When I try to divide numbers that would leave a remainder like 1 divided by 6 which should give me an answer like 0.16666666666 instead all it says for an answer is 1/6. it does say in the manual to input how many decimals i want and actually uses the 1 divided by 6 as an example but it shows .166666666 as what it should show up on my calculater

any numbers that are not dividable by 3
so take each number between 1 and 100 , divide it by 3 and if you don.t get a remainder that number is not what you are looking for
grade 2 maths really

Where to start. Most things we do in math are in base 10, probably because we have 10 fingers. Instead of base 10, binary is base 2. In computers, binary rules because a bit can be set to ON or OFF, 1 being ON, and 0 being OFF.

So the next thing is counting. 0 1 10 11 100 101 110 111 1000 1001
Decimal equivalent 0 1 2 3 4 5 6 7 8 9

How to convert from binary to decimal? Line up the columns and multiply down. For example, 1101. The columns, starting from the right are 2^0, 2^1, 2^2, 2^3, reading ^ as exponent.

See if I can line these up. 1 1 0 1
2^3 2^2 2^1 2^0
8 4 2 1
multiplying down 8 4 0 1

Next step is to add them up. 8+4+1 = 13!

Going the other way, we have to divide by the largest 2^x number, so what is left over and continue dividing until nothing is left.

For example, 56. 2^5 (32) is the largest 2^x number that goes into 56, so we put a 1 in this column. Our remainder is 24. 2^4 (16) goes into this once, so we put a 1 in this column. Our remainder is 8. 2^3(8) goes into this once, so we put a 1 in this column. Now the remainder is 0 so we put a 0 in the 2^2 (4) column, 2^1(2) column, and 2^0 (1) column.

We end up with 111000.

Let me know if you have any questions.

Good luck.

Paul

First, find out how many whole times 24 will go into 67
67/24 = 2 with a remainder that we don't care about. Write down 2
Next, multiply that answer (2) by 24 2 X 24=48
Subtract that 48 from 67 67-48=19
bring down next digit (8) append to the19 198
divide 24 into 198 198/24=8 Write down 8
multiply that answer by 24 8 X 24 =192
Subtract that 192 from 198 198 - 192 = 6
bring down next digit (7) append to the 6 67
divide 24 into 67 67/24 =2 Write down 2
multiply that answer by 24 2 X 24 = 48
subtract 48 from 67 27-24=19
bring down next digit (2)append to the 19 192
divide 24 into 192 192/24 = 8 Write down 8
multiply that answer by 24 24 X 8=192
subtract 192 from 192 192-192=0 You're done!
Your answer is the numbers you wrote down after the divisions = 2848

2 8 4 8
----------------------
24 ) 67872
- 48
198
-192
67
- 48
192
- 192
0

There are an infinite number of numbers that qualify. The smallest such positive number is 125. The next such number is 251.

For any polynomial A(x) there is P(x)=A(x)(x-1)(x-2)+2 and this polynomial fulfills above requirements.

Then, remainder of P(x) divide by (x-1)(x-2) is also 2!

MOD simply calculates the remainder that is left over after dividing by the MOD parameter.
For example, 10MOD7 = 3 since you can only divide 7 into 10 once, leaving a remainder of 3.
For larger numbers, eg 389487MOD31 you start by dividing 389487 by 31, which gives 12564.09677 (on the display), however you only want the fractional part so subtract 12564 to give 0.0967741935 . This includes the lower value digits that were previously hidden. Then multiply this result by 31 which gives 2.999999999 on the display. Since the result of a MOD calculation has to be a whole number, the answer needs to be rounded to 3. If the answer doesn't easily round to a whole number, it is likely that you have gone beyond the calculators working capacity.
An alternative method is to take the integer part of the first result (12564) and multiply it by 31 to get 389484, then subtract this from 389487, giving the answer of 3.

Finding the remainder from a division calculation is often referred to as using the MOD function, though this isn't available directly on the fx-991ES.
To obtain the remainder, for example when dividing 10 by 7, you can only divide 7 into 10 once, leaving a remainder of 3.
For larger numbers, eg when dividing 389487 by 31 (equivalent to 389487MOD31) the division gives an answer of 12564.09677 (on the display), however you only want the fractional part so subtract 12564 to give 0.0967741935 . This includes the lower value digits that were previously hidden. Then multiply this result by 31 which gives 2.999999999 on the display. Since the result of a MOD or remainder calculation involving integers has to be a whole number, the answer needs to be rounded to 3. If the answer doesn't easily round to a whole number, it is likely that you have gone beyond the calculator's working capacity.
An alternative method which avoids the rounding of the fractional part is to take the integer part of the first result (12564) and multiply it by 31 to get 389484, then subtract this from 389487, giving the answer of 3.

Yes, it's possible because not all numbers are multiples of ten. For example if you divide 63 by 30 (which is multiple of 10) you get 2 and remainder 3 (because 63=2*30+3).

Every associate leaves 4/5(n-1) molecules of a pile of n molecule. This result in an awful formula for the complete process (because every time one molecule must be taken away to make the pile divisible by 5):

4/5(4/5(4/5(4/5(4/5(4/5(p-1)-1)-1)-1)-1)-1), where p is the number of molecule in the original pile, must be a whole number.

The trick is to make the number of molecule in the pile divisible by 5, by adding 4 molecules. This is possible because you can take away those 4 molecules again after taking away one fifth part of the pile: normally, 4/5(n-1) molecules are left of a pile of n molecules; now 4/5(n+4) =4/5(n-1) +4 molecules are left of a pile of n+4 molecules. And because of this, the number of molecule in the pile stays divisible by 5 during the whole process. So we are now looking for a p for which the following holds:

4/5×4/5×4/5×4/5×4/5×4/5× (p+4) = (46/56) × (p+4), where p is the number of molecule in the original pile, must be a whole number.

Therefore, the smallest (p+4) for which the above holds, is 56.

So there were p=5 power 6 - 4 = 15621 molecule in the original pile.