## Formulas:

Divisibility test: to test a number that is divisible by a particular number.

Number | Divisibility Test |
---|---|

Div by 2 | If the unit place is even i.e. divisible by a multiple of 2 or 0. |

Div by 3 | If the sum of all the numbers is divisible by 3. |

Div by 4 | If the last two digits are divisible by 4. |

Div by 5 | If the unit digit is either 5 or 0. |

Div by 6 | If the given number is divisible by both 2 and 3. |

Div by 7 | If twice the number at the unit place is subtracted from the rest of the digits and the remainder is divisible by 7. e.g. 875 = 87-2(5) = 77/ 7 = 11 hence the number is divisible by 7ss |

Div by 8 | If the last three digits are divisible by 8. |

Div by 9 | If the sum of digits is divisible by 9. |

Div by 11 | If the difference between the sum of digits at even places and odd places is either ‘0’ or divisible by 11. |

**Point to Remember: **If N is a composite number of the form N= a^{p} .b^{q}…, where a and b are primes then the number of divisors of N is given by n=(p+1)(q+1)…

**Point to Remember: **The sum of the divisors of N, represented by S is: S= ((a^{p+1 }– 1) . (b^{q+1} – 1)…) / (a-1)(b-1)…

## Practice Problems:

**Problem**: One of the factors of (7^{2k} + 3^{2k}), where k is odd is an odd number, is:

**Solution:**

Putting k=1, as k is odd we got (49 + 9) = 58 so 58 is the number which divides the system.

**Problem:** The sum of the digits of two digits is 1/7 of the number. The unit digits are 4 less than the tens digit. if the number obtained by reversing its digits is divided by 7. The remainder will be:

**Solution:**

1/7(10x + y) = x + y

10x + y = 7x + 7y

3x = 6y

x/y = 2*m/1*m , x – y = 4

2m – m = 4, m=4

So the digits are x = 2m = 8 and y = m = 4

So the number is 84. Reverse is 48

So 48/7 remainder is 6.

**Problem: **If p is a prime number and p divides ab i.e. ab/p, where ‘a’ and ‘b’ are integers, then.

(a) a/p or b/p | (b) a/p and b/p |

(c) (a-b) / p | (d) none. |

**Solution:**

The concept used here is that as p is a prime it can only divide its multiples and as a, b is a product which is divisible by p so p must divide one of these two so option (a) is the correct answer.

**Problem**: The product of a rational number and an irrational number will always be.

**Solution:**

The product will always be an irrational number.

**Problem**: What is the value of x for which x, x + 1, x + 3 are all prime numbers?

(a). 0 | (b). 1 |

(c). 2 | (d). 101 |

**Solution:**

There is only one option which satisfies this which is option (c) as substituting x = 2 we get 2, 3, and 5 which are all primes.

**Problem:** In a division operation, the divisor is 5 times the quotient and twice the remainder. if the remainder is 15, then what is the dividend?

**Solution:**

Dividend = d * q + r.

Given, d = 5q and d=2r

When r=15, d= 2 * 15 = 30

Q=d/5 =30 / 5 = 6

Dividend = 30 * 6 + 15 = 195

**Problem**: The unit digit in the product (127)^{170} is

**Solution:**

The unit digit in 7^{4 }is 1.

Take the unit digit 127 which is 7

(7^{42})^{4} * 7^{2} = unit digit of 7^{42} is 1 and 1 power 4 is also 1

So the remaining is 7 square which is 49 so the unit digit is 9.

**Problem:** The unit digit in the product(7^{71 }* 6^{59} * 3^{65}) is

**Solution:**

As the unit digit of 7^{4 }= 3^{4 }=1

And unit digit of 6 raised to any power is 6

So the unit digit of the system is (7^{68 + 3 } * 6 * 3^{64+1}) = (3 *6*3) =54

So the unit digit of the system is 4.

**Problem**: When n is divided by 4, the remainder is 3. What is the remainder when 2n is divided by 4 ?

**Solution:**

n = 4q + 3

2n = 8q + 6

2n = (8q + 4) + 2 = 4(2q + 1) + 2

So if 2n is divided by 4 the remainder will be 2.

**Problem:** If x/y = 3/5, then the value of x-y / x+y is.

**Solution:**

put x = 3 and y = 5 in eqn

3 – 5 / 3 + 5 = -2/8 = -1/4.

