Number System — Part 2 MCQs with Explanations

$$\eqalign{ & \frac{{{{65}^{203}}}}{7} \cr & Or,\,\frac{{{{\left( {63 + 2} \right)}^{203}}}}{7} \cr & 63\,{\text{is}}\,{\text{divisible}}\,{\text{by}}\,7, \cr & {\text{so}}\,{\text{remainder}}\,{\text{will}}\,{\text{depend}}\,{\text{on}}\,{\text{the}}\,{\text{powers}}\,{\text{of}}\,2 \cr & Or,\,\frac{{{2^{203}}}}{7} \cr & {\text{Its}}\,{\text{remainder}}\,{\text{would}}\,{\text{be}}\,{\text{same}}\,{\text{as}} \cr & \frac{{{2^3}}}{7} \cr & {\text{Now,}}\,{\text{Required}}\,{\text{Remainder}}\,{\text{wold}}\,{\text{be}}, \cr & \frac{8}{7} = 1\cr & {\text{Required}}\,{\text{remainder}} = 1 \cr} $$ Note: We have manipulated the powers in the form of (4x + n). It means 203 is taken as, 203 = 4x + n = 4 × 50 + 3. We neglect power which is in the multiple of 4.

$$\eqalign{ & \frac{{{{67}^{107}}}}{7} \cr & Or,\,\frac{{{{\left( {7 \times 9 + 4} \right)}^{107}}}}{7} \cr & {\text{The}}\,{\text{remainder}}\,{\text{will}}\,{\text{be}}\,{\text{same}}\,{\text{as}} \cr & \frac{{{4^{107}}}}{7} \cr & Or,\,\frac{{{4^3}}}{7} \cr & Or,\,\frac{{64}}{7} \cr & {\text{Required}}\,{\text{Remainder}} = 1 \cr} $$

$$\eqalign{ & \frac{{{{54}^{124}}}}{{17}} \cr & Or,\,\frac{{{{\left( {17 \times 3 + 3} \right)}^{124}}}}{{17}} \cr & {\text{The}}\,{\text{remainder}}\,{\text{would}}\,{\text{be}}\,{\text{same}}\,{\text{as}}, \cr & \frac{{{3^{124}}}}{{17}} \cr & Or,\,\frac{{{3^{4 \times 31}}}}{{17}} \cr & {\text{Remainder}}\,{\text{will}}\,{\text{be}}\,{\text{same}}\,{\text{as}}, \cr & \frac{{{3^4}}}{{17}} \cr & Or,\,\frac{{81}}{{17}} \cr & \text{Remainder} = 13 \cr} $$

To find unit digit of a number or an Expression, We have to divide the number or expression by 10 and the remainder obtained by this operation would be the required unit digit. $$\eqalign{ & \frac{{ {{{\left( {173} \right)}^{45}} \times {{\left( {152} \right)}^{77}} \times {{\left( {777} \right)}^{999}}} }}{{10}} \cr & {\text{Remainder}}\,{\text{would}}\,{\text{be}}\,{\text{same}}\,{\text{as}}, \cr & \frac{{ {{3^{45}} \times {2^{77}} \times {7^{999}}} }}{{10}} \cr & Or,\,\frac{{ {3 \times 2 \times {7^3}} }}{{10}} \cr & Or,\,\frac{{ {6 \times 343} }}{{10}} \cr & {\text{Remainder}}\,{\text{would}}\,{\text{be}}\,{\text{same}}\,{\text{as}}\,\frac{{ {6 \times 3} }}{{10}} \cr }$$ Thus, Required remainder and unit digit will be 8.

