25 divided by 3 is 8 with a remainder of 1. 8 1/3
(x1000 - x500 - 1)/(x - 1) (x1000 + 0x999 + ... + 0x501 - x500 + ... + 0x - 1)/(x - 1) Now you can find the remainder, if you like...
Thrice with a remainder of sixteen
If x ≡ 39 mod 357 then: x = 357k + 39 for some integer k. Now 357 = 21×17, and 39 = 2×17+5 → x = 21×17×k + 2×17 + 5 → x = 17(21k + 2) + 5 → x = 17m + 5 where m = 21k + 2 (is an integer) → x ≡ 5 mod 17 → the remainder when the number is divided by 17 is 5.
The remainder is the amount that cannot be divided into the equation, in other words, the leftovers. Example: 7 / 2 = 3 So now to figure out the remainder we multiply the divider by the answer. [in other words: 7 / (divider) = (answer)] [divider * answer] 2 * 3 = 6 So now to calculate the remainder we subtract this new answer by the number we started out with, in this case 7. 7 - 6 = 1 So the remainder is 1 [3 remainder 1]
The dividend is the number you are dividing so you know it is 2 divided by something. Now to get a remainder of 2, you need the divisor to be larger than than 2, so: 2/3 works as does 2/4, 2/5,2/6,2/7 and so on
There is no such number. For suppose x was a number such that it keft the same remainder r, when divided by 99 and 204. Then x = 99a + r and x = 204b +r for some integers a and b. Consider y = x + 99*204 then y > x Now y = (99a + r) + 99*204 = 99*(a+204) + r so y leaves a remainder of r when divided by 99. Also y = (204b + r) + 99*204 = 204*(b+99) + r so y leaves a remainder of r when divided by 204. This contradicts the proposition that x is the maximum such number.
If you are familiar with converting numbers to binary it works the same only using 6. The trick is to divide the number by six, write down the remainder, and keep the quotient. Repeat that until the quotient reaches zero. Then, take the remainders you've written down, and reverse their order. That will give you the number in the base to which you're converting. In the case of converting 200 into base six notation, you would do the following: 200 divided by 6 equals 22, with a remainder of 2 22 divided by 6 equals 3, with a remainder of 3 5 divided by six equals 0, with a remainder of 5 Now we can take the remainders and read them in the reverse order that we wrote them in: 532. That is our value expressed in base six.
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When you divide a number (like in long division) and end up with a remainder that keeps coming up the same remainder as you carry the numbers into the decimal portion. For example: 1 divide by 3: You start out 1.000 divide by 3, and 1 divided by 3 is zero, bring down the 1 and a zero and then you have 10 divided by 3 is 3 remainder 1, so now you have 0.3 remainder 1. If you continue this process you will keep getting more 3's and more remainder 1, until you realize that the 3's will repeat forever. (You will always get a 3 remainder 1, no matter how far you carry it out) Some numbers give 2 or more digits that repeat, like dividing by 11 gives a pair of digits that repeat, while dividing by 7 gives a set of six digits that repeat.
If x ≡ w mod 357 then: x = 357k + w for some integer k. Now 357 = 21×17, and w = 17n + c for some integers n ≥ 0 and 0 ≤ c < 17 → w ≡ c mod 17 This gives: x = 21×17×k + 17n + c → x = 17(21k + n) + c → x = 17m + c where m = 21k + n (is an integer) → x ≡ c mod 17 → the remainder when the number is divided by 17 is the same as the remainder when the original remainder w is divided by 17.
