@ravi.theja 83
@sos2god said:are u sure we can use euler theorem here...as 78 and 100 share a common factor 2....
right...euler's theorm can't be used here.
@ravi.theja said:how many perfect squares less than 10^6 are multiples of 24?
@Estallar12 @pirateiim478 do this 😃 post the solution also
@ravi.theja said:how many perfect squares less than 10^6 are multiples of 24?
Is it 83??
24 = 2^3*3
so the squares will be of the form (12k)^2
hence, total = [1000/12] = 83
@getupsid said:1) N = 204*221*238*255*......*280. how many consecutive zeroes will be there at the end of this number N?? a. 8b. 10c. 11d. 122) Find the HCF of ((2^100)-1) and ((2^120)-1).a. (2^10)-1b. (2^20)-1c. 1d. none of these
1. Please check the Q again
2. 2^20-1
2. 2^20-1
@ravi.theja said:1) question is not clear..the first few terms follow a pattern of having difference as 17..but 280 cannot be der.2) 2^HCF(100,120) -1 = 2^20-1
what is the concept involved here man ??@gs4890
how is it the lcm of 100 n 120?? tried binominal too.. but cudnt derive it.. 
Ranjit invests Rs.5000 for 3 years at 5% p.a. compound interest reckoned yearly. Income tax at the rate of 20% on the interest earned is deducted at the end of each year. Find the amount at the end of the third year?
(a) 5624.32 (b) 5630.50 (c) 5788.125 (d) 5627.20
(a) 5624.32 (b) 5630.50 (c) 5788.125 (d) 5627.20
@catahead said:Ranjit invests Rs.5000 for 3 years at 5% p.a. compound interest reckoned yearly. Income tax at the rate of 20% on the interest earned is deducted at the end of each year. Find the amount at the end of the third year?(a) 5624.32 (b) 5630.50 (c) 5788.125 (d) 5627.20
5624.32?
5000(1.05)-5000=250
250*20/100=50
5000+250-50=5200
5200(1.05)-5200=260..260*20/100=52
5200+260-25=5408
5408(1.05)-5408=270.4....270.4*20/100=54.08
5408+270.4-54.08=5624.32
@ravi.theja said:1) question is not clear..the first few terms follow a pattern of having difference as 17..but 280 cannot be der.2) 2^HCF(100,120) -1 = 2^20-1
@ravi.theja said:how many perfect squares less than 10^6 are multiples of 24?
Similar to asking how many numbers less than 1000 are multiples of 12.
So ans is 83.....
So ans is 83.....
@The_Loser said:A = {1, 2, 3, 4, 5}, B Š† A and C Š† B. How many pairs of (B, C) are possible?
5C5*2^5+5C4*2^4+5C3*2^3+5C2*2^2+5C1*2+5C0*1
I am assuming empty set is also a subset
I am assuming empty set is also a subset
@getupsid said:1) N = 204*221*238*255*......*280. how many consecutive zeroes will be there at the end of this number N?? a. 8b. 10c. 11d. 122) Find the HCF of ((2^100)-1) and ((2^120)-1).a. (2^10)-1b. (2^20)-1c. 1d. none of these
1. Series is not correct
2. (2^20)-1
2. (2^20)-1
@getupsid said:1) N = 204*221*238*255*......*280. how many consecutive zeroes will be there at the end of this number N?? a. 8b. 10c. 11d. 122) Find the HCF of ((2^100)-1) and ((2^120)-1).a. (2^10)-1b. (2^20)-1c. 1d. none of these
1. Series is not correct
2. (2^20)-1
2. (2^20)-1
find the remiander when 7^7^7^7.......7000 times is divided by 17
@ravi.theja said:find the remiander when 7^7^7^7.......7000 times is divided by 17
E(17) = 16
7^7^.....6999 times/ 16 gives remainder of 7
thus, 7^7 / 17 => 12 should be the remainder.
7^7^.....6999 times/ 16 gives remainder of 7
thus, 7^7 / 17 => 12 should be the remainder.
@gs4890 said:E(17) = 167^7^.....6999 times / 16 gives remainder of 1thus, 7 mod 17 => 7 should be the remainder.
bhai OA is 12..!! 😃 check karo once again..even am nt getting it