Quant Boosters  Anubhav Sehgal, NMIMS Mumbai  Set 1

N/2 = x^2
N/3 = y^3
N^2 = 6.x^2.y^3
N = xy.root(6y)
For N to be a natural number,6y must be a perfect square => y = 6 => x = 2 * 3^2 and N = 2^3 * 3^4
=> Number of factors = 4 * 5 = 20

Q23) Let n be the smallest positive number such that the number S = 8^n * 5^600 has 604 digits. Then the sum of digits of S is ?

We know that 10^n has 1 followed by n zeroes which is a total of (n + 1) digits.
S = 8^n * 5^600 = 2^3n * 5^600
For n = 200. S would have been 10^600 with 601 digits but we need S to have 604 digits.
These digits can come only from additional power of 2 as power of 5 is fixed.
Hence we look for additional power of 2 that gives us those 3 extra digits.
2^10 = 1024 is the first number with 4 digits that gives us those 3 extra digits.But power of 2 is 3n that is a multiple of 3. Hence we move onwards to 2^11 and then 2^12 = 4096[12 = 3n and 600 already = n form]
Thus, S = 2^612 * 5^600 = 4096000..600 zeroes => Sum of digits = 4 + 0 + 9 + 6 = 19

Q24) How many integers between 1 and 1000 both inclusive can be expressed as difference of squares of two non negative integers?

Case 1 : Both numbers are even : 2p,2q
4p^2  4q^2 = 4(p^2 q^2) = 4k form
Case 2 : Both numbers are odd : 2p + 1 , 2q + 1
4p^2 + 4p + 1  4q^2  4q  1 = 4(p^2 + p  q^2  q) = 4k form
Case 3 : First even , second odd : 2p,2q + 1
4p^2  4q^2  4q  1 = 4k  1 = 4k + 3 form
Case 4 : First odd , second even : 2p + 1,2q
4p^2 + 4p + 1  4q^2 = 4k + 1 formHence, difference of squares of two non negative integers can be 4k,4k+1 or 4k + 3 form but never 4k + 2 form
So for 11000 => (3/4)(1000) = 750 cases it can be expressed as the difference.

Q25) A number N when divided by a divisor D gives a remainder of 52.The number 10N when divided by D gives a remainder of 8.How many values of D are possible?

N = aD + 52
10N = bD + 8
=>
10N = 10aD + 520
10N = bd + 8
(b  10a)D = 512
D is a factor of 512 and leaves a remainder of 52 => D is a factor of 512 greater than 52.
64,128,256,512 : 4 values

Q26) In how many ways can 7^17 be written as a product of 3 natural numbers?

abc = 7^17
a = 7^x , b = 7^y , c = 7^z
abc = 7^(x + y + z) = 7^17
x + y + z = 17 => C(17 + 3  1,3  1) = C(19,2) non negative integral solutions
But this includes solutions where any two of a,b,c are equal and permutated in 3!/2! = 3 ways while
we need ways to write it as product of 3 natural numbers hence should be counting them only once.
2a + b = 17 => 9 solutions
Unordering the solutions : (C(19,2)  3*9)/3! + 9 = 33 ways ; +9 to include them once.

Q27) Sum of 20 distinct numbers is 801.What is their minimum possible LCM ? (source TG)

Firstly it is a bit of hit and trial question..
LCM will be minimum when either maximum possible numbers are equal or when they are all factors of the same number as we need LCM,the Lowest common multiple to be minimum.
In any other case there will be extra power of primes or new primes introduced increasing the LCM.Part I :
If the numbers may or may not be distinct
We try to bring them as close to each other(or equal as possible)
801 = 19 * 42 + 3
So we take 19 numbers as 42 and 20th number as its factor so that LCM stays minimum.Part II :
When we need all numbers to be distinct
Now that we can't have any numbers equal we try to find a number which has at least 20 factors whose sum is 801.
Here there is a bit of hit and trial looking for numbers with 21,22,23 and eventually 24 factors to reach 360.
And then eliminating some of its factors to get our required sum of 20 numbers(out of those 24 factors) to be 801.
And their LCM will be 360 as they all were its factors and had no other lower common multiple even after removal of the said four extra factors.LOTS TO READ and UNDERSTAND but there is no alternate way for this. It is just for conceptual clarity and practice.

Q28) Let M and N be single digit numbers. If product 2M5 * 13N is divisible by 36.How many ordered pairs of (M,N) are possible?

2M5 is odd
=> 13N is divisible by 4
=> N = 2 or 6
i) 2M5 * 132
=> 2M5 is divisible by 3
=> M = 2, 5, 8
ii) 2M5 * 136
=> 2M5 is divisible by 9
=> M = 2

Q29) The number of integer solutions of the equation x^2 + 12 = y^4.

x^2 + 12 = y^4
y^4  x^2 = 12
(y^2 + x)(y^2  x) = 12 = 1 * 12 = 2 * 6 = 3 * 4
No solutions for 1 * 12 and 3 * 4 factor pairs as RHS(=12) is even hence we need just even * even factor pairs.
(y^2 + x)(y^2  x) = 2 * 6
y=+/ 2 and x=+/2
4 solutions

Q30) The number 2006! is written in base 42. How many zeroes are there at the end?

42 = 2 * 3 * 7
No of zeroes in 2006! will be the highest power of 7 in 2006!
[2006/7] + [2006/49] + [2006/343] = 331
as 42 in base 42 will be 10 and will be the provider of trailing zeroes