# Quant Boosters - Maneesh - Set 3

• Given LCM of 100 natural numbers = K
To find the LCM of first 105 natural numbers, factorize the first 5 numbers after 100.
101 - prime
102 - 2 * 3 * 17
103 - prime
104 - 8 * 13
105- 3 * 5 * 7

during the calculation of LCM of first 100 natural numbers, the numbers 2, 3, 17, 13,8,5,7 are already considered. Hence only numbers which will affect the LCM will be 101 and 103 only.
Hence LCM = K * 101 * 103 = 10403K

• Q21) Three thieves robbed a diamond shop independently but distributed their diamonds according to the rule that person or persons with maximum diamonds should give diamonds equal to HCF of their individual number of diamonds to the person or persons with the lowest number of diamonds. This process is repeated till all have equal diamonds.

a) If they have 39, 52, 26 diamonds ,how many times should the process be repeated?
a) 1
b) 3
c) 4
d) Not possible

b) If they have 51, 85, 102 diamonds how many times should the process be repeated?
a) 3
b) 4
c) 5
d) Not possible

• a) HCF (39, 52, 26) = 13
first step, 13 will be deducted from second person and 13 will be added to third person.
Then it will become (39,39,39)

b) Total number of diamonds = 51 + 85 + 102 = 238
But its not a multiple of 3. So we can’t divide equally among three people.
Hence option 4) not possible is the answer.

• Q22) Which one is bigger (9)^1/3 or (20)^1/4 ?

• 9^(1/3) = 9^(4/12) = (6561)^1/12
20^(1/4) = 20^(3/12) = (8000)^1/12
Hence 20^1/4 is greater.

• Q23) The factorial of a number can be decomposed in to a product of prime numbers.Let the maximum power of 2 in the decomposition of the factorial of a number (n) be k. Find n, if k = 9?

• K = n/2 +n/4 +n/8……..
when n= 10 k = 5+2+1 = 8
when n= 12 k = 6+3+1 = 10
Hence no such number exists.

• Q24) If a number is divided by M gives a remainder 17, when same number is divided by (8 * M) remainder comes 47. Then what would be the remainder when same number is divided by (4 * M)?

• MQ+17 = 8Mq+47
M (Q-8q) = 30
M > 17
so M will be 30.
Q-8q = 1
let Q = 9 and q = 1
Then number is 287.
287 mod 120 = 47.

• Q25) The last digit of a is same as the last digit of its square. How many such numbers exists from 0 to 90?
a) 34
b) 37
c) 9
d) 4

• Unit digit of a number and its square will be equal when unit digit is any of 0,1,5,6.
From 0 to 90 : unit digit 0 ->10 numbers
From 0 to 90 : unit digit 1,5,6 -> 9 each
Hence total 37 numbers.

• Q26) 94^3 – 23^3 - 71^3 is at-least divisible by?
a) 71 and 23
b) 23 and 74
c) 71 and 94
d) 23, 71 and 94

• 94^3-23^3 is divisible by 94-23 = 71
94^3-71^3 is divisible by 94-71 =23
23^3+71^3 is divisible by 23+71 = 94
Hence it is divisible by 23, 71 and 94.

• Q27) Find the least number which being divided by 5,6,8,9,12 leaves in each case a remainder 1 but when divided by 13 leaves no remainder?
a) 2987
b) 3601
c) 3600
d) 2986

• LCM [5,6,8,9,12] = 360
360A +1 = 13B
From option the number of the form 360+1 is 3601 only.

• Q28) Find the greatest number consisting of 6 digits which on being divided by 6,7,8,9,10 leaves 4,5,6,7,8 as remainders respectively.
a) 997920
b) 997918
c) 999999
d) 997922

• LCM [6,7,8,9,10] = 2^3 * 3^2 * 7 * 5 = 2520
Greatest 6 digit number which is a multiple of 2520 = 997920
Required number = 997920 – 2 = 997918.
PS : we are subtracting 2 coz 6-4 = 7-5 = 8-6 = 9-7 = 10-8 = 2

• Q29) If X = 13^3 + 14^3 + 15^3 + 16^3 then what is the remainder when x is divided by 58?
a) 0
b) 1
c) 57
d) 29

• A^n + B^n is divisible by A+B when n is odd.
So x is divisible by 29
X = 29(13^2+16^2-13 * 16) + 29(14^2+15^2-14 * 15)
= 29(13^2+14^2+15^2+16^2-13 * 16 - 14 * 15) = 29 * an even number
Hence x is a multiple of 58.

• Q30) A train moving at uniform speed takes 20 sec to pass a cyclist riding at 11 km/hr but only 9 sec to pass a post. Find the length of train.

58

61

61

61

71

67

61

46