# Quant Boosters - Maneesh - Set 1

• LCM(3, 4, 5, 6) = 60
Number of the form 60a + 2
60a + 2 = 7b
lowest such number -> 182 when a = 3 & b = 26

• Q12) For how many pairs (a,b) of natural numbers is the LCM of a and b 2^3 * 5^7 * 11^13?

• Let one of the number contains 2^3
Other may contain 2^0,2^1,2^2,2^3 -> 4 possibilities
Number of ordered pairs -> 2 * 4 – 1 = 7 (since (2^3,2^3) required only once)
Similarly powers of 5 -> 2 * 8 – 1 = 15
Powers of 11-> 2 * 14 – 1 = 27
Total 7 * 15 * 27 = 2835

• Q13) An Egyptian fraction has a numerator equal to 1 and its denominator is a positive integer. What is the maximum number of Egyptian fractions such that their sum is equal to 1 and their denominators are equal to 10 or less?

• Possible numbers 1/2 ,1/3,1/4,1/5,1/6,1/7,1/8,1/9
Out of which 1/5, 1/10, 1/7, 1/9 can be eliminated since it wont give sum of 1
Similarly 1/4 and 1/8 can be eliminated.
So remaining 1/2, 1/3 and 1/6 = > 1/2 + 1/3 + 1/6 = 1
So maximum number -> 3

• Q14) What is the minimum number of square marble tiles required to tile a floor of length 2 meters 56 cm and width 3 m and 36 cm ?

• Length = 256 cm
Width = 336 cm
Square side -> HCF[256,336] = 16
No of tiles required = 256 * 336/16 * 16 = 336

• Q15) Three containers having 9/2, 13/4 and 19/3 litres of three different cold drinks are to be served equally to all guests in a party such that each one gets maximum. How many maximum number of guests could be entertained ?

• HCF [9/2 , 13/4 , 19/3] = 1/12
Maximum number of guests = 9/2 * 12 + 13/4 * 12 + 19/3 * 12 = 54 + 39 + 76 = 169

• Q16) Find the number of trailing zeros for P = 10 * 10! + 11 * 11! + 12 * 12! + … 99 * 99!

• P = 10!(11-1)+11!(12-1)+……+99!(100-1)
= 11!-10!+12!-11!+13!-12!+…..100!-99!
= 100!-10!
Number of trailing zeros p contain -> number of trailing zeros in 10! = 10/5 = 2

• Q17) Let n! = 1 * 2 * 3 … n for integer n > = 1 If P = 1 + ( 2 * 2! ) + ( 3 * 3!) + …. +(12 * 12!). Then P+3 divided by 13! leaves a remainder of?

• n * n! = (n + 1 - 1) * n! = n!(n+1) – n! = (n+1)!-n!
1 * 1! + 2 * 2! + 3 * 3! +…12 * 12! = 2! - 1! + 3! - 2! + … + 13! - 12! = 13!-1!
13! - 1 + 3 mod 13! = 2 mod 13! =2

• Q18) 64329 is divided by a certain number while dividing the number 175, 114 and 213 appear as 3 successive remainders. Divisor is

• D | 64329
xxx
---------
1752
yyyy
---------
1149
zzzz
--------
213

So 643 - 175 = xxx = 468
1752 – 114 = yyyy = 1638
1149 - 213 = zzzz = 936
Since D is the factor of all these D = HCF[468,1638,936] = 234

• Q19) For how many integers n is n^4 + 6n < 6n^3 + n^2

• n^4+6n < 6n^3+n^2
n^4-6n^3+6n-n^2 < 0
n(n^2-1)(n-6)< 0
n^2 – 1 always positive
So n (n-6) is less than 0
n-6 less than 0
Possible n values 2,3,4,5(n cant b zero since it will become 0)

• Q20) How many numbers between 1 and 400 both included are not divisible either by 3 or 5?

• Multiples of 3 = 399-3/3 +1 = 133
Multiples of 5 = 400-5/5 +1 = 80
Multiples of 15 = 390-15/15 +1 = 26
Multiples of 3 or 5 = 133+80-26 = 187
Numbers which are not multiples of 3 or 5 -> 400-187 = 213

• Q21) Find the remainder when 12345689101112……40 is divided by 36

61

64

55

61

61

61

61