@sujamait said:anyone ?
Couldn't get the sequence even after filling half the page 
1.Start with n = 2. Two players A and B move alternately by adding a proper divisor of n to the current n. The player in whose move 1990 or a greater number is achieved will win. Both the players are intelligent and good at maths. Who will win and how??
Note: Except the number itself all other factors are proper divisors.
Note: Except the number itself all other factors are proper divisors.
@sujamait d??the whole sequence has 4odd numbers followed by even..in other words sequence mod 2 yields 11110..the 1st three options are not in sync with the sequence but the fourth one seems to be..just an observation..wats the OA
@deedeedudu said:Couldn't get the sequence even after filling half the page
The sequence seems to be this:
T(n + 1) = [2*T(n) - T(n-4)] mod 10 (for n>=5) 😃 , but I don't know till now what to make of it 😛 ...
What is the maximum possible sum of the number of Tuesdays and Wednesdays in two consecutive
years?
(a) 209 (b) 210 (c) 208 (d) 212
years?
(a) 209 (b) 210 (c) 208 (d) 212
Two identical marked dices are brought together and kept with one of their faces in full contact. How many different arrangements are possible?
(1) 36 (2) 60 (3) 72 (4) 84 (5) none of these
(1) 36 (2) 60 (3) 72 (4) 84 (5) none of these
@sparklingaubade said:What is the maximum possible sum of the number of Tuesdays and Wednesdays in two consecutiveyears?(a) 209 (b) 210 (c) 208 (d) 212
210?
@sparklingaubade said:Two identical marked dices are brought together and kept with one of their faces in full contact. How many different arrangements are possible? (1) 36 (2) 60 (3) 72 (4) 84 (5) none of these
4*21=84
The numbers a, b, c are in arithmetic progression, while a - k, b, c + k are geometric progression, with common ratio r. Which of the following is true ?
r -1 ‰¤ r 0 1
r -1 ‰¤ r 0 1
@sparklingaubade said:What is the maximum possible sum of the number of Tuesdays and Wednesdays in two consecutiveyears?(a) 209 (b) 210 (c) 208 (d) 212
210..
@sparklingaubade
@sparklingaubade said:What is the maximum possible sum of the number of Tuesdays and Wednesdays in two consecutiveyears?(a) 209 (b) 210 (c) 208 (d) 212
210?
If a.b.c=2^5.3^4*5^7. How many Natural No Solutions?
@vbhvgupta said:The numbers a, b, c are in arithmetic progression, while a - k, b, c + k are geometric progression, with common ratio r. Which of the following is true ?r -1 ≤ r 0 1
0
@sparklingaubade said:What is the maximum possible sum of the number of Tuesdays and Wednesdays in two consecutiveyears?(a) 209 (b) 210 (c) 208 (d) 212
210 ?
@sparklingaubade said:What is the maximum possible sum of the number of Tuesdays and Wednesdays in two consecutiveyears?(a) 209 (b) 210 (c) 208 (d) 212
210?
@krum said:7c2*6c2*9c2=11340 ?
i don't know the ans yet but can u plz tell ur approach as how u came to this ?
@sparklingaubade said:What is the maximum possible sum of the number of Tuesdays and Wednesdays in two consecutiveyears?(a) 209 (b) 210 (c) 208 (d) 212
52*7=364
1jan=Tuesday
52 weeks = > 52 tuesdays and wednesdays
total 104
1+364=365
ie 365th day will be tuesday
Total 105
1jan of next year=wednesday
again 52 weeks = 104
but there will be 1 tuesday less
therefore
in all=105+104-1
=>208
OA?
1jan=Tuesday
52 weeks = > 52 tuesdays and wednesdays
total 104
1+364=365
ie 365th day will be tuesday
Total 105
1jan of next year=wednesday
again 52 weeks = 104
but there will be 1 tuesday less
therefore
in all=105+104-1
=>208
OA?
@freebull said:If a.b.c=2^5.3^4*5^7. How many Natural No Solutions?
7C2*6C2*9C2
Let a=2^a1*3^a2*5^a3
b=2^b1*3^b2*5^b3
c=2^c1*3^c2*5^c3
a1+b1+c1=5
Total solutions (5+3-1)C(3-1)=7C2
Likewise for powers of 3 and 5
Let a=2^a1*3^a2*5^a3
b=2^b1*3^b2*5^b3
c=2^c1*3^c2*5^c3
a1+b1+c1=5
Total solutions (5+3-1)C(3-1)=7C2
Likewise for powers of 3 and 5