fsb by pt for number system 2

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For PT Faculty use only

Pinnacle Regular - Maths FSB - 2008-09

Pinnacle RegularMaths - Faculty Support Booklet (2008-09)

Number Systems - I (Chapter 2)

Additional Theory / Concepts

Ø Prime Numbers: Any prime number greater than 3 is always of the form 6k + 1 and 6k – 1

Ø Twin PrimesThe Number 1 more than the product of the number in any pair of twin primes is always a perfect square.e.g. 3 × 5 = 15, 5 × 7 = 35, 11 × 13 = 143.

Ø Fibonacci Sequence : The sequence is defined by the following recurrence relation:f(n) = f(n – 1) + f(n – 2) for n > 1That is, after two starting values, each number is the sum of the two preceding numbers. The first Fibonaccinumbers, also denoted as Fn, for n = 0, 1, … , are: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610,987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, ........ .

Ø The sum of the reciprocal of the divisors of a perfect number including that of its own is always = 2.

e.g. for the perfect number 28, 11

12

14

17

114

128

2+ + + + + = .

Ø Every even perfect number is of the form 2n – 1 (2n –1) where 2n – 1 is a prime number.e.g., n = 2, 3, 5, 7, 11, 13, 17, 19, 31, 61, 89, 107, 127, 257, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9941,11213 and 19937. 27 perfect number are known so far.

Ø Triangular Number : More formally, a triangular number is a number obtained by adding all positive integersless than or equal to a given positive integer , i.e.,

1, , , , ..., so the first few triangle numbers are 1, 3, 6, 10, 15, 21, ...

Ø An amicable pair consists of two integers for which the sum of proper divisors (the divisors excluding thenumber itself) of one number equals the other. Amicable pairs are occasionally called friendly pairs. Thesmallest amicable pair is (220, 284) which has factorizations. The first few amicable pairs are (220, 284),(1184, 1210), (2620, 2924) (5020, 5564), (6232, 6368), (10744, 10856), (12285, 14595), (17296, 18416),(63020, 76084), .

Ø An almost perfect number, also known as a least deficient or slightly defective (Sine 1997) number, is apositive integer n for which the divisor function satisfies sigma(n)=2n – 1. The only known almost perfectnumbers are the powers of 2, namely 1, 2, 4, 8, 16, 32, ...

Ø Value of Pie - 3.14159265358979323846264...

Pinnacle Regular - Maths FSB - 2008-09 (2) of (8)

For PT Faculty use onlyØØ If X = 123456123 × 123456567 and Y = 123456234 × 123456456, then is X > Y or X < Y?

Sol.X < Y

ØØ XYZ denotes a 3 digit number. If X and Y are interchanged, the value of the number decreasesby 90. Then how many values exist for X and Y?

Sol. (100X + 10Y + Z) – (100Y + 10X + Z) = 90Hence 90(X – Y) = 90∴ 9 values of X and Y are possible that are(1, 0), (2, 1), (3, 2), (4, 3), (5, 4), (6, 5), (7, 6), (8, 7) and (9, 8)

ØØ A boy has a certain number of marbles. When he divides them in groups of two unequal numbers,the sum of squares of the two numbers is 150 times the difference of the two. Which of thefollowing could be the total number of marbles?(1) 27 (2) 35 (3) 42 (4) 53

Sol.x2 + y2 = 150 (x – y) so x2 + y2 is even. This is possible when both x and y are even or both odd. In any casex + y is even. Only one option is even number. The parts are 18 and 24. Ans.(3)Alternative Solution: As the sum is a multiple of 150, the unit’s digits of the squares of two unequalnumbers should be 9, 1 or 6, 4 or 5, 5. Hence the numbers should end in 3, 9; 2, 4; 2, 6; 8, 6; 5, 5; 8, 4;7, 1; 3, 1; 3, 9. Hence the unit’s digit of the total should be 0, 2, 4, 6 or 8.

