Categories

## Sum of divisors and Integers | TOMATO B.Stat Objective 99

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Sum of divisors and Integers.

## Sum of divisors and Integers (B.Stat Objective Question)

The sum of all positive divisors of 1800, where 1 and 1800 are also considered as divisors of 1800, is

• 104
• 6045
• 1154
• none of these

### Key Concepts

Integers

Sum of divisors

Exponents

B.Stat Objective Problem 99

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

here 1800=(2)(2)(2)(3)(3)(5)(5) where n=$(p_1^a)(p_2^b)(p_3^c)$

Second Hint

sum of divisors of 1800

=$(\frac{2^{4}-1}{2-1})$$(\frac{3^{3}-1}{3-1})$$(\frac{5^{3}-1}{5-1})$ where sum of divisors of n=$(\frac{p_1^{a+1}-1}{p_1-1})(\frac{p_2^{b+1}-1}{p_2-1})(\frac{p_3^{c+1}-1}{p_3-1})$

Final Step

=6045.

Categories

## Divisibility and Integers | TOMATO B.Stat Objective 89

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Integers and divisibility.

## Divisibility and Integers (B.Stat Objective Question )

300 digit number with all digits equal to 1 is

• divisible neither by 37 nor by 101
• divisible by both 37 and 101
• divisible by 37 and not by 101
• divisible by 101 and not by37

### Key Concepts

Integers

Remainders

Divisibility

Answer: divisible by 37 and 101

B.Stat Objective Problem 89

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

here we take 300 digit number all digit 1s

Second Hint

111…11=$\frac{999…99}{9}$(300 digits)

=$\frac{10^{300}-1}{9}$=$\frac{(10^{3})^{100}-1}{9}$=$\frac{(10^{3}-1)X}{9}$

since $10^{3}-1$=999 is divisible by 37 then 111…11(300 digits) is divisible by 37

Final Step

111…11=$\frac{999…99}{9}$(300 digits)

=$\frac{10^{300}-1}{9}$=$\frac{(10^{4})^{75}-1}{9}$=$\frac{(10^{4}-1)Y}{9}$

since $10^{4}-1$=9999 is divisible by 101 then 111…11(300 digits) is divisible by 101.

Categories

## Sets and Probability | B.Stat Objective Problems

Try this problem from I.S.I. B.Stat Entrance Objective Problem from TOMATO based on Sets and Probability. You may use sequential hints to solve the problem.

## Sets and Probability (B.Stat Objective problems)

Sixty students appeared in a test consisting of three papers I ,II, and III. Of these students, 25 passed in paper I, 20 in paper II and 8 in paper III. Further 42 students passed in at least one of papers I and II, 30 in at least one of papers I and III, 25 in at least one of papers II and III. Only one student passed in all the three papers. Then the number of students who failed in all the three papers is

• 17
• 15
• 45
• 33

### Key Concepts

Sets

Probability

Algebra

B.Stat Objective Problem

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

Let failed P I=P(A)=35, p II=P(B)=40, p III =P(C)=52, P (IandII)=$P(A \bigcap B)$= 18 P (I and III)=$P(A \bigcap C)$=30 P (II and III)=$P(B \bigcap C)$=35 P (I or II or III)=$P(A \bigcup B \bigcup C)$=59

Second Hint

Then by Poincare Theorem, $P(A \bigcup B \bigcup C)$=P(A)+P(B)+P(C)-$P(A \bigcap B)$-$P(A \bigcap C)$-$P(B \bigcap C)$+x

Then 59=35+40+52-18-30-35+x where x is the required value

Final Step

then x=59-44=15.

Categories

## Problem – Number Series ( B.Stat Objective Problem )

We are going to discuss about Number Series from B.Stat Objective Problem .

A student studying the weather for d days observed that(i) it rained on 7 days morning or afternoon, (ii) when it rained in the afternoon it was clear in the morning, (iii) there were five clear afternoons (iv) there were six clear mornings. Then d equals

• 8
• 9
• 11
• 10

### Key Concepts

Series

Algebra

Integers

B.Stat Objective Problem

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

Here A1, A2, A3, A4, A5, A6, A7, A8, A9 are mornings of day 1, day 2, day 3, day 4, day 5, day 6, day 7, day 8 and day 9 again B1, B2, B3, B4, B5, B6, B7, B8, B9 are afternoons of day 1, day 2, day 3, day 4, day 5, day 6 and day 7, day 8 and day 9

A1 A2 clear morning B1 B2 rainy afternoon from first two conditions

Second Hint

A3 rainy A4 clear A5 rainy A6 clear A7 rainy and B3 clear B4 rainy B5 clear B6 rainy B7 clear from first two conditions

Final Step

So, from first 7 days, 4 clear mornings and 3 clear afternoons. Then A8 clear A9 clear B8 clear B9 clear from last two conditions .Then all condition satisfied that is 4+2=6 mornings clear 3+2 =5 afternoons clear. d=9 that is in 9 days.

