# Quant Boosters - Swetabh Kumar - Set 4

• If log2 4 * log4 8 * log8 16 * ……………nth term = 49, what is the value of n?

48
log 4 /log 2 * log 8/log 4 * log 16/log 8....
log 2^(n+1)/log 2 = 49
log_base 2_ 2^(n+1) = 49
n+1=49 so n=48

a+2b+3c+4d+5e+.......26z=?
Assume a=26; b=25, c=24, d=23..... and so on

3276
1 * 26 + 2 * 25 + 3 * 24 ... 26 * 1
= 27n-n^2 sigma from n=1 to 26
= 27 * 26 * 27/2 - 26 * 27 * 53/6 = 3276

If x and y are real numbers, then the minimum value of x^2+4xy+6y^2-4y +4 ?

By differentiation,
dF/dx = 2x+4y=0 so x=-2y
dF/dy = 4x+12y-4=0 x+3y=1
solving both, y=1 x=-2
put it in F, so 4-8+6-4+4 = 2

No of 5s in 1 to 240 in base 6.

till 215, it is 3 * 6^3/6 = 108
216 is 1000 and 240 is 1040
so 00 se 40 mein 05, 15, 25, 35 so 4. so 108+4=112

Remainder when 195! Is divided by 394?

394=2*197
195!= 1 mod 197 and 0 mod 2 so 2k and 197k+1 so 198

No. of ways in which 3^16 can be written as a product of 3 factors?

ordered=18C2 = 153
(aab) : (1,1,x), (3,3,x), (3^2,3^2,x)...(3^8,3^8,1) 9 cases
so (153-3*9)/3! = 21
plus 9 cases when (a,a,b) form

a)How many triangles can be formed by the vertices of a regular polygon of n sides? - nC3
b) how many triangles can be formed by the vertices of a regular polygon of n sides having one side common with that of the polygon? - n(n - 4)
c) how many triangles can be formed by the vertices of a regular polygon of n sides having two sides common with that of the polygon? - n
d) how many triangles can be formed by the vertices of a regular polygon of n sides having no side common with that of the polygon? - nC3 - n(n-4) - n

sum of all even factors of 2160?

2160=2^4 * 3^3 * 5
so (2+4+8+16)(1+3+9+27)(1+5) = 7200

Remainder of 37!/41?

39! =1 mod 41
39 * 38 * 37! = 42 mod 41
13 * 19 * 37! = 7 mod 41 =130 mod 41
19 * 37! = 10 mod 41 = 133 mod 41
37! = 7 mod 41

How many of the first 1200 natural no are either prime to 6 or to 15

coprime to 6: 1200(1/2)(2/3) = 400
coprime to 15: 1200(2/3)(4/5) = 640
coprime to 30: 1200(1/2)(2/3)(4/5) = 320
so 400+640-2*320 = 400

If the sum of a GP is given by 20767, the last term by 13851 and the first term by 19 then find the common ratio of the GP.

say N terms.
19r^(n-1) = 13851
r^(n-1) = 729.
19 (r^n-1)/(r-1) = 20767
19 (729r-1)=20767r-20767
729r-1=1093r-1093 so r=3

Looks like your connection to MBAtious was lost, please wait while we try to reconnect.