even numbers, then the next three odd numbers followed by the next four even numbers and so on.
What is the 2003rd term of the sequence?
(a) 3953 (b) 3943 (c) 3940 (d) 3950
The sequence 1, 2, 4, 5, 7, 9, 10, 12, 14, 16, 17, €Ś. has one odd number followed by the next two
even numbers, then the next three odd numbers followed by the next four even numbers and so on.
What is the 2003rd term of the sequence?
(a) 3953 (b) 3943 (c) 3940 (d) 3950
even numbers, then the next three odd numbers followed by the next four even numbers and so on.
What is the 2003rd term of the sequence?
(a) 3953 (b) 3943 (c) 3940 (d) 3950
@pathetic said:Rohan writes all three-digit numbers of base 6, one below the other in an order. Find the number of times the digit '1' is used by Rohan.
99 times??
@maroof10 said:@amresh_maverick 265
options are
@amresh_maverick said:No of integral sols of |x-3| + |y-5| =11
36
44
55
60
72
44
55
60
72
@Koushik98 said:The sequence 1, 2, 4, 5, 7, 9, 10, 12, 14, 16, 17, €Ś. has one odd number followed by the next twoeven numbers, then the next three odd numbers followed by the next four even numbers and so on.What is the 2003rd term of the sequence?(a) 3953 (b) 3943 (c) 3940 (d) 3950
1
2,4
5,7,9
10,12,14,16
17,19,21,23,25
observe the patter here...
Row 1 -> 1 element, ends in 1^2
Row 2 -> 2 elements, ends in 2^2
Row 3 -> 3 elements, ends in 3^2
.
.
and so on...
n(n+1)/2 = 2003
n ~ 62
62*63/2 = 1953
2003 - 1953 = 50 more terms
so it would be in the 63rd row
62nd row wld have ended with 62^2 = 3844
63rd row's first term = 3845
so 50th term = 3845 + 49*2 = 3943...
@pathetic said:Rohan writes all three-digit numbers of base 6, one below the other in an order. Find the number of times the digit '1' is used by Rohan.
96 times?
regards
scrabbler
regards
scrabbler
@pathetic said:Rohan writes all three-digit numbers of base 6, one below the other in an order. Find the number of times the digit '1' is used by Rohan.
is it 120 times?? not sure...
@pathetic said:Rohan writes all three-digit numbers of base 6, one below the other in an order. Find the number of times the digit '1' is used by Rohan.
can we treat this same as no of 3 digit no formed using 0 ,1,2,3,4,5 where 1 has to appear and repetition is allowed ?
@scrabbler
Edit: got 96 , as I forgot to count no of 1 and instead counted no of 3 digits no formed
@scrabbler
Edit: got 96 , as I forgot to count no of 1 and instead counted no of 3 digits no formed
OA is 96 .
@amresh_maverick said:can we treat this same as no of 3 digit no formed using 0 ,1,2,3,4,5 where 1 has to appear and repetition is allowed ?@scrabbler
that's exactly what we need to do.... 😃
meine bekar ki harkat kari thi...so got 120...96 hona chahiye...
If a > 0; b > 0; c > 0 and a + b + c = 1, then the maximum value of (a/3+a )+(b/3+b)+(c/3+c) is ?
@viewpt :
@ScareCrow28 said:Check out the attachment! Samajh ajana chahiye barring my drawing skills
gt it properly..thnk u!!!
@pathetic said:Rohan writes all three-digit numbers of base 6, one below the other in an order. Find the number of times the digit '1' is used by Rohan.
total no's =5*6*6
100 to 155 1 will come in tens place 36 times
00 to 55 36 no's will probability of all 6 same hence 1 will come 6 times
6*5=30
30+36=66?
@amresh_maverick said:can we treat this same as no of 3 digit no formed using 0 ,1,2,3,4,5 where 1 has to appear and repetition is allowed ?@scrabbler
That's what it is na? That's how I did...
My logic: Total numbers = 180 (6^3 -6^2). Now 1st digit is 1, 2, 3, 4, 5 so 1 will be in 1/5th of cases = 36. Second and third digit mein 0 ka option bhi hai to 1/6th of cases = 30 + 30. hence 96.
regards
scrabbler
My logic: Total numbers = 180 (6^3 -6^2). Now 1st digit is 1, 2, 3, 4, 5 so 1 will be in 1/5th of cases = 36. Second and third digit mein 0 ka option bhi hai to 1/6th of cases = 30 + 30. hence 96.
regards
scrabbler
@pathetic said:If a > 0; b > 0; c > 0 and a + b + c = 1, then the maximum value of (a/3+a )+(b/3+b)+(c/3+c) is ?
a=b=c=1/3
(1/9+1/3)*3 = 4/3...??
i took the expression as ((a/3) + a)
if it is (a/(3+a)) then it wld be 3/10....
@pathetic said:If a > 0; b > 0; c > 0 and a + b + c = 1, then the maximum value of (a/3+a )+(b/3+b)+(c/3+c) is ?
4/3 ?
a = b = c = 1/3
OA 3/10
@scrabbler said:That's what it is na? That's how I did...My logic: Total numbers = 180 (6^3 -6^2). Now 1st digit is 1, 2, 3, 4, 5 so 1 will be in 1/5th of cases = 36. Second and third digit mein 0 ka option bhi hai to 1/6th of cases = 30 + 30. hence 96.regardsscrabbler
Second and third digit mein 0 ka option bhi hai to 1/6th of cases
ye to detail mein samjhana
ye to detail mein samjhana
Sorry it is (a/(3+a))