Call the number we are looking for x. Since the number has a remainder of 1 when divided by 6 or 7, x-1 is divisible by 6 and 7, which means that x-1 is divisible by 42. Similarly, x-7 is divisible by 88. The distance between x-7 and x-1 is 6. Therefore, we are looking for a number which is divisible by 88 and, when added to 6, is divisible by 42 (since when we find this number we can find the number we were originally looking for). So far we have that x-7 is divisible by 88 and that (x-7)+6 is divisible by 42. Since 42 is divisible by 6, (x-7)+6 is also divisible by 6. This means that x-7 is also divisible by 6. This means that, since x-7 is also divisible by 88, it must be divisible by 264. Now the restrictions for (x-7)+6 are that it must be divisible by 264 and that (x-7)+6 must be divisible by 7. However, if (x-7)+6 is divisible by 7, then so is (x-7)-1. At this point we only have to test at most 7 cases to find a number that works. However, there is a way to deal with smaller numbers. The remainder when 264 is divided by 7 is 5. (To do this you can use the divisibility rule for 7 (double the last digit and subtract the number from the number formed when you take all of the digits of the number other than the units digit) or you could say that, since 21 is a multiple of 7, 210 is a multiple of 7. Therefore, the remainder when 264 is divided by 7 is the same as the remainder when 264-210, or 54, is divided by 7. Since 49 is a multiple of seven, the remainder when 54 is divided by 7 is the same as the remainder then 54-49, or 5, is divided by 7, which is 5.) We want to find the number which when multiplied by 264, gives a remainder of 1 when divided by 7. In other words, we want to to find a multiple of 7 that is 1 less than a multiple of 5. Since all multiples of 5 end in 0 or 5, the number we are looking for must end in 4 or 9. Going through the multiples of 7, we see that 7*2, or 14, ends in a 4. This is one less than 15, which is divisible by 5. 15/5 is 3, the number of times you have to multiply 264 to get a number which is one more than a multiple of 7 is 3. Let's find the number. 264*3=792. To test, 792-1=791, which is 7*113. Remember that this is x-7. Therefore, to find x, all we have to do is add 7 to 792, which gives 799. It's always a good idea to check unless you have a time constraint, so let's check. To find the remainder when 799 is divided by 6, first notice that 666 is divisible by 6. Therefore, the remainder when 799 is divided by 6 is the same as the remainder when 799-666=133 is divided by 6. Next, note that 120 is divisible by 6. Now we have to find the remainder when 133-120=13 is divided by 6. Since 12 is a multiple of 6, we take 13-12 to find that 799 has a remainder of 1 when divided by 6. Check! Now, for 7. Since 700 is divisible by 7 and 91 is divisible by 7, 791 must be divisible by 7. This is pretty close to 799, so let's see what happens when we add one more 7 to 791. We get 798, which is one less than 799, so the remainder when 799 is divided by 7 is 1 as well. Check! Now, for 8. 800 is divisible by 8, so the remainder when 799 is divided by 8 is the same as when 799-800, or -1 is divided by 8. -1 is equivalent to 7 when dividing by 8, so the remainder is 7. Check! Finally, we check 11. Remember the rule that, to multiply a 2-digit number whose digits sum to less than ten by 11, take the sum of the two digits, and put it in the middle. 7+2=9, so 792 is divisible by 11. 799-792=7. Check! Since our number satisfies all of the requirements, the number 799 is an answer. To find other answers, take the 3 we found earlier and add 7 as many times as you want to it. Now, multiply that by 264 and add 7. Any of these numbers should work because they are 7 more than a multiple of 264 and the number 6 more than 7 less than the number is a multiple of 7 (because we added a multiple of 7 to it, so the remainder when divided by 7 should stay the same).
7 times 3=21, so 21 is part of the answer somehow. However, 21 is an odd number. Now, if we multiply 21 X 2= 42 The answer is an even number. 42/3=14 has no remainder 42/7=6, again no remainder So 42, and any whole multipleof 42, would be a satisfactory answer.
Since binary is about 2 digits do the calculatio asmentioned below: We have 146,right.... 1.)Divide it with 2 so is there any remainder ,no, so 0,quotient is 71 2.)Now divide the quotient with 2, now remainder is 1, quotient is,35 3.) Again repeat the same procedure, now remainder is 1, quotient is 17 4.) same process continues,remainder now 1, quotient is 8, 5.)Same, remainder is 0,quotient is 4 6.)Same, remainder is 0,quotient is 2 7.)Same, remainder is 0,quotient is 1 8.)Same, remainder is 1,quotient is 0. Now atlast consider all the remainders from bottom... You will get 10001110 is your answer in binary.
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Say for example you have 16/5.In remainders, the answer would be 3r1. Now you have to divide the remainder by the divisor So for example: 1/5=0.2 Now you have to place the 3 from the answer we got from the remainder in front of the 0.2 answer: 3.2
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Work from left to right... 5 wont divide by 9, so add the 3... 53 divided by 9 is 5, with 8 remainder. Now - 8 won't divide by 9, so add the 1... 81 divided by 9 is 9. Finish by adding the unused zero. This gives you the answer 590.
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To turn a remainder into a fraction you just put the remainder over the dividend. To turn a remainder into a decimal put a decimal in the divisor and in the answer so far, put a zero, bring the zero down and divide what you now have. If that does not come out evenly add another zero in the dividend and do that until it come out evenly
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