@sails said:9?
Yar it is 3
@vbhvgupta said:Explain?
Sure...
single digit numbers total =9
2-digit numbers = 90. => total digits used = 90*2 =180
3 digit numbers (100 -999) = 900 => total digits used =900*3 =2700
total terms so far = 9 +180 +2700 =2889
terms left = 28383 - 2889 = 25494
25494 digits will all will be 4 digit nos starting from 1000
=> 25494 / 4 = 6373.xx
=> 2nd last 4 digit term will be 6372, since 4 digit nos are starting from 1000.
since remainder was 2, hence series will end with 3.... 6371637263.......
@nramachandran said:(1/x) + (1/y) = 1/2 ; x and y are positive integers, What is the maximum value of x+y? Please post the approach
max value is obtained wen x=y ==> 2/x = 1/2 ==> x=y=4 x+y =8???
@pirateiim478 said:@Pratishruti said:x=5 or -3?No try again!!Find the sum of all 4 digit numbers formed by digits 1,2,3 and 4 when repetitions of digits is allowed??
66660??
@vbhvgupta said:Define a no K such that it is the sum of the squares of the first M natural numbers ( k = 1^2 + 2^2 +3^2+...+ M) where M
13??
@somnathbhatta said:@vbhvgupta 7 values of M .. if 55 is not considered.
@ravi.theja said:13??
Ans is 12 ....explain?? did u use the sum of SQ formula???
@nramachandran said:(1/x) + (1/y) = 1/2 ; x and y are positive integers, What is the maximum value of x+y? Please post the approach
2x + 2y = xy
=> xy -2x -2y =0
=> x(y-2) - 2(y-2) =4
=> (x-2)(y-2) =4
now 4 = 4*1; or 2*2
case 1. x-2 =4, y-2 =1;
=> x=6, y=3
x+y =9
or,
case 2. x-2 =2; y-2 =2
=>x=4, y=4
x+y =8
hence max value of x+ y=9= answer
@vbhvgupta
k = M(M + 1)(2M + 1)/6
For k to be divisible by 4, one of M or (M + 1) should be divisible by 8 (as we have an extra two in denominator), as (2M + 1) will always be odd.
For M = 8, 16, 24, 32, 40, 48; M will be divisible by 8 and
for M = 7, 15, 23, 31, 39, 47; (M + 1) will be divisible by 8
So, for total 12 values of M
For k to be divisible by 4, one of M or (M + 1) should be divisible by 8 (as we have an extra two in denominator), as (2M + 1) will always be odd.
For M = 8, 16, 24, 32, 40, 48; M will be divisible by 8 and
for M = 7, 15, 23, 31, 39, 47; (M + 1) will be divisible by 8
So, for total 12 values of M
@vbhvgupta said:@krum Can u please solve this?
C. 63
Numbers from 23-29
then 32-39
42-49
52-59
...
..
92-99
Total= 7 + 8*7 = 63
Numbers from 23-29
then 32-39
42-49
52-59
...
..
92-99
Total= 7 + 8*7 = 63
@vbhvgupta said:28383th termof the series 1234567891011121314151617181920.............
9 +90*2 +900* 3 =2889 ==> 4 digits numbers used are 28,383 - 2889 = 24,494/4 = 6122 + rem 2 ie., 71227123..ans 1??
@ravi.theja said:9 +90*2 +900* 3 =2889 ==> 4 digits numbers used are 28,383 - 2889 = 24,494/4 = 6122 + rem 2 ie., 71227123..ans 1??
Ans 3
@ravi.theja said:66660??
the answer u got is considering only cases where number is not repeated...same method follow for repition case...and find out..answer is 711040
@vbhvgupta said:Ans 63
ya.it has to be 63 i.e., 8*8 ways and the condition of a=2 and b=2 has be to be subtracted 64-1=63 ways @vbhvgupta sorry 12 solutions. the solutions should be 7, 8, 15, 16, 23, 24, 31, 32, 39, 40, 47,48. used the formula Sum of squares = n(n+1)(2n+1)/6 .