ØØ If a is a positive integer, and if the unit digit of a2 is 9 and the unit digit of (a + 1)2 is 4 thenwhat is the unit digit of (a + 2)2 ?(1) 1 (2) 3 (3) 5 (4) 7

Sol.Unit digit of a can be 3 or 7, but unit digit of (a + 1)2 = 4unit digit of a = 7. Hence units digit of (a + 2)2 = 1. Ans.(1)

ØØ How many times do you have to write the digit ‘3’, while writing numbers from 1 to 1000?(1) 189 (2) 300 (3) 201 (4) 430

Sol. In every 100 numbers 3 comes 10 times in the unit place and 10 times in the tens place = 20 times.∴Till 1000 we have 3 coming (20 × 10 ) 200 times.Also 3 comes at hundreds place for 100 times (300 to 399).Hence total number of times 3 is written between 1 to 1000 = 200 + 100 = 300. Ans.(2)

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ØØ If (pqr)2 = (ijkpqr), where i, j, k, p, q, r ∈∈ W, and pqr and ijkpqr are 3 digit and 6 digit numbersrespectively. Then the value of i × j × k × p × q × r is(1) 0 (2) 720 (3) can’t say (4) None of these

Sol.There are only 4 digits whose square also gives the same unit digit which are 0, 1, 5 and 6.But the number whose unit digit is 0 (zero), this number after squaring makes the ten’s digit zero also.Again it means the tens digit of the original number be zero. Now if we square such a number whose tensand unit digits are zero, this number gives 4 zeros (at least) on the end of the square which is not possiblesince 000 can not be considered a relevant number. Thus zero cannot be the unit digit.Now, if we consider unit digit 1, and any digit except zero.Note: Zero shows the same behaviour as in the first case, then unit digit remains the same on beingsquared the given number but tens digit gets changed so, 1 as a unit digit is also impossible.Now if we consider 6 as probable unit digit similar behaviour is seen with 6 also.Thus, if we consider 5 as a unit we always get 5 as a unit digit after squaring any number whose unit digitis 5. Also we always get the 2 as the tens digit when we square any number whose unit digit is 5. Now tomaintain the tens digit ‘2’ we consider a number whose tens and unit digits be respectively 2, 5. Now if wesquare to 25, we get a three digit number 625. It means if it is possible, then the three digit given numbermust be 625.Now, (625)2=390625Which satisfies the given conditions.Now 3 × 9 × 0 × 6 × 2 × 5 = 0Hence (1) is correct. Ans.(1)

ØØ If ab, cd, ba and dc are two digit numbers then the maximum value of (ab × cd) – (ba × dc) is,where a, b, c, d are distinct non-zero integers(1) 7938 (2) 7128 (3) 6930 (4) None of these

Sol. (10a + b) × (10c + d) – (10b + a) × (10d + c)= (100a.c + 10b.c + 10a.d + b.d) – (100b.d + 10a.d + 10b.c + a.c)= 99 (a.c – b.d)Now, in order to the difference be maximum so a.c will be maximum and b.d will be minimum, thusa × c = 9 × 8 = 72and b × d = 1 × 2 = 2Hence 99 (ac – bd) = 99(72 – 2) = 99 × 70 = 6930. Ans.(3)

ØØ Let a, b, c be distinct digits. Consider a two-digit number ‘ab’ and a three-digit number ‘ccb’,both defined under the usual decimal number system, if ab2 = ccb > 300, then the value of b is(1) 1 (2) 2 (3) 5 (4) 6

Sol. (ab)2 = ccb, the greatest possible value of ‘ab’ to be 31. Since 312 = 961 and since ccb > 300, 300 < ccb < 961,so 18 < ab < 31. So the possible value of ab which satisfies (ab)2 = ccb is 21. So 212 = 441,

∴ a = 2, b = 1, c = 4. Ans.(1)


For PT Faculty use onlyØØ (BE)2 =MPB, where B, E, M and P are distinct integers, then M ?

