Required number which leaves remainder 5 is, 240 + 5 = 245

$$\eqalign{ & {7^{6n}} - {6^{6n}} \cr & = {7^6} - {6^6} \cr & = {\left( {{7^3}} \right)^2} - {\left( {{6^3}} \right)^2} \cr & = \left( {{7^3} - {6^3}} \right)\left( {{7^3} + {6^3}} \right) \cr & = \left( {343 - 216} \right) \times \left( {343 + 216} \right) \cr & = 127 \times 559 \cr & = 127 \times 13 \times 43 \cr} $$ Clearly, it is divisible by 127, 13 as well as 559

As the Number gives a remainder of 4 when it is divided by 7, then the number must be in form of (7x + 4) The same gives remainder 1 when it is divided 4, so the number must be in the form of {4 × (7x + 4) + 1} Also, the number when divided by 3 gives remainder 2, thus number must be in form of [3 × {4 × (7x + 4) + 1} + 2] Now, On simplifying, [3 × {4 × (7x + 4) + 1} + 2] = 84x + 53 We get the final number 53 more than a multiple of 84 Hence, if the number is divided by 84, The remainder will be 53

$$\eqalign{ & {\text{Given}}, \cr & \frac{{{2^{256}}}}{{17}} \cr & {\text{We}}\,{\text{can}}\,{\text{write}}\,{\text{it}}\,{\text{as}}:\,{\left( {{2^4}} \right)^{64}} \cr & \frac{{{{16}^{64}}}}{{17}} \cr & {\text{Individually, when 16 is divided by 17,}} \cr & {\text{gives a negative reminder of - 1}}{\text{.}} \cr & {\text{Required Remainder}}, \cr & {\left( { - 1} \right)^{64}} = 1 \cr & {\text{Alternatively}}, \cr & \frac{{{{16}^{64}}}}{{17}},\,{\text{can}}\,{\text{be}}\,{\text{written}}\,{\text{as}} \cr & \frac{{\left( {16 \times 16 \times 16 \times 16 \times 16\,......\,64\,{\text{times}}} \right)}}{{17}} \cr & {\text{Now,}}\,{\text{we}}\,{\text{take}}\,{\text{the}}\,{\text{negative}}\,{\text{remainder}}\,{\text{of}}\,{\text{each,}} \cr & {\text{16}}\,{\text{divided}}\,{\text{by}}\,{\text{17}}\,{\text{gives}}\,{\text{negative}}\,{\text{remainder}}\,{\text{ - 1}}{\text{.}} \cr & {\text{So,}}\,{\text{the}}\,{\text{Remainder}}\,{\text{will}}\,{\text{be}} \cr & \left( { - 1 \times - 1 \times - 1 \times - 1 \times - 1\,.......\,64\,{\text{times}}} \right) \cr & = 1 \cr} $$

$$\frac{{{4^{96}}}}{6},$$ we can write it in this form $$\frac{{{{\left( {6 - 2} \right)}^{96}}}}{6}$$ Now, Remainder will depend only the powers of -2. So, $$\frac{{{{\left( { - 2} \right)}^{96}}}}{6},$$ it is same as $$\frac{{{{\left( {{{\left[ { - 2} \right]}^4}} \right)}^{24}}}}{6},$$ it is same as $$\frac{{{{\left( {16} \right)}^{24}}}}{6}$$ Now, $$\frac{{\left( {16 \times 16 \times 16 \times 16{\kern 1pt} ......{\kern 1pt} 24{\kern 1pt} {\text{times}}} \right)}}{6}$$ On dividing individually 16 we always get a remainder 4. $$\frac{{\left( {4 \times 4 \times 4 \times 4{\kern 1pt} ......{\kern 1pt} 24{\kern 1pt} {\text{times}}} \right)}}{6}$$ Hence, Required Remainder = 4 NOTE: When 4 has even number of powers, it will always give remainder 4 on dividing by 6.

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 andCommon Difference, d = 7. Sum, $$\eqalign{ & = \left\{ {{\text{n}} \times \frac{{{\text{a}} + {\text{L}}}}{2}} \right\} \cr & = 13 \times 52 \cr & = 676 \cr} $$

NOTE: a n + b n is always divisible by (a + b) when n is odd. So, (15 23 + 23 23 ) always divisible by 38. And 38 is a multiple of 19, So, the number which is divisible by 38, is divisible by 19 too. (15 23 + 23 23 ) 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, = 2 6 , 3 4 , 4 3 , 6 2 , 12 = 64, 84, 64, 36, 12 Hence, greatest number would be $${3^{\frac{1}{3}}}$$

