the cube of the first no exceed their product by 2,the cube of the second no is smaller than their product by 3,and the cube of the third no exceeds their product by 3

A)3^(1/3) B)9^(1/3) C)2 D)any of these

How many factors of 8! are of the form 3K+2?

8! = 2^{7*} 3^{2} *5 *7. Total number of factors = 96.

From this, first remove factors which are multiple of 3 (of the form 3k) which are 64 in number.

You also have to remove factors of the form 3k + 1. Note that every perfect square is of the form 3k + 1. Since you have already removed multiples of 3, remove all the numbers which are of the form 3k + 1. Remember that multiplication of numbers of the form 3k + 1 also gives you a product of the form 3k + 1. The powers of 2 you can take are 2^{0,} 2^{2}, 2^{4}, 2^{6}. Powers of 5 and 7 will be 5^{0}, 7^{0} and 7^{1}. Total such numbers = 4 * 1 * 2 = 8.

Also, multiplication of two numbers of the form 3k + 2 also gives you a product of the form 3k + 1. The powers of 2 you can take are 2^{1}, 2^{3}, 2^{5}, 2^{7}. Powers of 5 and 7 will be 5^{1}, 7^{0} and 7^{1}. Total such numbers = 4 * 1 * 2 = 8.

Removing these numbers, left factors = 96 - 64 - 16 = 16.

How many four-digit numbers are there with less than 6 different prime factors?

- 1224 b. 8476 c. 9000 d. 7613

Hello Vidisha, Please find the solution below.

2x3x5x7x11x13=30030 so that means all the number of 4 digit form will have less than 6 distinct prime numbers

**Find the highest power of 12 that will divide ( 5 ^{36} - 1)is?1) 12) 23) 34) 4**

Hello Vidisha

Find the solution:

5^{36} - 1 = 25^{18} - 1 = (24 + 1)^{18} - 1 = 24^{18} + 18 x 24^{17} + ... + 18 x 24 + 1 - 1 = 12^{2}k.

OR

5^{36} - 1 = (4 + 1)^{36} - 1 = 12^{2}k.

So highest power of 12 that will divide the expression is 2

Q: 523abc is divisible by 7, 8 and 9. Then a x b x c is equal to?

Hello Richa,

Since 7,8 and 9 are co-prime, the number which is divisible by 7,8 and 9, must also be divisible by their L.C.M i.e. by 504.

Now let's have a look at the given number, 523abc. It should lie between 523000 and 523999.

The multiples of 504 lying in this range are 523152 and 523656. Two possible values of abc are 152 and 656.

a,b and c are distinct, so abc must be 152, hence a x b x c = 1 x 5 x 2 = 10

Find the smallest number with 20 divisors.

Hi Surbhi,

If N = a^{x} * b^{y } * c^{z} -----, the total number of factors of N = (x+1) * (y+1) * (z+1)--------,

where a, b, c------ are prime factors of N.

Now, the number of factors of N is 20,

so 20 = 20 * 1 = (19+1), therefore N = a^{19}, and Min value of N = 2^{19} = 524,288

or 20 = 10 * 2 = (9+1) * (1+1), therefore N = a^{9} * b, and Min value of N = 2^{9} * 3 = 15,36

or 20 = 5 * 4 = (4+1) * (3+1), therefore N = a^{4} * b^{3}, and Min value of N = 2^{4} * 3^{3} = 432

or 20 = 5 * 2 * 2 = (4+1) * (1+1) * (1+1), therefore N = a^{4} * b * c, and Min value of N = 2^{4} * 3 * 5 = 240

So, the minimum number with 20 factors = 240

sir, please help

abc = 7^{13}

a = 7^{x} , b = 7^{y} , c = 7^{z}

x + y + z = 13

Total number of non negative integral solutions : ^{15}C_{2}

Solutions when two of them are equal :

(0,0,13) , (1,1,11) ,…….., (6,6,1) => 7

which are permuted 3 times each in ^{15}C_{2}

Hence unordering we get ,

(^{15}C_{2} - 3 x 7)/6 + 7 = 21 ways

Sir, can you explain the last step? How do we do the unordering?

For how many integers x^{4} + x^{3} + x^{2} + x + 1 is a perfect square?

If the sum of two natural numbers is multiplied by each number separately, the products so obtained are 2418 and 3666. What is the difference between the numbers?

(x + y ) x = 2418 = x² + xy ... ( 1)

( x + y)y = 3666= y² + xy ... (2)

on adding (1) and (2)

x² + 2xy + y² = 6084

( x + y)² = 78²

x + y = 78

x = 2418/78 = 31

y = 3666/78 = 47

y - x = 47 - 31 = 16 .

How many numbers less than 100 have exactly 7 composite factors?

0

1

2

3

I think we can count manually

nos are 48,80,84 so answer is 3

first of all we have to think that no would be greater than around 30 then check those no. which have many factors

any other approach?

How many natural numbers from 1 to 2011 has a sum of digits which is multiple of 5?

Hi Pranshu,

The sum of the digits of all the numbers of the form 9k + 1 and 9k + 5 are divisible by 5. Where k is a natural number.

Total : 401 numbers

The sum of 20 numbers(may or may not be distinct)is 801. What is their minimum LCM?

has to go with the options what were the options Simran ?

Options were

A. 36

B. 42

C. 56

D.60

Just try with options.

Start with 36.

801 = 36*22 + 9.

But we need 20 numbers only. So rule out 36.

Take 42.

801 = 42*19 + 3.

LCM(42,3) = 42. (answer)

(Data sufficiency question)

Is X divisible by 15?

I. X is a number formed by writing ten consecutive natural numbers side by side.

II. X is the number formed by multiplying the factorials of three consecutive natural numbers.

Sir, I think both the statements combined are sufficient to answer the question, since the number of digits in X is at least 10. So the three consecutive natural numbers taken must have been greater than 5? But the answer is that even together they are insufficient. Why so?

Find the total number of even factors of the L.C.M. of first 25 natural numbers.

2304

The highest power of 2 in LCM of first 25 natural number is 4, as 16 = 2^4

The highest power of 3 in LCM of first 25 natural number is 2 , as 9 = 3^2

Highest power of 5 , 5^2

Similarly highest power of 7, 11, 13 , 17 , 19 , and 23 = 1

LCM : 2^ 4 x 3^2 x 5^2 x 7 x 11 x 13 x 17 x 19 x 23 .

No of even factors : 4 x 3 x 3 x 2 x 2 x 2 x 2 x 2 x 2 = 2304