In such cases remainder will always be 1.

$$\eqalign{ & {\text{Required}}\,{\text{answer}}, \cr & = {\frac{{146}}{5}} + {\frac{{146}}{{{5^2}}}} + {\frac{{146}}{{{5^3}}}} \cr & = 29 + 5 + 1 \cr & = 35 \cr} $$ Note: We have taken integral value only, not the fractional. For example $$\frac{{146}}{5}$$ = 29.2 but we have taken 29 and so on.

$$\eqalign{ & {\text{Number of mint before Eswari has taken}}, \cr & = \left( {x - {\frac{x}{2}} } \right) + 2 = 17 \cr & Or,\,x = 30 \cr & {\text{Number of mint before Fatima has taken}}, \cr & = \left( {x - {\frac{x}{4}} } \right) + 3 = 30 \cr & Or,\,x = 36 \cr & {\text{Number of mint before Sita has taken}}, \cr & = \left( {x - {\frac{x}{3}} } \right) + 4 = 36 \cr & Or,\,x = 48 \cr & {\text{Hence,}}\,{\text{there}}\,{\text{were}}\,{\text{48}}\,{\text{mints}}\,{\text{originally}}{\text{.}} \cr} $$

When the birds sat one on a branch, there was one extra bird. When they sat Two to a branch one branch was extra. Now, we go through the options; Checking option (A); If there were 3 branches, there would be 4 birds. (This would leave one bird without branch as per the question.) When 4 birds would sit Two to a branch there would be one branch free (as per the question.). Hence, the option (A) is correct.

We go through the options, Checking option (C) $$\frac{{71}}{{17}}$$ = 4 quotient and remainder 3. $$\frac{{71}}{8}$$ = 8 quotient and remainder 7 as remainder.

If we multiply the last three digits of each terms we get the last three digits. 345 × 321 = 110745 So, the last three digits are 745

At last thief is left with one diamond. Hence, the number of diamonds before he gave some diamonds to the third watchman, $$\eqalign{ & \Rightarrow x - \left( { {\frac{x}{2}} + 2} \right) = 1 \cr & {\text{or, }}{\kern 1pt} \frac{{ {x - 4} }}{2} = 1 \cr & {\text{or, }}{\kern 1pt} x = 6 \cr} $$ Hence, he had 6 diamonds before he gave 5 to the third watchman. Similarly number of diamonds before giving to second watchman, $$\eqalign{ & \frac{{ {x - 4} }}{2} = 6 \cr & {\text{or,}}\,{\kern 1pt} x = 16 \cr} $$ And number of diamonds before giving to the first watchman, $$\eqalign{ & \frac{{ {x - 4} }}{2} = 16 \cr & {\text{or, }}{\kern 1pt} x = 36 \cr} $$ ∴ The thief has stolen 36 diamonds originally

Using options, We find that the sum of numerator and denominator of 97 : 84 is (97 + 84) = 181 which is a prime number. Hence, it is the appropriate answer

a n + b n is always divisible by (a + b) when n is odd. So, (2222 5555 + 5555 2222 ) is always divisible by (2222 + 5555) = 7777 And 7777 is multiple of 7, so (2222 5555 + 5555 2222 ) is divisible by 7

If the girl counts the way as given in the question, The counting serial for thumb will be 1, 9, 17, 25 . . . . . . Hence, Number 1992 ( as 1992 is divisible by 8, common difference of series formed by counting) will also fall on thumb. Hence, Number 1994 will end on her middle finger.

(x 2 - y 2 ) = 45 Or, (x - y)(x + y) = 45 Thus, the factors of 45 possibles are 15, 3, 9, 5, 1 and 45 Hence, numbers are 9 and 6 or 7 and 2 or 23 and 22 So, number of such pairs = 3