(1) 2 (2) 3 (3) 9 (4) None of these

Sol.Since MPB is a three-digit number, and also the square of a two-digit number, it can have a maximum valueof 961 viz. 312. This means that the number BE should be less than or equal to 31. So B can only take thevalues 0, 1, 2 and 3. Since last digit of MPB is also B, it can only be 0 or 1 (as none of the squares ends in2 or 3). The only squares that end in 0 are 100, 400 and 900. But for this to occur the last digit of BE alsohas to be 0. Since E and B are distinct integers, both of them cannot be 0. Hence, B has to be 1. BE can bea number between 11 and 19 (as we have also ruled out 10), with its square also ending in 1. Hence, thenumber BE can only be 11 or 19.112 = 121. This is not possible as this will mean that M is also equal to 1.Hence, our actual numbers are 192 = 361. Hence, M = 3. Ans.(2)

ØØ P, Q and R are three consecutive odd numbers in ascending order. If the value of three times Pis 3 less than two times R, find the value of R.(1) 5 (2) 7 (3) 9 (4) 11

Sol.As P, Q and R are consecutive odd numbers, Q = P + 2 and R = P + 4. Now 3 P = 2 (P + 4) – 3. On solvingthis equation, we get P = 5. Therefore, R = 5 + 4 = 9. Ans.(3)

DIRECTIONS: A boy is supposed to put a mango into a basket if ordered 1, an orange if ordered 2 and an appleif ordered 3. He took out 1 mango and 1 orange if ordered 4. He was given the following sequence of orders.

12332142314223314113234

ØØ At the end of the sequence, what will be the number of oranges in the basket?(1) 2 (2) 3 (3) 4 (4) 6

Sol.1 2 3 3 2 1 4 2 3 1 4 2 2 3 3 1 4 1 1 3 2 3 4Number of oranges put in = Number of times for orders 2 = 6Number of oranges taken out = Number of times for 4 = 4Required answer = 6 – 4 = 2. Ans.(1)

ØØ At the end of the sequence, what will be the total number of fruits in the basket?(1) 10 (2) 11 (3) 13 (4) 17

Sol.Number of fruits put in = Number of times for orders 1, 2 or 3 = 19Number of fruits taken out = 2 × 4 = 8Required answer = 19 – 8 = 11. Ans.(2)

ØØ Let x, y and z be distinct positive integers satisfying x < y < z and x + y + z = k. What is thesmallest value of k that does not determine x, y, z uniquely ?(1) 9 (2) 6 (3) 7 (4) 8

Sol. If x + y + z = 8, (x, y, z) could be (1, 2, 5) or (1, 3, 4). Ans.(4)

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DIRECTIONS: A, B, C and D collected one rupee coins following the given pattern.

Together they collected 100 coins.Each one of them collected even number of coins.Each one of them collected at least 10 coins.No two of them collected the same number of coins.

ØØ The maximum number of coins collected by any one of them cannot exceed(1) 64 (2) 36 (3) 54 (4) None of these

Sol.For anyone of them to collect maximum number of coins, the remaining three should collect minimumnumber of coins. And from the conditions given, this has to be 10, 12 and 14. So if the three of them collect(10 + 12 + 14) = 36 coins, the fourth one has to collect (100 – 36) = 64 coins, which has to be the maximumby any one person. Ans.(1)

ØØ If A collected 54 coins, then the difference in the number of coins between the one who collectedmaximum number of coins and the one who collected the second highest number of coins mustbe at least(1) 12 (2) 24 (3) 30 (4) None of these

Sol.Since A has collected 54 coins out of 100, he should obviously be the person who collected the maximumnumber of coins. For the difference between him and the second highest person to be minimum, the secondhighest person should collect maximum number of coins possible under given conditions. And for this tohappen, the remaining two should collect minimum number of coins. So if the two of them collect 10 and 12coins, i.e., 22 coins between themselves, the third person would to collect (100 – 54 – 22) = 24 coins.Hence, the difference between him and the highest person should at least be (54 – 24) = 30. Ans.(3)

ØØ If A collected 54 coins and B collected two more coins than twice the number of coins collectedby C, then the number of coins collected by B could be(1) 28 (2) 20 (3) 26 (4) 22

