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## 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

Number theory

Algebra

Prime

## Check the Answer

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$(3,5,7)$

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## 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

Number theory

Algebra

Prime numbers

## Check the Answer

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.

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## 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

Integer

Divisor

Number theory

## Check the Answer

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 $20$

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## 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

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 $1+1=2$

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## Problem on Prime Numbers | SMO, 2012 | Problem 20

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2012 based on Prime numbers.

## Problem on Prime Numbers – (SMO Test)

Let A be a 4 – digit integer. When both the first digit (leftmost) and the third digit are increased by n, and the second digit and the fourth digit are decreased by n, the new number is n times A. Find the value of A.

• 1201
• 1551
• 1818
• 2000

Algebra

Prime Number

## Check the Answer

Challenges and Thrills – Pre – College Mathematics

## Try with Hints

If you got stuck you can follow this hint:

We can assume the 4 digit number to be A = $\overline {abcd}$

If we expand it into the equation

1000(a+n) + 100(b – n) + 10(c+n) + (d-n) = nA

Try the rest of the sum ………..

After the previous hint :

If we compare the equation it gives :

A + 909 n = nA or

(n-1)A = 909 n

Now one thing we can understand that n and (n-1) are relatively prime and 101 is a prime number . So n= 2 or n= 4.

We have almost got the answer .So try to do the rest now ……….

If n = 4 then A = 1212, which is impossible right?

as b<n given .so

n=2 and A = $909 \times 2$ = 1818