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QUESTION: 1

LCM of first 100 natural numbers is N. What is the LCM of first 105 natural numbers?

Solution:

N is the LCM of first 100 natural numbers.

Now 101 is a prime number, so 101 and N do not have any common factors. Hence their LCM = 101 X N

102= 2 X 51, since 2 < 100 and 51 < 100, 102 divides N

103 is a prime number, no common factors between 103 and N

104= 52 X 2; since 2 < 100 and 52 < 100, 105 divides N

105 = 21 X 5; since 5 < 100 and 21 < 100, 105 divides N

Hence, LCM of first 105 natural number is N X 101 X 103

QUESTION: 2

The total number of 3 digit numbers which have two or more consecutive digits identical is:

Solution:

In each set of 100 numbers, there are 10 numbers whose tens digit and unit digit are same. Again in the same set there are 10 numbers whose hundreds and tens digits are same. But one number in each set of 100 numbers whose Hundreds, Tens and Unit digit are same as 111, 222, 333, 444 etc.

Hence, there are exactly (10 + 10 - 1) = 19 numbers in each set of 100 numbers. Further there are 9 such sets of numbers.

Therefore such total numbers = 19 * 9 = 171.

**Alternatively,**

9 * 10 * 10 - 9 * 9 * 9 = 900 - 729 = 171.

QUESTION: 3

Find the last non zero digit of 96!

Solution:

**Correct Answer :- D**

**Explanation : 4*z(96/5)*z(6)**

**=4*z(19)*8**

**=8*(4*z(19/5)*z(9))**

**=8*(4*z(3)*8)**

**=8*(4*6*8)**

**=1536 i.e last digit is 6**

QUESTION: 4

Two players A and B are playing a game of putting ‘+’ and signs in between any two integers written from 1 to 100. A starts the game by putting a plus sign anywhere between any two integers. Once all the signs have been put, the result is calculated. If it is even then A wins and if it is odd then B wins, provided they are putting signs by taking turns one by one and either of them can put any sign anywhere between any two integers. Who will win at the end?

Solution:

Whatever is the sign between two consecutive integers starting from 1 to 100, it will be odd. So, we are getting 50 sets of odd numbers. Now, whatever calculation we do among 50 odd numbers, result will always be even. So, A will win always.

QUESTION: 5

How many divisors of 10^{5} will have at least one zero at its end?

Solution:

In order for a divisor (or any number) to have a zero at its end, it must have a 10 as a factor, i.e., a 2-and-5 pair. Notice that

10^5 = 2^5 x 5^5 = (2^4 x 5^4) x (2 x 5)

Therefore, any divisors of 2^4 x 5^4 will have a zero at its end when multiplied by 2 x 5. Since 2^4 x 5^4 has (4 + 1)(4 + 1) = 25 divisors, 10^5 has 25 divisors that have at least one zero at its end.

QUESTION: 6

Find the remainder when 4^{96} is divided by 6.

Solution:

4^{96}/6, We can write it in this form

(6 - 2)^{96}/6

Now, Remainder will depend only the powers of -2. So,

(-2)^{96}/6, It is same as

([-2]^{4})^{24}/6, it is same as

(16)^{24}/6

Now,

(16 * 16 * 16 * 16..... 24 times)/6

On dividing individually 16 we always get a remainder 4.

So,

(4 * 4 * 4 * 4............ 24 times)/6.

Hence, Required Remainder = 4.

**NOTE:** When 4 has even number of powers, it will always give remainder 4 on dividing by 6.

QUESTION: 7

Tatto bought a notebook containing 96 pages leaves and numbered them which came to 192 pages. Tappo tore out the latter 25 leaves of the notebook and added the 50 numbers she found on those pages. Which of the following is not true?

Solution:

When we are adding the sum of page numbers of 25 pages, it will always be an odd number. So, she could not have found the sum of pages as 1990.

QUESTION: 8

There are 50 integers a1, a2 … a50, not all of them necessarily different. Let the greatest integer of these 50 integers be referred to as G, and the smallest integer be referred to as L. The integers a1 through a24 form sequence S1, and the rest form sequence S2. Each member of S1 is less than or equal to each member of S2.

**Q. Elements of S1 are in ascending order and those of S2 are in descending order. a _{24} and a_{25} are interchanged then which of the following is true?**

Solution:

Assume S1 = {1, 2, 3,..., 24} and S2 = { 50, 49, 48, 47.. .., 25}.

Now even if we have interchanged 24 and 25, S1 continues to be in ascending order and S2 continues to be in descending order.

However, by choosing negative values of a_{24} and a_{25}, we can show that S1 continues to be in ascending order, but S2 is no longer in descending order.

QUESTION: 9

Srini wrote his class 10th board examination this year. When the result came out he searched for his hall ticket to see his roll number but could not trace it. He could remember only the first three digits of the 6 digit number as 267. His father, however, remembered that the number was divisible by 11. His mother gave the information that the number was also divisible by 13. They tried to recollect the number when all of a sudden Srini told that the number was a multiple of 7. What was the unit digits of the number?

Solution:

His roll number is divisible by 1001 and the first three digits are 267. Hence the last three digits will also be 267.

QUESTION: 10

When 7179 and 9699 are divided by another natural number N , remainder obtained is same. How many values of N will be ending with one or more than one zeroes?

Solution:

When 7179 and 9699 are divided by another natural number N, remainder obtained is same.

Let remainder is R, then (7179 — R) and (9699 — R) are multiples of N and {(9699 — I?) — (7179 — R)} is multiple of N. Then 2520 is multiple of N or the largest value of N is 2520. Total factors of N which are multiples of 10 is 18.

