??
?
Find the remainder for the following expression
[(110!) / (107^2)]?
The enclave of/ conservative jewish families/ rarely interacts with its neighbours/ NE......FE pls explain , i do not know the OA..
What is El Nino?
A man borrows a sum of 50000 for 2 years at rate of 20% pa ci compounded annually. If he repays 24000 at the end of second year and clears his debt, how much did he paid at the end of first year?
In a triangle angle bisector lies closer to angle containing smaller side while perpendicular bisector to larger side...ie if AD AM are angle bisector and perpendicular bisector then BD < DC or MC < MB
Trigonometry me ek formulae hai...asin@+bcos@=c.......then acos@-bsin@=+-rt (a^2+b^2-c^2)....plz consider lhs part....isko yaad kaise karna hai...kaun sa interchange ho rha..sign, coefficient ya angle wala part....2)reverse mtlb ...asin@-bcos@=c.......then acos@+bsin@=+-rt (a^2+b^2-c^2) me bhi valid hai kya???
which is correct? i dont know answer
- both are correct
- ram has seen the movie already
- ram had seen the movie already
- none of them is correct
- skip
0 voters
Divisibilty Methods For checking divisibility by any 'prime/odd' number except for factors of 5', you have the concept of base number.
Number Add Base Number Subtract Base Number
---------------------------------
3: 1 2
7: 5 2
9: 1 8
11: 10 1
13: 4 9
17: 12 5
19: 2 17
21: 19 2
23: 7 ?
27: ? 8
29: 3 ?
...
...
And so on....(i'll describe a method to get these below)
Now for checking divisibility either add the last digit*add base number to the number "formed by removing last digit" or you can subract last digit*subtract base number from the "formed by removing last digit"
for e.g. check for 51/17 either 5- 1*5 =0 or 5 + 1*12 =17 hence divisible.
to check 312/13 we can 31+2*4 =39 hence dvisible or we can 31-2*9 = 13 ence divisible.
to check 61731/19 = 6173 + 1*2 = 6175 = 617 + 5*2 = 627 = 62 + 7*2 = 76 hence divisible.
to check 357976/29 = 35797 + 6*3 = 35815 = 3581 +5*3 = 3596 = 359 + 6*3 = 377 = 37 + 7*3 = 58 hence divisible..
to check 382294/11 = 38229-4*1 =38225 = 3822-5 = 3817 = 381 - 7 = 374 = 37 -4 =33 Hence divisible..
The Subract base Number for a number can be obtained as the {(samllest multiple of number which ends in one)-1}/10
i.e. for 3 or 7 it is (21-1)/10 =2
for 13 it is 91-1/10 = 9.
The AddbaseNumber for a number can be obtained as the {(samllest multiple of number which ends in nine)+1}/10
i.e. for 13 it is (39+1)/10 =4.
for 7 it is 49+1/10 = 5
Proof:
For SubtactBaseNumber say the number abcde...
I want to check divisibility by 17 where subtractbasenumber is 5
I can always write abcde... as 10X+Y (where Y is last digit and X is number formed by removing last digit)
Now X-Y*5 = (10X -50Y)/10 = (10X + Y -51Y)/10 = (OriginalNumber - 51 Y ) / 10
The number '51 Y' is a multiple of 17 so if "OriginalNumber" is divisible by 17 then "OriginalNumber - 51*Y" got to be.. i.e. "10X - 50Y"
as 10 and 17 are co-prime if "10X- 50Y" is divisible the "X-5Y" got to be.....
same theory hold's for addbasenumbers too....
This also defines why it is so easy to check divisibility by 3 or 9 just keep on adding the digits...
And you can check divisibility by 11 just by keeping on subrating digits form previous number.. (which is same as taking sum of even/odd location separately..) Divisibility by 7
Only for those interested in Number theory (Not a Cat short-cut)
say the number is :
38,391,787
Separate into pairs of digits
38 39 17 87
Consider the difference between each pair of digits and the nearest multiple of seven, beginning for the first pair at right, lower (upper) for the first, upper (lower) for the second and so on, alternating for each new pair.
4 -----4 (21-17)
38 39 17 87
---4 ------3 (87-84)
The resulting digits, read from right are 3444 (which is also a number multiple of 7).
Proceed in the same way with 3,444
1
34 44
----2
The final pair 21 is a multiple of seven, so is the original number 38,391,787.
ANOTHER EXAMPLE
Look how fast this method is.
Consider the 15-digit number 531,898,839,909,822
2 ----2--- 3 ----0
5 31 89 88 39 90 98 22
---3 ---4 ----6 ----1
Now we have 10,634,232
4 -----0
10 63 42 32
----0 ----4
And now 4,004
2
40 04
---4
Which gives 42, a multiple of 7.
We only need three steps for a 15-digit number. This is called TOJA's method of divisibility. Incidentally this also works for 11 and 13. Just a little manupulation is required, (in case you get a remainder of more than 9)
Let A = 5,962
7
59 62 77 which is a multiple of 1
--7
EXAMPLE 2
Let A = 5, 971,845
6---- 4
5 97 18 45
--9 ----1
8
14 96 -> 88 ->divisible
----8
EXAMPLE 3
Let A = 80,714,546
8 ----10
80 71 45 46
----5-------2
The resulting numbers ( 2 10 5 8 ) don’t form a decimal number, so proceed in this way: Put the exceeding number 1 from 10, below the 2 and sum.
2 0 5 8 -> 3 0 5 8
1
3
30 58 33
---3
https://www.facebook.com/groups/632112500304610/?ref=bookmarks
tehri dam is on which river?
largest fresh water lake of india
loktak lake is in which state
Pitot tube is used to measure?
next winter olympics will be held at which city
find a number which when divided by 15, 17, 19 leaves a remainder 1, 5, 9 respectively.
A bag contain total of 105 coins of 1rs, 50p 25p. Find no of 1 rs coin if there is 50.5 rs in a bag and it is known that no of 25p coin is 133.33% more than no of 1rs coin....
If the points (0,0), (2,0), (0,-2) and (k,-2) are concyclic then k = ?
ODD one .....14,19,29,40,44,51,59,73 (don't know ans )
- 44
- 59
- 29
- 51
0 voters