Factors and Multiples

This is today's Question of the Day. Please check the link  http://tathagat.mba/quant-question-of-the-day-29-06-2018/  

tomorrow for the answer.   

Thank you, Sir 🤩 

2⁴ + 2⁷ + 2ⁿ

= 2⁴ [ 1 + 2³ + 2 ^{(n -4)} ]

= 2⁴  [ 9 +  2^{(n -4)} ]

9 +  2^{(n -4)} has to be a perfect square .

n = 8 is only possibility.  

9 +  2^{(n -4)} has to be a perfect square

Sir, From this how do we conclude that n=8 is the only possibility from all the natural numbers, unless range is not given to us in the question?

9 + 2^(n-4) is a perfect square 

Two cases :
(i)When ( n -4) is a perfect square  

9 + x² = y²

9 = y² – x²
(x-y) (x+y) = 1 x 9 or 3 x 3

Solving this we get x = 4 is a possible value .

(ii) When ( n -4) is  not a perfect square 

9 + 2^(n-4) is of the form 3k + 2 Hence , Never a perfect square .

[All perfect squares are of the form 3n or 3n + 1.]

Hello Ritika , 

We know , Product of an odd natural number and 5 always ends with the unit digit   5. 

So we can select 

2^0 ( 1 way ) 

3^0 , 3^1 , 3^2 , 3^3 , 3^4 , 3^5 ( in 6 ways ) 

similarly powers of 5 in 4 ways 

hence 1 * 6 * 4 = 24 factors of N have unit digit equal to 5. 

Hello Vidisha , we know sum of squares of first n natural number = 

n ( n +1 )( 2n +1 )/6 

1² + 2² + 3² + ... + 2005²

= 2005 × 2006 × 4011/6

= 2005 × 1003 × 1337

= 5 × 401 × 17 × 59 × 7 × 191

= 5 × 7 × 17 × 59 × 191 × 401

Hence, required sum of primes

= 5 + 7 + 17 + 59 + 191 + 401 = 680.

Hello Apoorva ! 

Smallest number which has five distinct prime factors is 2 × 3 × 5 × 7 × 11 = 2310 . 

All three digit natural numbers are smaller than 2310 . 

Hence there are  900 three digit natural numbers which have less than 5 prime factors.  🙂

Hello Apoorva ! 

For a natural number N the number of factors of N below √N is equal to number of factors of N  above √N . 

So if N has five factors between √N and 1 [ 6 below √N ] it must have 6 factors above it's square root. 

If it is a perfect square then it's square root is also a factor . 

Hence , 6 + 1 + 6 = 13 for first question and 6 + 6 = 12 is the answer for the second question.  🙂

8! = 2^7* 3² *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, 2², 2⁴, 2⁶. Powers of 5 and 7 will be 5⁰, 7⁰ and 7¹. 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 2¹, 2³, 2⁵, 2⁷. Powers of 5 and 7 will be 5¹, 7⁰ and 7¹. Total such numbers = 4 * 1 * 2 = 8.

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

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

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 = a¹⁹, and Min value of N = 2¹⁹ = 524,288

or 20 = 10 * 2 = (9+1) * (1+1), therefore N = a⁹ * b, and Min value of N = 2⁹ * 3 = 15,36

or 20 = 5 * 4 = (4+1) * (3+1), therefore N = a⁴ * b³, and Min value of N = 2⁴ * 3³ = 432

or 20 = 5 * 2 * 2 = (4+1) * (1+1) * (1+1), therefore N = a⁴ * b * c, and Min value of N = 2⁴ * 3 * 5 = 240

So, the minimum number with 20 factors = 240

abc = 7¹³

a = 7^x , b = 7^y , c = 7^z

x + y + z = 13

Total number of non negative integral solutions : ¹⁵C₂

Solutions when two of them are equal :

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

 which are permuted 3 times each in ¹⁵C₂

Hence unordering we get ,

(¹⁵C₂  - 3 x 7)/6 + 7 = 21 ways

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

(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 . 

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?

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