Quant Boosters  Anubhav Sehgal, NMIMS Mumbai  Set 4

1080 = 2^3 * 3^3 * 5
10 = 2 * 5
30 = 2 * 3 * 5For, the factor to be divisible by it must have 2 * 5 in it. So taking that out from 1080, we are left with: 2^2 * 3^3.
Now with 2 * 5 taken out, we must ensure that a 3 is not taken out else the factor will become divisible by 30.
So we can only take 3 factors namely 2^0, 2^1 and 2^2 out of the (2 + 1)(3 + 1) = 12 factors of this remaining number.
Answer: 3 factors = 2 * 5 * (2^0), 2 * 5 * (2^1), 2 * 5 * (2^2) = 10, 20, 40

Q26) How many natural numbers less than 200 will have 12 factors?

12 = 2 * 6 = 3 * 4 = 2 * 2 * 3
Case 1
2 * 6 => p * q^5
Take q = 2 and out values of other primes.
32 * 3, 32 * 5 only are less than 200: 2 values.
q = 3 alone takes above 200 so no need to check further.Case 2
3 * 4 => p^2 * q^3
q = 2 => 8 * 9 only.
q = 3 => 27 * 4 only.Case 3
2 * 2 * 3 => p * q * r^2
r = 2 => 4 * 3 * 5, 4 * 3 * 7, 4 * 3 * 11, 4 * 3 * 13, 4 * 5 * 7: 5 values.
r = 3 => 9 * 2 * 5, 9 * 2 * 7, 9 * 2 * 11: 3 values.
r = 5 => 25 * 2 * 3: 1 value.
None further.
Total: 2 + 1 + 1 + 5 + 3 + 1 = 13 values.

Q27) How many three digit numbers will have a total of 3 factors?

Only squares of primes have exactly three factors namely 1, p, p^2 where p is a prime.
11^2 = 121
13^2 = 169
…
31^2 = 961
11, 13, 17, 19, 23, 29, 31: 7 numbers

Q28) How many natural numbers less than 100 have exactly 2 factors?

Only primes have two factors which is 1 and the number itself.
There are 25 primes in the range 199.
Hence the answer will be 25.
{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}

Q29) N = 2^8 * 3^10 * 5^8 * 7^2.
a) How many factors of N are multiples of 360 but not a multiple of 540?
b) How many factors of N are factor of 360 but not a factor of 540?

a) N = 2^8 * 3^10 * 5^8 * 7^2.
360 = 2^3 * 3^2 * 5
540 = 2^2 * 3^3 * 5Let us call 360 the Important (I) and 540 the Unimportant (UI).
In such questions, we look at each prime’s individual powers.
Deficiency: The primes that have power less in I as compared to UI must be restricted for the factor of N to be power of that prime in I to less than power of that prime in UI.
Surplus: While, the primes that have power more in I as compared to UI must have the power of that prime greater than power of that prime in UI but less than equal to the power of that prime in N.So, here prime 3 has a deficiency case and prime 2 has a surplus case for 360.
So any factor of N that is a multiple of 360 but not a multiple of 540 must be of the form:
2^w * 3^x * 5^y * 7^z where w = 3 to 8, x = 2, y = 1 to 8 and z = 0 to 2.
So number of such factors = (8 – 3 + 1) * (1) * (8 – 1 + 1) * (2 – 0 + 1) = 6 * 1 * 8 * 3 = 144b) Looking in the similar way,
w = 3, x = 0 to 2, y = 0 to 1, z = 0
1 * 3 * 2 * 1 = 6 such factors.
Note here the idea would be to keep power of each prime greater than UI and equal to I for surplus case while less than equal to I in case of deficiency. Also, a prime not present in I but present in UI or N would not be considered as it’s a case of being a factor.Shorter approach for above 2 questions:
360 = 2^3 3^2 5 and 540 = 2^2 3^3 5.
So their LCM = 2^3 3^3 5 and HCF = 2^2 3^2 5.
(i) Required number factors is = number of factors of 360  number of factors of both 360 and 540 (i.e. their HCF) = 4 * 3 * 2  3 * 3 * 2 = 6.
(ii) Required number of factors = factors of N which are multiple of 360  factors of N which are multiples of both 360 and 540 (i.e. LCM) = 6 * 9 * 8 * 3  6 * 8 * 8 * 3 = 144.

Q30) Rahul had 2n marbles numbered from 1 through 2n. He removed n marbles that were numbered consecutively. The sum on the remaining marbles was 1615. Find the sum of all possible values of n

Let the marble numbers removed be (a + 1), (a + 2)… (a + n).
Then, 1615 = 2n * (2n + 1)/2 – n * (2a + n + 1)/2
3230 = 2n * (2n + 1) – n * (2a + n + 1)
3230 = n * (4n + 2 – 2a – n – 1) = n * (3n – 2a + 1)
2 * 5 * 17 * 19 = n * (3n – 2a + 1)
n = 34 or 38 satisfies.
Sum = 72