what is the Remainder when 333555777888999 is divided by 3 or 9 or 11 (give short cut to solve it?

Answer 1

Any number is divisible by 3 if the sum of its digits are divisible by 3. Same for 9 - if the sum of digits is divisible by 9, then the number itself is a multiple of 9.

In this case, sum = 3(3 + 5 + 7 + 8 + 9) = 3*32 = 96, so the number is divisible by 3 but not 9.

For 11s, you can use a process of dividing the number up into 2-digit increments to determine its divisibility.

For your number, starting from the right:

99 = no remainder
89 = 1 remainder; carry over
88 + 1 = 1 remainder; carry over
77 + 1 = 1 remainder; carry over
57 + 1 = 3 remainder; carry over
55 + 3 = 3 remainder; carry over
333 + 3 = 6 remainder; carry over

The whole number, then, when divided by 11, leaves a remainder of 6 (or 0.5454545454... on the calculator). This is confirmed by use of the calculator:

333555777888999 / 11 = 30323252535363.5454545454...

Answer 2

most persons answers when divided by 3 and by 9 so i would not. the remainder when divided by 11 is |(3+3+5+7+7+8+9+9)-(3+5+5+7+8+8+9)| which is 6. if the difference was greater than 11 then we divided the difference by 11 and the remainder of the second calculation is the same as the first.

Answer 3

3 and 9 have been given already, you add up all the digits, if that sum is divisible by 3 or 9 then so is the base number. The remainder from that exercise will also give you the remainder of base division.

For 11, we have a slightly more difficult method. First add up all the odd place digits (1st, 3rd, 5th, etc) 3+3+5+7+7+8+9+9 = 51
The the even places 3+5+5+7+8+8+9 = 45
Then subtract those two 51-45 = 6
so the base number is not divisible by 11, but once again the remainder from this division is out remainder

Answer 4

When that number is divided by 3 there is no remainder.

The law of divisibility of 3 says that if you add up all the digits and THAT number is divisible by 3, then the original number is also divisible by 3.

ex) 3+3+3 =9 (3 x3 =9)
5+5+5 = 15 (3 x5=15)
7+7+7 = 21 (3 x7 =21)
8+8+8 = 24 (3 x8 =24)
9+9+9 = 27 (3 x 9 = 27)
you can already see that since each of these numbers appears three times, it will be divisible by 3.

But just to check:
9 + 15 + 21 + 24 +81 = 96 and 96 is divisible by 3 (3 x 32 =96)

And the final check:

333555777888999 / 3 = 111185259296333

Answer 5

It's pretty easy with a calculator - divide by 3, or 9, or 11. You'll get a number followed by some fraction. Multiply the whole number (ignoring the stuff after the decimal point) by 3 (or 9, or 11) and subtract the answer from your original number. This will give you the remainder.

ie - 333555777888999/3 = 111185259296333 - exact, no remainder
333555777888999/7 = 47650825412714.142...
47650825412714*7 = 333555777888998
333555777888999-333555777888998=1 so this is the remainder.

Answer 6

Sum all the digits in that 3+3+3+5+5+5+7+7+7+8+8+8+9+9+9=96
then keep 96 in ur hand and use the following
1)96/3=(remainder 0) >>>which tells that it is divisible by "3" and remainder=0
(for suppose if it is difficult to do calculations with 96 do the same thing "9+6=15" and keep 15 in u r hand )<<<
2)96/9 not divisible by "9" and gives remainder as "6" (or)15/9 gives remainder =6
3)for this u have to do 1st(3) +last no.(9) sum and keep it, after that 2nd(3) and last but one(9) no. and so on..........
and check that no . is divesible by 11
eg:121/11=11
1+1=2(i.e., the middle no.)

Answer 7

for 3 is easy add all the digit other than 3 and 6 and 9
5*3+7*3+8*3 and divide by 3 remainder = 0
9 is easy 3*3+5*4+7*3+8*3 devide by 9 remained = 0
for 11 add altenate terms and find the diff.
3+3+5+7+7+8+9+9
3+5+5+7+8+8+9

taake the diff it is 2

Answer 8

3 is easy, add up all the digits, if it equals to a number that is also divisible by 3, then it is. So something like 999 adds up to 9 + 9 + 9 = 27 which is equal to 2 + 7 = 9 which is divisible by 3

same thing for 9

Answer 9

If a number occurs thrice then that number is divisible by 3,11,37.If the sum of the number is 9 then that number is divisible by 9

Answer 10

use the tests for divisibility