Prime Number for ISI BStat | TOMATO Objective 70

Try this beautiful problem from Integer based on Prime number useful for ISI BStat Entrance.

Prime number | ISI BStat Entrance | Problem no. 70


The number of integers \(n>1\), such that n, n+2, n+4 are all prime numbers is ......

  • Zero
  • One
  • Infinite
  • More than one,but finite

Key Concepts


Number theory

Algebra

Prime

Check the Answer


Answer: One

TOMATO, Problem 70

Challenges and Thrills in Pre College Mathematics

Try with Hints


taking n=3, 5, 7, 11, 13, 17....prime numbers we will get

Case of n=3

n= 3

n+2=5

n+4=7

Case of n=5

then \(n\)=5

n+2=7

n+4=9 which is not prime....

Case of n=7,

then n=7

n+2=9 which is not prime ...

n+4=11

Can you now finish the problem ..........

We observe that when n=3 then n,n+2,n+4 gives the prime numbers.....other cases all are not prime.Therefore any no can be expressed in anyone of the form 3k, 3k+1 and 3k+2.

can you finish the problem........

If n is divisible by 3 , we are done. If the remainder after the division by 3 is 1, the number n+2 is divisible by 3. If the remainder is 2, the number n+4 is divisible by 3

The three numbers must be primes! The only case n=3 and gives\((3,5,7)\)

Subscribe to Cheenta at Youtube


Hundred Integers | ISI-B.Stat Entrance | TOMATO 82

Try this beautiful hundred integers problem based on Remainder useful for ISI B.Stat Entrance.

Hundred Integers | ISI B.Stat Entrance | Problem-82


Let \(x_1,x_2,......,x_100\) be hundred integers such that the sum of any five of them is 20. Then..

  • the largest \(x_i\) equals 5
  • the smallest \(x_i\) equals to 3
  • \(x_{17} = x_{83}\)
  • none of the foregoing statements is true

Key Concepts


Number theory

Divisor

integer

Check the Answer


Answer:\(x_{17} = x_{83}\)

TOMATO, Problem 82

Challenges and Thrills in Pre College Mathematics

Try with Hints


Let us take the numbers be \(x_i , x_j , x_k ,x_l , x_m \)

Now \(x_i + x_j + x_k + x_l + x_m = 20\) and again \(x_i + x_j + x_k + x_l + x_n = 20\)

Can you now finish the problem ..........

From the above relation there are three case arise that....

1)\(x_m = x_n\)

2)All the integers are equal.

3)\(x_{17} =x_{83}\)

So the correct answer is \(x_{17} =x_{83}\)

Subscribe to Cheenta at Youtube


Prime number Problem | ISI BStat | TOMATO Objective 96

Try this beautiful problem from Integer based on Prime number useful for ISI B.Stat Entrance.

Prime number | ISI B.Stat Entrance | Problem no. 96


The number of different prime factors of 3003 is.....

  • 2
  • 15
  • 7
  • 16

Key Concepts


Number theory

Algebra

Prime numbers

Check the Answer


Answer: 16

TOMATO, Problem 96

Challenges and Thrills in Pre College Mathematics

Try with Hints


At first, we have to find out the prime factors. Now \(3003\)=\(3 \times 7 \times 11 \times 13\). but now it can be expressed as another prime number also such as \(3003=3 \times 1001\). So we have to find different prime factors.

Can you now finish the problem ..........

Now, if you have a number and its prime factorisation, \(n={p_1}^{m_1} {p_2}^{m_2}⋯{p_r}^{m_r}\) you can make divisors of the number by taking up to \(m_1\) lots of \(p_1\), up to \(m_2\) lots of \(p_2\) and so on. The number of ways of doing this is going to be\( (m_1+1)(m_2+1)⋯(m_r+1)\).

can you finish the problem........

for the given case \(3003\) has \(2^4=16 \)divisors.