Number of terms in the series = 10. (We can get it easily by pointing the number of zeros in power of terms. In 1 st term number of zero is 1, 2 nd term 2, and 3 rd term 3 and so on.) $$\frac{{{{10}^{10}}}}{7},$$ Written as, $$\frac{{{{\left( {7 + 3} \right)}^{\left( {4 \times 2 + 2} \right)}}}}{7}$$ The remainder will depend on $$\frac{{{3^2}}}{7}$$ So, remainder will be 2 $$\eqalign{ & \frac{{{{10}^{1000}}}}{7},\,{\text{remainder}} = 2 \cr & \frac{{{{10}^{10000}}}}{7},\,{\text{remainder}} = 1 \cr} $$ So, we get alternate 2 and 1 as remainder, five times each. So, required remainder is given by $$\eqalign{ & \frac{{\left( {2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 + 2 + 1} \right)}}{7} \cr & = \frac{{15}}{7} \cr} $$ Remainder when 15 is divided by 7 = 1

Let the number of red beads = x Hence, Blue beads = (x - 3) and Green beads = (x - 8) According to the sum, x + x - 3 + x - 8 = 154 or, 3x - 11 = 154 or, x = 55

Let there be x people in the village so, then 4x sheep must be there. No, number of person leaving the village,= (x - 8) [as 8 person died in Tsunami] And the number of sheep to be carried with the survived people,4x - (47 + 21) = x - 8. [Here we are equating No of alive sheep and no. of alive person. As 1 person has 4 sheep means 1 person = 4 sheep.] Or, 3x = 60 Or, x = 20 Or, 4x = 80 Hence, 80 sheep

Option method: The best way to go through options. You will find that when you consider 150 questions you cannot get 75 marks anyhow. Alternatively: Since 75 is an integer, he must have to do (75 + 4x) problems, where x is a whole number. Now we put the value of x (0, 1, 2, 3, 4, 5 .........) to get the correct answer. As he gets the net marks in integer it means when he does 3n question wrong, he losses actually 4n marks.

Total number of terms in expression = 9 (18 2 - 1) = (18 - 1) × (18 + 1) (18 4 - 1) = (18 2 + 1) × (18 2 - 1) = (18 2 + 1) × (18 - 1) × (18 + 1) (18 6 - 1) = [(18 3 ) 2 -1] = (18 3 + 1) (18 3 - 1) = 17 × k and so on. Hence, there will be 9 times 17 in the whole expression as each term of expression gives one 17. Therefore for maximum power of 17 will be 9.

In each set of 100 numbers, there are 10 numbers whose tens digit and unit digit are same. Again in the same set there are 10 numbers whose hundreds and tens digits are same. But one number in each set of 100 numbers whose Hundreds, Tens and Unit digit are same as 111, 222, 333, 444 etc Hence, there are exactly (10 + 10 - 1) = 19 numbers in each set of 100 numbers. Further there are 9 such sets of numbers Therefore such total numbers = 19 × 9 = 171 Alternatively, 9 × 10 × 10 - 9 × 9 × 9 = 900 - 729 = 171

Since, the magazine has 84 pages it means that there are 21 sheets of paper, which are folded in the middle. The pattern of pages will be this way, Left Right 1, 2 83, 84 3, 4 81, 82 - - - - - - - - 29, 30 55, 56

$$\eqalign{ & {\text{Since}}, \cr & \frac{6}{{10}} \to \,{\text{remainder}}\,{\text{is}}\,{\text{6}}{\text{}} \cr & \frac{{{6^6}}}{{10}} \to \,{\text{remainder}}\,{\text{is}}\,{\text{6}}{\text{}} \cr & \frac{{{6^{{6^6}}}}}{{10}} \to \,{\text{remainder}}\,{\text{is}}\,{\text{6}}{\text{}} \cr & \frac{{{6^{{6^{{6^6}}}}}}}{{10}} \to \,{\text{remainder}}\,{\text{is}}\,{\text{6}}{\text{}} \cr & \frac{{{6^{{6^{{6^{{6^6}}}}}}}}}{{10}} \to \,{\text{remainder}}\,{\text{is}}\,{\text{6}}\,{\text{and}}\,{\text{so}}\,{\text{on}}. \cr & \cr & {\text{Thus,}}\,{\text{in}}\,{\text{all}}\,{\text{the}}\,{\text{such}}\,{\text{cases}}\, \cr & {\text{The}}\,{\text{remainder}}\,{\text{will}}\,{\text{always}}\,{\text{be}}\,6 \cr} $$