QUESTION: 11

Twenty-five boxes of sweets are delivered to Mr Roy’s home. Mr Roy had ordered sweets of three different types. What is the minimum number of boxes of sweets which are having sweets of same type?

Solution:

This is one classic example of pigeon-hole principle. Since Mr Roy has ordered for 25 boxes and three different types of sweets, so minimum 9 boxes of sweets will have the same type of sweets.

QUESTION: 12

Find the remainder when 73 * 75 * 78 * 57 * 197 * 37 is divided by 34.

Solution:

Remainder,

(73 * 75 * 78 * 57 * 197 * 37)/34 ===> (5 * 7 * 10 * 23 * 27 * 3)/34

[We have taken individual remainder, which means if 73 is divided by 34 individually, it will give remainder 5, 75 divided 34 gives remainder 7 and so on.]

(5 * 7 * 10 * 23 * 27 * 3)/34 ===> (35 * 30 * 23 * 27)/34 [Number Multiplied]

(35 * 30 * 23 * 27)/34 ===> (1 * -4 * -11 * -7)/34

[We have taken here negative as well as positive remainder at the same time. When 30 divided by 34 it will give either positive remainder 30 or negative remainder -4. We can use any one of negative or positive remainder at any time.]

(1 * -4 * -11 * -7)/34 ===> (28 * -11)/34 ===> (-6 * -11)/34 ===> 66/34 ===R===> 32.

Required remainder = 32.

QUESTION: 13

Three distinct prime numbers, less than 10 are taken and all the numbers that can be formed by arranging all the digits are taken. Now, difference between the largest and the smallest number formed is equal to 495. It is also given that sum of the digits is more than 13. What is the product of the numbers?

Solution:

Prime numbers less than 10 = 2, 3, 5, 7 So, option (d) cannot be the answer.

Sum of digits is more than 13 - so, set of [2, 3, 5] is not possible.

So, option (a) cannot be the answer.

Now check for options (b) and (c) by taking the values.

QUESTION: 14

What will be remainder when 1212121212... 300 times, is being divided by 99?

Solution:

This number 1212121212... 300 times is divisible by 9. So, we can write 1212121212...300 times = 9 N, where N is the quotient obtained when divided by 9. Now this question is like -

1212121212... 300 times / 99

= Remainder 9 / 9 x 134680...written 50 times / 11

[121212 = 13468 x 9]

Now we will have to find the reminder obtained when 134680134680.. . 50 times is divided by 11.

For this, we are supposed to use the divisibility rule of 11 from right hand side. [Using the divisibility rule from left hand side might give us the wrong remainder, like if we find out the remainder obtained when 12 is divided by 11, remainder = 1 = (2-1)≠(1 - 2)]

Remainder 134680...written 50 times / 11 = 2

So, the total remainder = 18

Alternatively, divisibility rule of 10" - 1, n = 2 can be used to find the remainder in this case.

QUESTION: 15

Which of the following would always divide a six-digit number of the form ababab?

Solution:

Number = ababab

=ab×10000+ab×100+ab

=ab(10000+100+1)

=ab(10101)

QUESTION: 16

Find the unit digit:

(76476756749)^{8754874878}

Solution:

**Correct Answer :- a**

**Explanation : The unit digit of the number will depend on the last digit.**

**As we know that 91 = 9**

**92 = 81**

**93 = 729**

**94 = 6561**

**The unit digit of the number is 1 and 9, from the options we can pick the answer**

**Hence option a) is correct**

QUESTION: 17

Find the unit digit:

17^{19∧13}

Solution:

17 is raised to the power of 19 and 19 is raised to the power of 13.

To find the last digit of the number of this kind we will start with the base, and the base here is 17.

To get the unit digit of a number our only concern is the digit at the unit place i.e.7.

The cyclicity of 7 is 4.

Dividing 19^{13 }by 4.

Remainder will be 3.

7 raised to power 3 (7^{3}), the unit digit of this number will be 3.

QUESTION: 18

If a number when successively divided by 7, 5, and 3 leaves the remainders 4, 3, and 1 respectively. Find the largest such 3-digit number.

Solution:

This is a problem of successive division.

Divisors 7 5 3

Remainders 4 2 1

The smallest such numbers will be {(5 * 1) + 2} * 7 + 4} Which will be equal to 53.

Now, the general form of the numbers will be:

(7 * 5 * 3) k + 53

= 105k + 53

When we put k = 9

Then the number we obtain is 105 * 9) + 53 = 998

The largest such 3-digit number is 998.

QUESTION: 19

In a school when certain number of chocolates were distributed equally among a group of 24 children, 7 chocolates got left. When the same number of chocolates got distributed among a group of 36, then 19 chocolates were remaining. If the number of chocolates is a 3-digit number. Find the largest number of chocolates that can possibly be. ** **

Solution:

Here the remainder is different but the difference between the divisors and the respective remainders is same in both the cases.

24 - 7 = 17 & 36 - 19 = 17

∴ Here, we can use LCM – Model 1

LCM (24, 36) = 72

The largest 3 digit number which is a multiple of 72 is 936

Therefore, the required number will be 936 - 17 = 919

QUESTION: 20

What would be the greatest number that divides 14, 20, and 32 leaving the same remainder?

Solution:

Here, the number which divides 14, 20, and 32 leaves the same remainder.

∴ We will be using HCF model 2

The required number will be the HCF of (20 - 14), (32 - 12), and (32 - 14).

i.e. HCF (6, 12, 18)

which will be 6.

Therefore, the required number is 6.

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