@vbhvgupta said:Q
All the numbers of form (9k+1) ,(9k+2),(9k+3),(9k+4) and 1 number of 9k form can be taken.
We just need to count now.
So, 333 + 334 + 333 + 333 + 1 = 1334 numbers.
EDITED : Including 1 and 3000 => 1336 Numbers. :splat:
@Estallar12 said:All the numbers of form (9k+1) ,(9k+2),(9k+3),(9k+4) and 1 number of 9k form can be taken.We just need to count now.So, 333 + 334 + 333 + 333 + 1 = 1334 numbers.
Ans given is 1336
@vbhvgupta said:Q
1332?? 9k+1 ,9k+2,9k+3,9k+4... 9k+4 are the four numbers to be in subset...now 9k+4 1332+4 =1336
@vbhvgupta said:@Logrhythm 65*29*37*63*71*87*85 last 2 digit?
Keep reducing N.
65*29*37*63*71*87*85
Multiplying each of the last two terms in short and keeping the last two digits -
25*23*27*63
75*21 = 75.
Multiplying each of the last two terms in short and keeping the last two digits -
25*23*27*63
75*21 = 75.
To keep it short, the given N has two numbers ending with 5. So, 25 should be the last two digits. :)
@vbhvgupta said:Q
9n+1 form - 10,....,2998 -> (2998-10)/9 + 1 = 333 terms
9n+2 form - 2,11,....,2999 -> (2999-1)/9 + 1 = 334 terms
9n+3 form - 3,12,...,2991 -> (2991-1)/9 + 1 = 333 terms
9n+4 form - 4,13,...,2992 -> (2992-1)/9 + 1 = 333 terms
now we can't take 9n+5 as (9n+5)+(9n+4) would make 9k form
9n+2 form - 2,11,....,2999 -> (2999-1)/9 + 1 = 334 terms
9n+3 form - 3,12,...,2991 -> (2991-1)/9 + 1 = 333 terms
9n+4 form - 4,13,...,2992 -> (2992-1)/9 + 1 = 333 terms
now we can't take 9n+5 as (9n+5)+(9n+4) would make 9k form
but we can throw in a 9n form number.
hence total = 333*3 + 334 + 1 = 999 + 334 + 1 = 1334
@vbhvgupta said:Q
NUMBERS can be of the form 9k+1, 9k+2, 9k+3.... 9k+8
we can take 9k+1, 9k +2, 9k+3, 9k+4 nos..
9k+1 = > total 334 nos...
9k+2 => total 334 nos
9k+3 => total 334 nos
9k+4 => total 333 nos
and one no of the form 9k.
total 1336 nos
@vbhvgupta said:@Logrhythm 65*29*37*63*71*87*85 last 2 digit?
65 has a 5, 85 also has a 5 which makes a 25
editing.......
@Estallar12 said:Keep reducing N.45*21*71*3545*85= 25.To keep it short, the given N has two numbers ending with 5. So, 25 should be the last two digits.
the bold part i got sir..
how u reduced it to 45*21*71*35..plz explain..
how u reduced it to 45*21*71*35..plz explain..
@allan89 said:NUMBERS can be of the form 9k+1, 9k+2, 9k+3.... 9k+8we can take 9k+1, 9k +2, 9k+3, 9k+4 nos..9k+1 = > total 334 nos...9k+2 => total 334 nos9k+3 => total 334 nos9k+4 => total 333 nosand one no of the form 9k.total 1336 nos
the question says "between" 1 and 3000, you have included them while counting. common mistake 😃
@rhinorocks said:the bold part i got sir..how u reduced it to 45*21*71*35..plz explain..
Woh method chodho.
Short waala explanation hi dekh lo. :splat:
@Logrhythm said:65 has a 5, 85 also has a 5 which makes a 2565/5 = 1385/5 = 17if i take out this 25 the expressrion reduces to 13*29*37*63*71*87*17 and odd*25 always ends in 25
not always.....for example 25*3=75,25*7=175
@Logrhythm said:the question says "between" 1 and 3000, you have included them while counting. common mistake
yeah... thanks.. but i did notice it.. but then options indicate that 3000 has to be included.. so I included.... lol
@Logrhythm said:the question says "between" 1 and 3000, you have included them while counting. common mistake
Twas specifically mentioned in the question in Brackets (Include 1 and 3000). :P
@allan89 said:yeah... thanks.. but i did notice it.. but then options indicate that 3000 has to be included.. so I included.... lol
:D