$$\eqalign{ & \frac{{\left[ {\left( {888 \times 888 \times 888} \right) - \left( {222 \times 222 \times 222} \right)} \right]}}{{\left[ {\left( {888 \times 888} \right) + \left( {888 \times 222} \right) + \left( {222 \times 222} \right)} \right]}} \cr & = \frac{{\left( {{{888}^3} - {{222}^3}} \right)}}{{\left[ {{{888}^2} + \left( {888 \times 222} \right) + {{222}^2}} \right]}} \cr & = 888 - 222 \cr & = 666 \cr} $$

$$\eqalign{ & {32^{{{32}^{32}}}}\,{\text{means}}\,{32^{\left( {32.32.32.32.......32\,\text{times}} \right)}} \cr & {\text{and}}\,{\text{the}}\,{\text{remainder}}\,{\text{of}}\,\frac{{32}}{7}\,{\text{is}}\,4 \cr & {\text{So}}, \cr & \frac{{{4^{\left( {32.32.32.32.......32\,\text{times}} \right)}}}}{7} \cr & \frac{{{4^{\left( {2.2.2.2.2........32\,\text{times}} \right)}}}}{7} \cr & {\text{Remainder}} = 4 \cr & {\text{Since}},\,\frac{4}{7} \to \,{\text{Remainder}}\,4 \cr & \frac{{{4^2}}}{7} \to \,{\text{Remainder}}\,2 \cr & \frac{{{4^3}}}{7} \to \,{\text{Remainder}}\,1 \cr & \frac{{{4^4}}}{7} \to \,{\text{Remainder}}\,4 \cr} $$

Remainder of $$\frac{{ {{{888}^{222}} + {{222}^{888}}} }}{3}$$ = 0 Since, 888 and 222 both (bases) are divisible by 3

The remainder when 10 is divided by 6 is 4. The remainder when 10 2 is divided by 6 is 4. The remainder when 10 3 is divided by 6 is 4. The remainder when 10 4 is divided by 6 is 4. Thus, remainder is always 4. So, the required remainder, $$\eqalign{ & = \frac{{4 + 4 + 4 + 4 + \,.....\,99{\text{ times}}}}{6} \cr & = \frac{{396}}{6} \cr} $$ Thus, Required remainder is 0.

Since, $${3^4} = 81$$ gives remainder 1 on divided by 80 So, $$\frac{{{3^{4n}}}}{{80}}$$ gives remainder 1 Thus, $$\frac{{{3^{67!}}}}{{80}}$$ will also give the remainder as 1 Since, 67! = 4n for a positive integer n

Since, $$\frac{{{{\text{a}}^{\text{n}}}}}{{{\text{a}} + 1}}$$ gives remainder 1 when 'n' is even. Now since, 97! is an even number so remainder will be 1. Note:- Factorial of any no. is even.

$$\eqalign{ & {2^{59}}\,{\text{can}}\,{\text{be}}\,{\text{expressed}}\,{\text{as}}\,{2^{8n}} \cr & {\text{So}}, \cr & {\text{Remainder}}\,\frac{{{2^{59}}}}{{255}} \cr & = {\text{Remainder}}\,\frac{{{2^{8n}}}}{{255}} \cr & = {\text{Remainder}}\,\frac{{{{256}^n}}}{{255}} = 1 \cr & {\text{Required}}\,{\text{remainder}} = 1 \cr} $$

6 4 × 8 6 × 10 8 × 12 10 = (2 × 3) 4 × (2 3 ) 6 × (2 × 5) 8 × (2 2 × 3 ) 10 = 2 4 × 3 4 × 2 18 × 2 8 × 5 8 × 2 20 × 3 10 = 2 50 × 3 14 × 5 8 Thus, the total prime factors, = 50 + 14 + 8 [By adding maximum power of prime factors.] = 72

Let the digits of x be x = abcde According to the question, x = 1bcde [Given ten thousands place is 1.] Now we can check the options as given that the sum of the all digit is divisible by 3. 10080 is the only number given in the option which satisfies all the given conditions.

