Quant Boosters  Anubhav Sehgal, NMIMS Mumbai  Set 2

Note here that 5 – 1 is not equal to 7 – 2 is not equal to 9 – 3 is not equal to 11 – 4.
So you cannot apply LCM (5, 7, 9, 11) – N where N is the constant difference.
N = 5a + 1
N = 7b + 2
N = 9c + 3
N = 11d + 4Multiply each by 2 so that remainders also count by a gap of 2 like the divisors are doing.
2N = 5(2a) + 2
2N = 7(2b) + 4
2N = 9(2c) + 6
2N = 11(2d) + 8Now divisors and remainders share a common difference of 3. Designating them as new variables
M = 5l + 2 = 7m + 4 = 9n + 6 = 11o + 8
Least M = LCM (5, 7, 9, 11) – 3
Least N = [LCM (5, 7, 9, 11) – 3]/2 = 1731 since M = 2N.Seeing the procedure you can now directly solve such questions in one line.
Take LCM of divisors. Subtract the would be common remainder after you have figured out the number to multiply with to make the count of divisors and remainders of the same step.
[Here Count of divisors was of 2 : 5, 7, 9, 11 and remainders was of 1. So we multiplied by 2/1 = 2]Another example : Divisors are 5,8,11 ; Remainders are 3,5,7. Multiplier = 3/2. See common remainder and then, Answer : [LCM(5,8,11) – 0.5]/(3/2) = 440 – 0.5/ (1.5) = 439.5/1.5 = 879/3 = 293
NOTE : General form of number will be = Smallest number + k * LCM(Divisors). You may be asked to find the kth smallest number or some other variant. So you may add multiples of LCM to the smallest number to get the required answer.

Q27) A positive integer p is called almost prime, when it has only 1 divisor aside from 1 and p. Find the sum of the 6 smallest almost primes.

Only squares of primes have exactly 3 factors.
Hence your required sum
= 2^2 + 3^2 + 5^2 + 7^2 + 11^2 + 13^2 = 377

Q28) Find n if sqrt(17^2 + 17^2 + ... n times) = 3 * 17^2

3 * 17^2 = sqrt(9 * 17^4) = sqrt(9 * 17^2 * 17^2)
i.e. 9 * 17^2 times 17^2 inside the square root.
i.e 51^2 times
i.e 2601

Q29) a!b! = a! + b! and find (a + b). (Positive integers)

General approach to solve such equations
a!b!  a!  b! + 1 = 1
(a!  1)(b!  1) = 1
a!  1 = b!  1 = 1
a! = b! = 2
a = b = 2
a + b = 4

Q30) Which is the third smallest number when subtracted from 6300 results in a perfect square ?

6300  n = k^2
80^2 = 6400
79^2 will give us the first smallest number on subtracting..
77^2 will give us the third
6300  5929 = 371.

79............? ??