Let two numbers be N 1 and N 2 . Now, we have HCF of the numbers × LCM of the numbers = Multiplication of the numbers. 936 × 4 = 72 × N 1 Or, N 1 = $$\frac{{936 \times 4}}{{72}}$$ Or, N 1 = 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

Let the numbers are in the form of (10x + y), so when the digits of the number are reversed the number becomes (10y + x) According to question, (10y + x) - (10x + y) = 18 Or, 9(y - x) = 18 → y - x = 2 So, the possible pairs of (x, y) are (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8) and (7, 9) But, we need the number other than 13. Thus, there are 6 possible numbers i.e. 24, 35, 46, 57, 68, 79 So, total numbers of possible numbers are 6

Using options, We find that four consecutive odd numbers are 37, 39, 41 and 43 The sum of these 4 numbers is 160, when divided by 10 we get 16 which is a perfect square. Thus, 41 is one of the odd numbers

Any four digit number in which first two digits are equal and last two digits are also equal will be in the form 11 × (11a + b) i.e. it will be the multiple of 11 like 1122, 3366, 2244, . . . . Now, let the required number be aabb. Since aabb is a perfect square, the only pair of a and b that satisfy the above mentioned condition is a = 7 and b = 4 Hence, 7744 is a perfect square

Let the 1 st , 2 nd , 3 rd and 4 th digits be a, b, c and d respectively. Then, a + b = c + d ----------(i) a + d = c ----------(ii ) b + d = 2(a + c) ----------(iii) from eqn. (i) and (ii), a + b = a + 2d →b = 2d and eqn (iii); 2d + d = 2(a + a + d) → 3d = 2(2a + d) → d = 4a or, a = $$\frac{{\text{d}}}{4};$$ → Now, from eqn. (ii), a + d = $$\frac{{\text{d}}}{4}$$ + d = $$\frac{{{\text{5d}}}}{4}$$ = c Or, c = $$\frac{5}{{4{\text{d}}}}$$ The value of d can be either 4 or 8. If d = 4, then c = 5 If d = 8, then c = 10 But the value of c should be less than 10 Hence, value of c would be 5

They will ring together after, LCM of 48 and 50 secs. 48 = 2 × 2 × 2 × 2 × 3; 50 = 2 × 5 × 5 LCM = 2 × 2 × 2 × 2 × 3 × 5 × 5 = 1200 secs = 20 min. They will beep together at 12:20

LCM of 36, 42 and 72, 36 = 2 × 2 × 3 × 3 42 = 2 × 3 × 7 72 = 2 × 2 × 2 × 3 × 3 LCM = 2 ×2 × 2 × 3 × 3 × 7 = 504 seconds. LCM of 36, 42 and 72 is 504 Hence, the lights will change simultaneously after 8 minutes and 24 seconds.

We have, HCF of the numbers × LCM of the numbers = Multiplication of the numbers. (12) × (2 3 × 3 2 × 5 × 103 × 107) = 2472 × 1284 × N Hence, N = $$\frac{{\{ \left( {12} \right) \times \left( {{2^3} \times {3^2} \times 5 \times 103 \times 107} \right)\} }}{{2472 \times 1284}}$$ Or, N = 3 × 5

$$\eqalign{ & 2.5 = \frac{{25}}{{10}}, \cr & 0.5 = \frac{5}{{10}}, \cr & 0.175 = \frac{{175}}{{1000}}, \cr} $$ Now, LCM of two or more fractions is given by: $$\eqalign{ & \frac{{{\text{LCM}}\,{\text{of}}\,{\text{Numerators}}}}{{{\text{HCF}}\,{\text{of}}\,{\text{Denominators}}}} \cr & {\text{Thus,}} \cr & \frac{{{\text{LCM}}\,{\text{of}}\,25,\,5,\,175}}{{{\text{HCF}}\,{\text{of}}\,10,\,10,\,1000}} \cr & = \frac{{175}}{{10}} \cr & = 17.5 \cr} $$

The solution of this question is based on the rule, The HCF of (a m - 1) and (a n - 1) is given by (a HCF of m, n - 1) Thus for this question the answer is (3 5 - 1) Since, 5 is the HCF of 35 and 125

