Naga, jus applying the theorem to check for lower primes 2, 3, 5, 7 and 11
for 2, 8 is perfectly divisible by 2
for 3, 88 is not perfectly divisible by 3..
for 5, 8888 is not perfectly divisible by 5
for 7, 888888 is perfectly divisible by 7
for 11, 8888888888 is perfectly divisible by 11
are we missing something in the theorem??
Regards
SK
jain_ashu SaysThis theorem is applicable only for prime numbers greater than 5.
naga25french Saysvalid for prime number greater than 5... oops i missed that
One from my side:
Q)In an integer d1d2...dk from left satisfy didi+1 for i even. How many integers from 1000 to 9999 have four distinct digits?
Sahana Kavitha SaysWhere do you guys refer all these theorems?
.. no real source of info
One from my side:
Q)In an integer d1d2...dk from left satisfy didi+1 for i even. How many integers from 1000 to 9999 have four distinct digits?
naga25french Saysits quite to difficult to tell without option.. pls let us know the options if u have
One from my side:
Q)In an integer d1d2...dk from left satisfy didi+1 for i even. How many integers from 1000 to 9999 have four distinct digits?
Sahana Kavitha Saysis it 5040?
Originally Posted by jain_ashu View Post
One from my side:
Q)In an integer d1d2...dk from left satisfy di
jain_ashu Saysthe options are a) 630 b) 778 c) 882 d) none of these
Originally Posted by jain_ashu View Post
One from my side:
Q)In an integer d1d2...dk from left satisfy didi+1 for i even. How many integers from 1000 to 9999 have four distinct digits?
Draft for your comments, ans 7(every 10)*8(every 100)*90 = 5040
1023-1032-1042-1052-----
1024-1034-1043-1053-----
1025-1035-1045-1054-----
1026-1036-1046-1056-----
1027-1037-1047-1057-----
1028-1038-1048-1058-----
1029-1039-1049-1059-----
7 + 7 + 7 + 7 +++++720 times
A few doubts :-P
1.Form a subset of integers choosen b/w 1 to 3000 such that no 2 integer add up to amultiple of 9.wt can be the max no: of element in subset.
a.1668 b.1332 c.1333 d.1334
2.a triangular no:is defined as a no:which has a property of being expressed as a sum of consecutive natural nos starting with 1.how many trngular no: less than 1000 have the property that they r the difference of squire of 2 consecutive natural nos
a.20 b.21 c.22 d.23
3. highest power of 3 in exp 58!-38!
A few doubts :-P
3. highest power of 3 in exp 58!-38!
the highest power of 3 is 3^17
A few doubts :-P
1.Form a subset of integers choosen b/w 1 to 3000 such that no 2 integer add up to amultiple of 9.wt can be the max no: of element in subset.
a.1668 b.1332 c.1333 d.1334
3. highest power of 3 in exp 58!-38!
A few doubts :-P
2.a triangular no:is defined as a no:which has a property of being expressed as a sum of consecutive natural nos starting with 1.how many trngular no: less than 1000 have the property that they r the difference of squire of 2 consecutive natural nos
a.20 b.21 c.22 d.23
1.
option d. 1334
9a+1,9b+2,9c+3,9d+4 = 3000
a = 333
b = 334
c = 333
d = 333
total = 1333 + 1 = 1334
3.
38!*(3k-1)==>3^17
max power = 17
Originally Posted by BIJAY PRASAD View Post
A few doubts 
1.Form a subset of integers choosen b/w 1 to 3000 such that no 2 integer add up to amultiple of 9.wt can be the max no: of element in subset.
a.1668 b.1332 c.1333 d.1334
1.
option d. 1334
9a+1,9b+2,9c+3,9d+4 = 3000
a = 333
b = 334
c = 333
d = 333
total = 1333 + 1 = 1334
Not able to understand your solution, pls explain.. the highlighted part..
Originally Posted by BIJAY PRASAD View Post
A few doubts
1.Form a subset of integers choosen b/w 1 to 3000 such that no 2 integer add up to amultiple of 9.wt can be the max no: of element in subset.
a.1668 b.1332 c.1333 d.13341.
option d. 1334
9a+1,9b+2,9c+3,9d+4 = 3000
a = 333
b = 334
c = 333
d = 333
total = 1333 + 1 = 1334
Not able to understand your solution, pls explain.. the highlighted part..
wat i ve understood is that we need a subset of integers where no 2 integers add upto be a multiple of 9.....
so if we take integers of the form 9a+1,9b+2,9c+3,9d+4 nd choose any 2 integers it's sum would never be a multiple of 9
eg 9a+1+9b+2=9(a+b)+3 which is not a multiple of 9 similarly for the rest of the numbers
now if 9a+1 is taken as an integer then a can have a value where 1since the integer has to be less than 3000 nd more than 1
similarly we get for the rest of d value.......
in d end just add all d values........
hope it is clear now............
Originally Posted by BIJAY PRASAD View Post
A few doubts
1.Form a subset of integers choosen b/w 1 to 3000 such that no 2 integer add up to amultiple of 9.wt can be the max no: of element in subset.
a.1668 b.1332 c.1333 d.1334
wat i ve understood is that we need a subset of integers where no 2 integers add upto be a multiple of 9.....
so if we take integers of the form 9a+1,9b+2,9c+3,9d+4 nd choose any 2 integers it's sum would never be a multiple of 9
eg 9a+1+9b+2=9(a+b)+3 which is not a multiple of 9 similarly for the rest of the numbers
now if 9a+1 is taken as an integer then a can have a value where 1since the integer has to be less than 3000 nd more than 1
similarly we get for the rest of d value.......
in d end just add all d values........
hope it is clear now............
we can have one more number of the form 9k,
total = 1333 +1 = 1334
One from my side:
Q)In an integer d1d2...dk from left satisfy didi+1 for i even. How many integers from 1000 to 9999 have four distinct digits?
Puys pls try this...
the options are a) 630 b) 778 c) 882 d) none of these