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=48a+2b+3c+4d+5e+.......26z=?
Assume a=26; b=25, c=24, d=23..... and so on3276
1 * 26 + 2 * 25 + 3 * 24 ... 26 * 1
= 27nn^2 sigma from n=1 to 26
= 27 * 26 * 27/2  26 * 27 * 53/6 = 3276If x and y are real numbers, then the minimum value of x^2+4xy+6y^24y +4 ?
By differentiation,
dF/dx = 2x+4y=0 so x=2y
dF/dy = 4x+12y4=0 x+3y=1
solving both, y=1 x=2
put it in F, so 48+64+4 = 2No 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=112Remainder when 195! Is divided by 394?
394=2*197
195!= 1 mod 197 and 0 mod 2 so 2k and 197k+1 so 198No. 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 (1533*9)/3! = 21
plus 9 cases when (a,a,b) forma)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(n4)  nsum of all even factors of 2160?
2160=2^4 * 3^3 * 5
so (2+4+8+16)(1+3+9+27)(1+5) = 7200Remainder 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 41How 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+6402*320 = 400If 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^(n1) = 13851
r^(n1) = 729.
19 (r^n1)/(r1) = 20767
19 (729r1)=20767r20767
729r1=1093r1093 so r=3