Subscribe to Cheenta at Youtube


Integers and Divisors | ISI-B.Stat Entrance | TOMATO 98

Try this beautiful problem from Integer based on Integers and Divisors useful for ISI B.Stat Entrance.

Integers and Divisors | ISI B.Stat Entrance | Problem-98


The number of positive integers which divide 240 (where both 1 and 240 are considered as divisors) is

  • 20
  • 18
  • 16
  • 12

Key Concepts


Integer

Divisor

Number theory

Check the Answer


Answer: 20

TOMATO, Problem 98

Challenges and Thrills in Pre College Mathematics

Try with Hints


We have to find out the number of positive integers which divide 240.so at first we have to find out the factors of 240...

\(240=2 \times 120\)

\(240=3 \times 80\)

\(240=4 \times 60\)

\(240=5 \times 48\)

\(240=6 \times 40\)

\(240=8 \times 30\)

\(240=10 \times 24\)

\(240=12 \times 20\)

\(240=15 \times 16\)

\(240=20 \times 12\)

\(240=24 \times 10\) ..............

so we notice that the divisors are repeat........

Can you now finish the problem ..........

We notice that after \(240=15 \times 16\) this stape all the factors are repeats.....so we have to calculate up to \(240=15 \times 16\) step only....

can you finish the problem........

Therefore the total number of positive integers are \(1,2,3,4,5,6,8,10,12,15,20,24,30,40,48,60,80,120,240\) i.e \(20\)

Subscribe to Cheenta at Youtube


Remainder Problem | ISI-B.Stat Entrance | TOMATO 90

Try this beautiful problem from Integer based on Remainder useful for ISI B.Stat Entrance.

Remainder Problem | ISI B.Stat Entrance | Problem-90


The remainder when \( 3^{12} +5^{12}\) is divided by 13 is......

  • 2
  • 1
  • 4
  • 3

Key Concepts


Division algorithm

Divisor

Number theory

Check the Answer


Answer: 2

TOMATO, Problem 90

Challenges and Thrills in Pre College Mathematics

Try with Hints


The given number is \( 3^{12} +5^{12}\)

we have to check if it is divided by 13 what will be the remainder? if we express the number in division algorithm form then we have........\( 3^{12} +5^{12}=((3)^3)^4+((5)^2)^6)=(27)^4 +(25)^6\)=\(((13 \times 2+1)^4+(13 \times 2-1)^6)\)

Can you now finish the problem ..........

Remainder :

Clearly if we divide \(((13 \times 2+1)^4+(13 \times 2-1)^6)\) by 13 then from \((13 \times 2+1)^4\) , the remainder be 1 and from \((13 \times 2-1)^6)\), the remainder is 1

can you finish the problem........

Therefore the total remainder is \(1+1=2\)

Subscribe to Cheenta at Youtube


Number counting | TOMATO ISI BStat Objective Problem 56

Try this beautiful problem Based on Number counting useful for ISI B.Stat Entrance.

Number counting| ISI B.Stat Entrance |Problem 56


In a group of 120   persons there are 70 Bengalis,35 Gujaratis and 15 Maharashtrians.Further  75 persons in the group are Muslims and the remaining are Hindus.Then the number of Bengali Muslims in the group is

  • between 10 and 14
  • between 15 and 19
  • exactly 20
  • 25 or more

Key Concepts


Number counting

Algebra

Set

Check the Answer


Answer: 25 or more

TOMATO, Problem 56

Challenges and Thrills in Pre College Mathematics

Try with Hints


Find the numbers of Hindus

Can you now finish the problem ..........

Assume that all hindus are in bangalis to find the minimum numbers of muslims persons

can you finish the problem........

Given that total number of person =120

75 persons are muslims

Therefore number of Hindus are(120-75)=45

There are 70 Bengalis and we assume that 45 hindus are in Bengalis to find the minimum number of muslims.

Therefore Bengali Muslims =(70-45)=25

Hence the number of Bengali Muslims in the group is 25 or more

Subscribe to Cheenta at Youtube