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8.3333
Hermann Hilpert
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∙ 13y agoIt is 7.3333333333
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What is 304 divided by 3?
101.3333
What is 216 divided by 13?
16.6154
What is the algorithm of 648 divided by 4?
162
How do you find remainders in a division problem?
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]
A number when divided by 357 leaves a remainder 39 what will be the remainder when the number is divided by 17?
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.
How would you find the remainder when x1000 plus x500 - 1 is divided by x-1?
(x1000 - x500 - 1)/(x - 1) (x1000 + 0x999 + ... + 0x501 - x500 + ... + 0x - 1)/(x - 1) Now you can find the remainder, if you like...
What is the largest positive integer that leaves the same remainder when divided into each 99 and 204?
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.
what is the largest number when divided into 63 and 75 will leave a remainder of three?
To find the largest number that, when divided into both 63 and 75, leaves a remainder of three, you can use the concept of greatest common divisor (GCD) or greatest common factor (GCF). The GCD of 63 and 75 is the largest number that can evenly divide both numbers. To find it, you can use the Euclidean algorithm: Start with the two numbers: 63 and 75. Divide 75 by 63: 75 ÷ 63 = 1 with a remainder of 12. Now, replace the larger number (75) with the remainder (12) and keep the smaller number (63) as is: 63 and 12. Repeat the process: 63 ÷ 12 = 5 with a remainder of 3. Again, replace the larger number (63) with the remainder (3) and keep the smaller number (12) as is: 12 and 3. Repeat once more: 12 ÷ 3 = 4 with no remainder. Now that you have reached a point where the remainder is 0, the GCD is the last non-zero remainder, which is 3. So, the largest number that, when divided into both 63 and 75, leaves a remainder of three is 3.
How do you halve 3.0?
By dividing by 2:3 divided by 2 is 1 remainder 1.The decimal point now follows.The remainder 1 is ten times the value of the 0 in the next column, so add 10 x 1 to 0 to get 10.10 divided by 2 is 5 remainder 0.so half of 3.0 is 1.5:...... 1 . 5... -------2 | 3 . 0....... 2....... --....... 1 . 0....... 1 . 0....... -----............... 0
What division problems having a dividend and a remainder of 2?
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
On dividing a number by 357 we get w as remainder. On dividing the same number 17 what will be the remainder?
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.
What even number can be divided by 7 and 3 without a remainder?
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.