The best way to go through the options Let there are initially 3 boxes then, 1 st customer gets = $$\frac{3}{2}$$ + $$\frac{1}{2}$$ = 2 Remaining boxes = 3 - 2 = 1 2 nd customer = $$\frac{1}{2}$$ + $$\frac{1}{2}$$ = 1 So, option B is correct.

Given, x + y + z = 0 Cubing both side, (x + y + z) 3 = 0 x 3 + y 3 + z 3 - 3xyz = 0 [using formula] x 3 + y 3 + z 3 = 3xyz

From 1- 99 digit 1 is used 20 times. And From 100 - 199, 1 is used 120 times So, from 1 to 199, 1 is used, 20 + 120 = 140 times We need 136.So leave 199, 198, 197 and 196 Required pages = 195

For a no. to be divisible by 11, Sum(odd digit nos) - Sum(even digit nos) = 0 or divisible by 11 If we look at 9876543210, the difference we get is 5 i.e. [(9 + 7 + 5 + 3 + 1) - (8 + 6 + 4 + 2 + 0) = 5] The series is up to 1000 digit, That means, we get $$\frac{{1000}}{{10}}$$ = 100 time 5, then the difference will be 5 × 100 = 500 In order for the difference to be divisible by 11, we need to add 5 and the no will become 505 505 is divisible by 11

Let S be the sum of the expression, Then S = 4 0 + 4 1 + 4 2 + 4 3 + ........ + 4 40 S = (1 + 4 + 16 + 64) + 4 4 (1 = 4 + 16 + 64) + ...... + 4 36 + 4 40 Since, (1 + 4 + 16 + 64) = 85, is divisible by 17. Hence, except 4 40 remaining expression is divisible by 17. Thus, $$\frac{{{4^{40}}}}{{17}} \to \frac{{{{\left( {{4^4}} \right)}^{10}}}}{{17}} \to \frac{{{1^{10}}}}{{17}}$$ Hence, remainder = 1

$$\eqalign{ & {\text{The}}\,{\text{given}}\,{\text{expression}}\,{\text{is}}\,{\text{in}}\,{\text{GP}}\,{\text{series}} \cr & S = {3^0} + {3^1} + {3^2} + {3^3} + ........ + {3^{200}} \cr & S = {\frac{{ {{3^0} \times \left( {{3^{201}} - 1} \right)} }}{{ {3 - 1} }}} \cr & S = \frac{{ {{3^{201}} - 1} }}{2} \cr & S = \frac{{ {{{\left( {{3^3}} \right)}^{67}} - {1^3}} }}{2} \cr & S = \frac{{ {{{27}^{67}} - {1^3}} }}{2} \cr} $$ Since, (A n - B n ) is divisible by (A - B), So, (27 67 - 1 3 ) is divisible by (27 - 1) = 26 Hence, Expression is also divisible by 13 as it is divisible by 26 Thus given expression is divisible by 13 so the remainder will be 0

Rajan (100)____________________ (1000) Raman These numbers are: (101, 102), (111, 102), (121, 122), (131,132).........(202, 203), (212, 213), ....... (303, 304), (313, 314) ......... (808, 809)......... (989, 990) Thus, there are total 20 × 7 + 21 + 18 = 179, mile stones which are red, because the hundreds digit on the next milestone is same as the unit digit of previous milestone.

Since, he has covered twice the distance which yet he has to cover. It means he has covered $$\frac{2}{3}$$ of the whole journey and remaining journey is $$\frac{1}{3}$$ rd.