Sol. If A has collected 54 coins. The remaining three of them should collect (100 – 54) = 46 coins betweenthemselves. Let us assume that C has collected 10 coins. So, B will collect (2 × 10) + 2= 22. So, D willcollect (46 – 10 – 22) = 14 coins, which is a possible combination. Let us now assume that C picks up 12coins, so B should pick up (2 × 12) + 2 = 26. So, D will have to collect (46 – 12 – 26) = 8 coins. Thiscombination is not possible. It can be concluded that C cannot pick up more than 10 coins and hence B hasto pick up 22 coins to satisfy the given condition. Ans.(4)

ØØ If 223 + 233 + 243 + ... + 873 + 883 is divided by 110 then the remainder will be(1) 55 (2) 1 (3) 0 (4) 44

Sol. (22)3 + (23)3 + (24)3 + ... + (87)3 + (88)3

(223 + 883) + (233 + 873) + (243 + 863) + (253 + 853 ...+(543 + 563) + 553

Now since we know that an + bn is divisible by (a + b) when n is an odd number. Therefore all the terms,except (55)3, is divisible by 110.Now the remainder when (55)3 is divided by 110 is 55.Hence the required remainder of the whole expression is 55. Ans.(1)


For PT Faculty use onlyØØ 1026 – 1 is divisible by

(1) 3 (2) 3 and 9 (3) 11 (4) 3, 9 and 11

Sol.1026 – 1 = 9999 ... (26 times). So 1026 – 1 is divisible by 3, 9, 11 and 99. Ans.(4)

ØØ N is the largest possible number divisible by 8 and consisting of all distinct digits in the decimalsystem. What is the remainder when N is divided by 1000?(1) 210 (2) 240 (3) 480 (4) 120

Sol.Divisibility by 8 depends on the last 3 digits. Now the largest number is 9876543210. The last three digitscan be written as 210, 120, 102, 201, 012, 021. Only 120 is divisible by 8, so the largest such number is9876543120. When divided by 1000, the remainder is surely 120. Ans.(4)

ØØ The remainder when 83 8312 14++ is divided by 169 will be

(1) 2 (2) 0 (3) 130 (4) 156

Sol.The given number = ( ) ( )− + +83 8313 1 13 1 . Now all terms except the last 2 terms will be divisible by 169.

Also the last terms of the two expressions will cancel each other as the power is odd. Hence the remainder

will be ( ) ( )[ ] 1691138311383 8282 +××+−×× , which is 130. Ans.(3)

ØØ Let N be an integer not divisible by 5. Which of the following numbers must leave 1 when dividedby 5?(1) N2 (2) N4 (3) N6 (4) None of these

Sol.Ans.(2)

ØØ 23PQ59 is a number of all distinct digits divisible by 11. Then the two digit number PQ will be(1) even (2) composite (3) prime (4) can’t say

Sol. (2 + P + 5) – (3 + Q + 9) = 0 or 11 or –11 etc. This gives P – Q = 5 and P – Q = –6. As all digits are distinct,the only possible solutions can be PQ = 61 or 17, i.e., prime number. Ans.(3)

ØØ What will be the remainder obtained when (1234567890123456789)24 is divided by 6561?

Sol. (1234567890123456789) is the number given to us. Sum of all the digits= 2 (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) = 2 (45) = 90, i.e., this number is divisible by 9. Thus we can write(1234567890123456789)24 as (9K)24. Similarly 6561 can be written as 94. This means that the given numberwill be completely divisible by 6561 and remainder will be 0.

ØØ Two number x and y are such that when divided by 6, they leave remainders 4 and 5 respectively.Find the remainder when x2 + y2 is divided by 6.

Sol.Take x = 10 and y = 11Therefore the remainder when 100+121 is divided by 6 is 5

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ØØ 4444 .... (999 times) is definitely divisible by(1) 22 (2) 44 (3) 222 (4) 2222

Sol.Out of the whole group, we can make a group of (444) 333 times and every group is perfectly divisible by222. Ans.(3)

ØØ 76n – 66n, where n is an integer > 0, is divisible by(1) 13 (2) 127 (3) 560 (4) both 1. and 2.