Categories

## Problem – Numbers and Group (B.Stat Objective Problem)

We are going to discuss about Numbers and Group from B.Stat Objective Problem .

In a group of 120 persons there are 80 Bengalis, 40 Gujratis, Further 71 persons in the group are Muslims and the remaining Hindus, Then the number of Bengali Muslims in the group is

• 8
• more than 30
• 11
• 10

### Key Concepts

number Series

Algebra

Integers

B.Stat Objective Problem

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

Assuming that out of 80 Bengalis, when 49 are Hindus then 31 are Muslims and when 9 are Hindus 71 are Muslims

Second Hint

Out of 71 Muslims when 40 are Gujratis 31 are Bengalis else 71 are Bengalis which contradicts with other options

Final Step

Then number of Bengali Muslims are from 31 to 71 that is more than 30.

Categories

## Greatest Integer and remainder | TOMATO B.Stat Objective 113

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Greatest Integer and remainder.

## Greatest integer and remainders (B.Stat Objective Question)

The greatest integer which, when dividing the integers 13511, 13903 and 14593 leaves the same remainder is

• 98
• 2
• 7
• 56

### Key Concepts

GCD

Greatest Integer

Remainder

B.Stat Objective Problem 113

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

here all numbers are odd

Second Hint

13511, 13903 and 14593 leaves different remainders when divided by 98, 56 and 7

Final Step

13511, 13903 and 14593 leaves same remainder when divided by 2

then greatest integer 2.

Categories

## Number of divisors and Integer | B.Stat Objective | TOMATO 83

Try this TOMATO problem from I.S.I. B.Stat Entrance Objective Problem based on Number of divisors and Integer.

## Number of divisors and Integer (B.Stat Objective)

The smallest positive integer n with 24 divisors (where 1 and n are also considered divisors of n) is

• 240
• 360
• 420
• 480

### Key Concepts

Number of divisors

Integer

Least positive integer

B.Stat Objective Question 83

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

number of divisors of 420=(2+1)(1+1)(1+1)(1+1)=24

Second Hint

number of divisors of 240=(4+1)(1+1)(1+1)=20

Final Step

number of divisors of 360=(3+1)(2+1)(1+1)=24 and number of divisors of 480=(5+1)(1+1)(1+1)=24 then required number =360.

Categories

## Combinatorics and Integers | TOMATO B.Stat Objective 93

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Combinatorics and Integers.

## Combinatorics and Integers (B.Stat Objective Question)

The highest power of 18 contained in ${50 \choose 25}$ is

• 104
• 1
• 1154
• none of these

### Key Concepts

Integers

Combinatorics

Exponents

B.Stat Objective Problem 93

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

here ${50 \choose 25}$=$\frac{50!}{(25!)^{2}}$=$\frac{(50)(49)(….)(26)}{(25)(24)(…)(1)}$

Second Hint

$=(2)^{13}(49)(47)(45)(43)(41)(39)(37)(35)(33)(31)(29)(27) \times \frac{1}{12!}$

Final Step

$=(2)^{10}(49)(47)(15)(43)(41)(13)(37)(35)(11)(31)(29)\times \frac{1}{(12)(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)}[(2)^{3}(27)(9)(3)]$

$=(2)^{10}(49)(47)(15)(43)(41)(13)(37)(35)(11)(31)(29)\times \frac{1}{(12)(11)(10)(8)(7)(5)(4)(1)}[(2)(9)]$gives a factor of $(18)^{1}$ then highest power of 18 is 1.

Categories

## Number of Factors | TOMATO B.Stat Objective 95

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Number of factors.

## Number of Factors (B.Stat Objective Question)

The number of different factors of 1800 equals

• 104
• 36
• 1154
• none of these

### Key Concepts

Integers

Number of divisors

Exponents

B.Stat Objective Problem 95

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

here 1800=(2)(2)(2)(3)(3)(5)(5)

Second Hint

number of divisors of 1800 =(3+1)(2+1)(2+1)

Final Step

=(4)(3)(3)=36.

Categories

## Number of divisors and Integers | TOMATO B.Stat Objective 97

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Number of divisors and Integers.

## Number of divisors and Integers (B.Stat Objective Question)

The number of different factors of 6000, where 1 and 6000 are also considered as divisors of 6000, is

• 104
• 40
• 1154
• none of these

### Key Concepts

Integers

Number of divisors

Exponents

B.Stat Objective Problem 97

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

Here 6000=(2)(2)(2)(2)(3)(5)(5)(5)

Second Hint

number of divisors of 6000 =(4+1)(1+1)(3+1) where number of divisors=(a+1)(b+1)(c+1) for n=$p_1^{a}p_2^{b}p_3^{c}$ as $p_1,p_2,p_3$ are primes

Final Step

=(5)(2)(4)=40.