[(1!)^(1!) + (2!)^(2!) + (3!)^(3!) + ............ + (1000!)^(1000!)] / [2^3 X 5^1].
Find the Remainder.
@The_Loser said:in hw many ways alphabet of letter be arranged so that dre are 7 letters bw a & b.
26^7*2?
@sunnychopra89 said:[(1!)^(1!) + (2!)^(2!) + (3!)^(3!) + ............ + (1000!)^(1000!)] / [2^3 X 5^1].Find the Remainder.
Use CRT
Let Nr=X
then Xmod8=5
also, Xmod5=2
8a+5=5b+2=37
a=4,b=7
ans=37
Let Nr=X
then Xmod8=5
also, Xmod5=2
8a+5=5b+2=37
a=4,b=7
ans=37
@Logrhythm i think there would be more cases instead of 50...
49+48 , 49+47, 49+46, 49+45 and so on... will not be complemented.
similaryly, 48 + 47, 48+46 and so on....
49+48 , 49+47, 49+46, 49+45 and so on... will not be complemented.
similaryly, 48 + 47, 48+46 and so on....
@The_Loser said:bhai kya kiya hai ye?
my mistake
a 7spaces b
due to a nd b=2
spaces 7 alphabets 26 due to repetition 26^7*2 ?
a 7spaces b
due to a nd b=2
spaces 7 alphabets 26 due to repetition 26^7*2 ?
@The_Loser said:in hw many ways alphabet of letter be arranged so that dre are 7 letters bw a & b.
26C7*2!*7!*18!
@The_Loser two cases
1. a _ _ _ _ _ _ _ b - 26^7
2. b _ _ _ _ _ _ _ a - 26^7 since each blank/alphabet can be filled with 26 letters..
Answer---> 26^7*2!
1. a _ _ _ _ _ _ _ b - 26^7
2. b _ _ _ _ _ _ _ a - 26^7 since each blank/alphabet can be filled with 26 letters..
Answer---> 26^7*2!
@sunnychopra89 said:[(1!)^(1!) + (2!)^(2!) + (3!)^(3!) + ............ + (1000!)^(1000!)] / [2^3 X 5^1].Find the Remainder.
37 ?
from 5!^5! remainder will be 0
[2^3 X 5^1]=x
(1!)^(1!)/x=1
(2!)^(2!)/x=4
(3!)^(3!)/x=16
4!^4!/x=16
total 37/x=R=37
from 5!^5! remainder will be 0
[2^3 X 5^1]=x
(1!)^(1!)/x=1
(2!)^(2!)/x=4
(3!)^(3!)/x=16
4!^4!/x=16
total 37/x=R=37
@
@bs0409 said:Use CRT
Let Nr=X
then Xmod8=5
also, Xmod5=2
8a+5=5b+2=37
a=4,b=7
ans=37
Could u plz telme in brief what is CRT/Chineese Remainer Theorem.
@sunnychopra89 said:@bs0409 Could u plz telme in brief what is CRT/Chineese Remainer Theorem.
Write Denominator as a*b where a and b are coprime to each other.
Find Nr mod a (=x say) and Nr mod b (=y say).
then solve am+x=bn+y for m and n in integers. the value will be the remainder....
Find Nr mod a (=x say) and Nr mod b (=y say).
then solve am+x=bn+y for m and n in integers. the value will be the remainder....
examination
fr 1st 3 papers = max marks 50
4th exm - max marks 100
find no f ways one can score 60% ?
Which among 2^1/2, 3^1/3, 4^1/4, 6^1/6 and 12^1/12 is the largest?
3^1/3.
@The_Loser said:examination fr 1st 3 papers = max marks 504th exm - max marks 100find no f ways one can score 60% ?
exactly 60% or more than 60%??
@bs0409 said:Write Denominator as a*b where a and b are coprime to each other.
Find Nr mod a (=x say) and Nr mod b (=y say).
then solve am+x=bn+y for m and n in integers. the value will be the remainder....
How to calculate x and y....I'm a bit confused...:(
If you hav a subset of integers to chosen bet. 1 to 3000, such that no integers add up to multiple of nine, what can be the max number of elements in the subset?
1. 1668
2. 1332
3. 1333
4. 1334
2. 1332
3. 1333
4. 1334
@shattereddream said:If you hav a subset of integers to chosen bet. 1 to 3000, such that no integers add up to multiple of nine, what can be the max number of elements in the subset?1. 16682. 13323. 13334. 1334
considering 1 and 3000 are not included
9k+1-333
9k+2-334
9k+3-333
9k+4-333
9k+5-333
9k+6-333
9k+7-333
9k+8-333
9k-333
we can have 9k+1,9k+2,9k+3,one among9k+4/9k+5 and one 9k no
334+333*3+1 =1334
@Subhashdec2 said:considering 1 and 3000 are not included9k+1-3339k+2-3349k+3-3339k+4-3339k+5-3339k+6-3339k+7-3339k+8-3339k-333we can have 9k+1,9k+2,9k+3,one among9k+4/9k+5334+333*3 =1333
i was getting 1336 first
@shattereddream said:If you hav a subset of integers to chosen bet. 1 to 3000, such that no integers add up to multiple of nine, what can be the max number of elements in the subset?1. 16682. 13323. 13334. 1334
9k+1 --> 333
9k+2 --> 334
9k+3 --> 333
9k+4 --> 333
one 9k no.
total --> 1334
9k+2 --> 334
9k+3 --> 333
9k+4 --> 333
one 9k no.
total --> 1334