Sol.Put n = 176n – 66n = 76 – 66 = (73 – 63) (73 + 63)This is a multiple of 73 – 63 = 127 and 73 + 63 = 559 which is divisible by 13 . Ans.(4)

ØØ Find the product of 0.31 42857• •

and 0.1590• •

.

(1) 0.0585 (2) 0.0624 (3) 0.05 (4) None of these

Sol.First convert each recurring decimal as a fraction. To convert 0 3142857.• •

, write the number without thedecimal as 3142857.The digit 3 is not recurring. So subtract 3 from the above number, to get 3142857 – 3 = 3142854. There aresix recurring digits and one non recurring digit. Hence write six 9’s and one zero to get 999999. Now divideto get

0 314285731428549999990

142857 22142857 70

2270

1135

.• •

= =××

= =

Similarly 015901590 15

990015759900

744

.• •

=−

= =

The required product = 1135

744

15 4

120

0 05× =×

= = . . Ans.(3)

ØØ The last digit of the expression 4 × 92 × 43 × 94 × 45 × 96 × ... × 499 × 9100 is(1) 4 (2) 6 (3) 9 (4) 1

Sol.The unit digit of each pair is 4, and there are 50 such pairs which are mutually multiplied together. Thusfinally we get 6 as unit digit. As

4 × 92 4 × 93 4 4 × 95 6 4 × 999 100

↓ ↓ ↓ ↓× ×

4 4 4 4

× ... ×Unitdigit

Again 4 × 4 × 4 × 4 ... 4 (upto 50 times)i.e., the unit of 450, which is 6[Since unit digit of 42n is 6 for n = 1, 2, 3, ... etc.] . Ans.(2)


For PT Faculty use onlyØØ The unit digit of 234 × 345 × 456 × 567 × 678 × 789 is

(1) 0 (2) 5 (3) can’t be determined (4)None of these

Sol.The unit digit of 234 is 2.

The unit digit of 345 is 1.





Therefore the unit digit of :234 × 345 × 456 × 567 × 678 × 789

= 0(Since 2 × 1 × 4 × 5 × 6 × 1 = 240). Ans.(1)Alternatively: 5 always gives unit digit 5 irrespective of its (positive) power and similarly 6 also gives 6 asa unit digit irrespective of its power. So, 5 × 6 = 30. Hence you will get the unit zero without solving thecomplete expression.

ØØ How many pairs of natural numbers satisfy the condition that the sum of their reciprocals is 1

12?

(1) 16 (2) 6 (3) 8 (4) None of these

Sol. + =1 1 1a b 12

or ab – 12a – 12 b + 144 = 144 or (a – 12) (b – 12) = 144. 144 has 15 factors.

So there can be 8 pairs. (The number of ordered pairs will be different) Ans.(3)

ØØ What will be number of zeroes at the end of (10! + 20! + 30! + 40! )(10! + 20! + 30! + 40!)

Sol.The above expression can be written as[10! (1 + 20 ×19 × ...... + 30 × 29 × ..... + 40 × 39 × .....)](10! + 20! + 30! + 40!) .10! ends in 2 zeros.Total number of zeros at the end = Number of zeros in the base × Intex.∴ Total number of zeros = 2 × (10! + 20! + 30! + 40!).

ØØ(144 1)(144 1)(1728 1)(1728 1)

11

8 12 2 6

yRemainder 0

− − − −LNM

OQP =

. What will be the maximum value of y?

Sol.We know that an – bn = divisible by (a – b) and (a + b) when n is even.

Given that 12 1 12 1 12 1 12 1

11

2 8 2 12 3 2 3 6

0

d i d i d i d i−RSTUVW −RST

UVW −RSTUVW −RST

UVWL

N

MMMM

O

Q

PPPP =

y

mainderRe

∴ Maximum value of y = 4.

fsb by pt for number system 2

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