**Problem**: Given a number 222222 is divisible by

(a). 3 | (b). 7 |

(c). 11 | (d). All the above |

**Solution:**

Any number which is repeated 6 times is always divisible by 3,7 and 11.

**Problem:** The last digit or unit digit in the expansion of (2457)^{754} is.

**Solution**:

The last digit in (2457)^{754} is the same as the (7)^{754} as 7^{4} = 1

We can write 754 in 188*4 + 2 so the unit digit will be the square power of 7

Which is 49 whose unit digit is 9 so the unit digit of the expansion is 9.

**Problem:** A number when divided by 2,3 or 5 gives the remainder 1. The smallest number is ?

**Solution:**

There is a formula for finding the smallest number when the remainder is the same for all i.e.

Required number = [LCM of 2,3 and 5] + remainder

= 30 + 1 = 31.

So the number is 31.

**Problem:** How many factors of 1080 are perfect squares?

**Solution:**

1080 = 2^{3} x 3^{3} x 5

For any perfect square, all the powers of the primes have to be even numbers. So, if the factor is of the form (2^{a} x 3^{b} x 5^{c})

The values a can take are 0 and 2, b can take 0 and 2, and c can take 0 only. So the total possibility is 2x2x1 = 4

**Problem:** What minimum number should be added to 131228 so that it is completely divided by 33

**Solution:**

131228 = 33(3976) + 20

Now 33 – 20 = 13

So 13 is the number which when added is completely divided by 33

**Problem:** What is the maximum value of m, if the number N = 90 x 42 x 324 x 55 is divisible by 3^{m} ?

**Solution:**

3^{2} x 10 x 3 x 7 x 2 x 3^{4} x 4 x 5 x 11 (by factorising)

= 3^{7} x 10 x 7 x 2 x 4 x 5 x 11

Hence it is divisible by 3^{7} so the m = 7.

**Problem:** 7^{10} – 5^{10} is divisible by

**Solution:**

As we know as the powers are even this will be divisible by (x+y) i.e. (7+5)

So this is divisible by 11.

**Problem:** The number of pairs (x,y), where x,y are integers satisfying the equation 21x + 48y = 5, is

**Solution:**

There is a concept of the Diophantine equation in which if the HCF(21,48) divides 5 then there are infinitely many solutions but here HCF is 7 which doesn’t divide 5 so there is 0 integral solution satisfying this equation.

**Problem:** The seven-digit number 876p37q is divisible by 225. The values of p and q can be respectively.

**Solution:**

As the number 225 is divisible by 5 and 9 then the seven-digit number also be divisible by 5 and 9.

If divisible by 9 then = 8 + 7 + 6 + p + 3 + 7 + q should also be divided by 9

= 31 + p + q so ,

p + q should be 5 or 14

Given number is divisible by 5 then q =5 so p = 0, 9.

**Problem:** What is the remainder when (17^{29} + 19^{29}) is divided by 18?

**Solution: **

= (17^{29} + 19^{29})/ 18

=(18 – 1)^{29}/18 + (18+1)^{29}/18

= (-1)^{29} + (1)^{29} = -1 + 1 = 0

**Problem:** The number 3^{521} is divided by 8. What is the remainder?

**Solution:**

according to the congruence theory

3^{2} ~ 1mod(8)

(3^{2})^{260} ~ 1 mod(8)

3^{520} * 3 ~ 3mod(8)(multiplying both sides of congruence by 3)

So the remainder will be 3.

**Problem: ** The number of divisors of a number 1080 excluding itself and 1, is.

**Solution:**

1080 = 2 x 2 x 2 x 3 x 3 x 3 x 5

= 2^{3} x 3^{3} x 5

So the number of divisors excluding 1 and itself is

= (3 + 1)(3 + 1)(1 + 1) – 2

=30

**Problem:** The number of prime numbers which are less than 100 is

**Solution:**

The number of prime numbers less than 100 are 25 and the numbers are 2, 3,5, 7, 11, 13, 17, 19, 23, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

**Problem:** √20-√20-√20-….infinite (the roots are under other roots from left to right) = ?

**Solution:**

Here we use the direct formula which is, x= (√(4a+1) – 1)/2

Putting a= 20 we get x = 9-1/4 => 4