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QUESTION: 1

How many factors of 2^{5} * 3^{6} * 5^{2} are perfect squares?

Solution:

Any factor of this number should be of the form 2^{a} × 3^{b} × 5^{c}

For the factor to be a perfect square a, b, c have to be even.

a can take values 0, 2, 4, b can take values 0, 2, 4, 6 and c can take values 0, 2

Total number of perfect squares =3 × 4 × 2 = 24

QUESTION: 2

The sum of the first 100 natural numbers, 1 to 100 is divisible by

Solution:

The sum of the first 100 natural numbers is:

= (n * (n + 1)) / 2

= (100 * 101) / 2

= 50 * 101

101 is an odd number and 50 is divisible by 2.

Hence, 50 * 101 will be divisible by 2.

QUESTION: 3

Convert the following binary number to decimal (010111)_{2}?

Solution:

To do this, at first translate it to decimal here so:

**(010111) _{2}**

= 0 * 2

= 0 + 16 + 0 + 4 + 2 + 1

= 23

Result of converting:

(010111)_{2} = 23_{10}

QUESTION: 4

Find the greatest five-digit number that is exactly divisible by 7, 10, 15, 21 and 28.

Solution:

Largest five digit number is 99999,

LCM of 7,10,15,21,28 is

7=7

10=2∗5

15=3∗5

21=3∗7

28=2∗2∗

so, 2∗2∗3∗5∗7=420

=> 99999/420

= remainder will be 39 so

99999 − 39 = 99960

QUESTION: 5

N! is having 37 zeros at its end. How many values of N is/are possible?

Solution:

The number of zero is determined by the no's. of 5's & 2's, whichever is less. (since 5 x 2 =10).

- In most of the cases number of 5's is less than that of 2's.

Now in the question, A/q,

- N / 5 + N / 5
^{2 }+ N / 5^{3 }+ N / 5^{4 }= 37

This gives the value of

N = 150,151,152,153,154

∴ Total N possible values are 5.

QUESTION: 6

P is a natural number. 2P has 28 divisors and 3P has 30 divisors. How many divisors of 6P will be there?

Solution:

If 2P has 28 divisors then 2P = (2^{6}) * (3^{3}).

If 3P has 30 divisors then 3P = (2^{5}) * (3^{4}).

∴ P = (2^{5}) * (3^{3})

Then 6P = 2 * 3 * (2^{5}) * (3^{3}) = (2^{6}) * (3^{4})

► 6P will have (6+1) * (4+1) = 35 divisors.

QUESTION: 7

What is the least number of soldiers that can be drawn up in troops of 12, 15, 18 and 20 soldiers and also in form of a solid square?

Solution:

In this type of question, We need to find out the LCM of the given numbers.

LCM of 12, 15, 18 and 20;

► 12 = 2 x 2 x 3;

► 15 = 3 x 5;

► 18 = 2 x 3 x 3;

► 20 = 2 x 2 x 5;

∴ LCM = 2 x 2 x 3 x 5 x 3

Since, the soldiers are in the form of a solid square. Hence, LCM must be a perfect square.

To make the LCM a perfect square, We have to multiply it by 5, hence, the required number of soldiers:

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

= 900.

QUESTION: 8

How many factors of 1080 are perfect squares?

Solution:

Factors of 1080 are :- 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 24, 27, 30, 36, 40, 45, 54, 60, 90, 120, 135, 180, 270, 360, 450, 540, 1080.

The factors of 1080 which are perfect square are 1,4,9,36 which is 4 in total.

QUESTION: 9

Rohan purchased some pens, pencils and erasers for his young brothers and sisters for the ensuing examinations. He had to buy atleast 11 pieces of each item in a manner that the number of pens purchased be more than the number of pencils, which is more than the number of erasers. He purchased a total of 38 pieces.

If the number of pencils cannot be equally divided among his 4 brothers and sisters, how many pens did he purchase?

Solution:

- Different possibilities for the number of pencils = 12 or 13.
- Since it cannot be divided into his 4 brothers and sisters, it has to be 13.
- In that case, number of pens = 14

QUESTION: 10

Write three rational numbers between 4 and 5?

Solution:

There are several rational numbers between 4 and 5.

The numbers are between 16/4 and 20/4.

Therefore, the answer is c, that is, 17 / 4, 18 / 4, 19 / 4.

QUESTION: 11

1 ’s are given 100 times, 2 ’s are given 100 times and 3’s are given 100 times. Now numbers are made by arranging these 300 digits in all possible ways. How many of these numbers will be perfect squares?

Solution:

__Solve this question step by step:__

- Any number formed by this method is clearly divisible by 3.
- Since it needs to be a square, it should be divisible by (3)
^{[2*k]}. k varies over the natural numbers. - Now consider, the original number. It has hundred 1’s, hundred 2’s and hundred 3’s. Sum of these digits is 600. This is not divisible by 9. Hence number is not divisible by 9.
- If a number is divisible by (3)
^{[2*k]}, it is divisible by 3^{k}. - This number is not divisible by 3
^{k}for any k>1.

Hence it is not a perfect square for any arrangement.

QUESTION: 12

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

Solution:

Remainder, (73 x 75 x 78 x 57 x 197 x 37) / 34

= (5 x 7 x 10 x 23 x 27 x 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 x 7 x 10 x 23 x 27 x 3) / 34

= (35 x 30 x 23 x 27) / 34

= (1 x (-4) x (-11) x (-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.]

Finally, (1 x -4 x -11 x -7) / 34

= (28 x -11) / 34

= (-6 x -11) / 34

= 66 / 34

∴ R = 32

QUESTION: 13

There are two integers 34041 and 32506, when divided by a three-digit integer n, leave the same remainder. What is the value of n?

Solution:

Find the difference between both 34041 and 32506.

► 34041 - 32506 = 1535

So, N should be a factor of 1535. Factors of 1535: 1, 5, 307, 1535

Here, 307 is a three digit factor of 1535. So, N = 307.

► 34041 and 32506 divided by 307 leaves remainder 271.

So, 307 is the answer.

QUESTION: 14

Teacher said that there were 100 students in his class, 24 of whom were boys and 32 were girls. Which base system did the teacher use in this statement?

Solution:

We are provided with the equation (32) + (24) = (100). Let us assume our base be 'b'

Then,we can say:

32 = 3 x b^{1 }+ 2 x b^{0} = 3b+2

Similarly,

► 24 = 2 x b^{1}+ 4 x b^{0 }= 2b+4

► 100 = 1 x b^{2 }+ 0 x b^{1 }+ 0 x b^{0} = b^{2}

Now , according to our question:

► 32 + 24=100

► (3b + 2) + (2b + 4) = (b^{2})

► 5b + 6 = b^{2}

► b^{2 }- 5b - 6 = 0

► b^{2 }- 6b + b - 6 = 0

► b(b - 6) + 1(b - 6) = 0

► (b - 6) * (b + 1) = 0

► b = 6,- 1

Base can't be negative. Hence b = 6.

∴ Base assumed in the asked question must be 6.

QUESTION: 15

Find the highest power of 24 in 150!

Solution:

24 = 8 × 3.

Therefore, we need to find the highest power of 8 and 3 in 150!

8 = 2^{3}

Highest power of 8 in 150!

= [(150 / 2) + (150 / 4) + (150 / 8) + (150 / 16) + (150 / 32) + (150 / 64) +(150 / 128)] / 3

= 48

Highest power of 3 in 150!

= [150 / 3] + [150 / 9] + [150 / 27] + [150 / 81]

= 72

As the powers of 8 are less, powers of 24 in 150! = 48.

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