The last digit of multiplication depends on the unit digit of (81 × 82 × 83 × 84 × 86 × 87 × 88 × 89) which is given by the remainder obtained on dividing it by 10. $$\eqalign{ & \frac{{\left( {81 \times 82 \times 83 \times 84 \times 86 \times 87 \times 88 \times 89} \right)}}{{10}} \cr & {\text{We}}\,{\text{take}}\,{\text{individual}}\,{\text{remainder}}\,{\text{of}}\,{\text{each}}\,{\text{digit,}} \cr & \frac{{\left( {1 \times 2 \times 3 \times 4 \times 6 \times 7 \times 8 \times 9} \right)}}{{10}} \cr & {\text{Numbers}}\,{\text{multiplied}}, \cr & \frac{{\left( {24 \times 42 \times 72} \right)}}{{10}} \cr & {\text{Individual}}\,{\text{Remainder}}\,{\text{has}}\,{\text{been}}\,{\text{taken}}, \cr & \frac{{\left( {4 \times 2 \times 2} \right)}}{{10}} \cr & Or,\,\frac{{\left( {16} \right)}}{{10}} \cr & Or,\,6 \cr & {\text{Remainder}} = 6 \cr & {\text{So,}}\,{\text{the}}\,{\text{last}}\,{\text{digit}}\,{\text{will}}\,{\text{be}}\,6 \cr} $$

Let number be (10x + y) According to question, (10x + y) - (10y + x) = 54 10x - 10y + y - x = 54 Or, 9x - 9y = 54 Or, x - y = 6 -------(i) Sum of digits, (x + y) = 10 ------- (ii) (i) - (ii) So, x - y - x - y = 6 - 10 Or, -2y = -4 Or, y = 2 and, x = 8 Then, the required number is = (10y + x) = 10 × 2 + 8 = 28

A number is divisible by 11 if the difference of the sum of the digits in the odd places and sum of the digits in even place is zero or divisible by 11. Hence, (A + 8) - (3 + 1) = 0 or multiple of 11. To get the difference 0 or multiple of 11, we need 7 at the place of A. So, sum of odd place - sum of even place= 15 - 4 = 11. And this is divisible by 11.

A number is divisible by 9 when the sum of its digit is divisible by 9 So, (3 + 8 + 1 + A) = must be divisible by 9 Thus, smallest natural number be 6 Or, (3 + 8 + 1 + 6) = 18, this is divisible by 9

$$\eqalign{ & {\text{In}}\,{\text{such}}\,{\text{case}},\, \cr & {\text{we}}\,{\text{first}}\,{\text{need}}\,{\text{to}}\,{\text{find}}\,{\text{the}}\,{\text{HCF}}\,{\text{of}}\,{\text{44,66,110}}. \cr & 44 = 2 \times 2 \times 11 \cr & 66 = 2 \times 3 \times 11 \cr & 110 = 2 \times 5 \times 11 \cr & {\text{HCF}} = 2 \times 11 = 22 \cr & {\text{Then,}}\,{\text{the}}\,{\text{required}}\,{\text{numbers}}\,{\text{of}}\,{\text{rows}}, \cr & = { {\frac{{44}}{{22}}} + {\frac{{66}}{{22}}} + {\frac{{110}}{{22}}} } \cr & = 10 \cr} $$

Rule- Greatest number with which if we divide P, Q, R and it leaves same remainder in each case. Number is of form = HCF of (P - Q), (P - R) Therefore, HCF of (4126 - 4003), (4249 - 4003) = HCF of 123, 246 = 41. [Taken for Positive result] . Detailed Explanation: The numbers can be written as, 4003 = AX + P where P = Remainder 4126 = BX + P 4249 = CX + P (B - A) × X = 123 (C - B) × X = 246 Thus the X is factor of 123 and 246

1420 = 2 × 2 × 5 × 71 = (2 0 + 2 1 + 2 2 ) × (5 0 + 5 1 ) × (71 0 + 71 1 ) Here, we count the number of terms in the expansion. In the above expansion, 1420 contains number of terms = 3 × 2 × 2 = 12 Then, the number of factors or divisors = 12

HCF of the numbers × LCM of the numbers = Multiplication of the numbers Or, 30 × 1890 = 270 × N Or, N = $$30 \times \frac{{1890}}{{270}}$$ Or, N = 210

Difference, 40 - 31 = 9 60 - 51 = 9 75 - 66 = 9 Difference between numbers and remainder is same in each case. Then, The answer = {(LCM of 40, 60, 75) - 9} 40 = 2 × 2 × 2 × 5 60 = 2 × 2 × 3 × 5 75 = 3 × 5 × 5 LCM = 2 × 2 × 2 × 5 × 5 × 3 = 600 But, the least number of 5 digits = 10000 $$\frac{{10000}}{{600}},$$ we get remainder as 400 Then, the answer = 1000 - (600 - 400) - 9 = 10191

576 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 448 = 2 × 2 × 2 × 2 × 2 × 2 × 7 HCF = 2 × 2 × 2 × 2 × 2 × 2 = 64 Hence, number of classes required $$\frac{{576}}{{64}} + \frac{{448}}{{64}}$$ = 9 + 7 = 16 Shortcut: The numbers (3 × 3) and 7 are not a part of HCF. And sum of multiplication of these number is the required answer.

A perfect square never ends with odd number of zeros.