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what is x^(4)+4x^(3)+6x^(2)+4x+5 divided by x^(2)+1

divide this and solve

Research, Knowledge and Information :


Long Division of Polynomials


Long Division of Polynomials ... 2 +4x−1 x+2 3x5 +5x4 −4x3 +7x+3 −3x5 −6x4 −x ... 6 2 4 −8 2 3 −1 −2 4 −1 5 The quotient is 3x4 −x3 −2x2 +4x−1 ...
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If c(x)=4-2 and d(x)=x^2+5x, what is (c•d)(x)? A. 4x^3+18x^2 ...


... (x + 1) = 6(x^3 + 3x^2 + 3x + 1) - 3(x^2 + 2x + 1) + 4x + 4 - 9 f(x + 1) = 6x^3 + 18x^2 + 18x + 6 - 3x^2 - 6x - 3 + 4x + 4 - 9 f(x + 1) ... (2x^2+x-3) divided by ...
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6x squared + 13x + 5 (divided by) 4x squared - 4x - 3 - Math ...


6x 2 x 13x + x = 6x 2 +10x xxxxx = xxxx x+5)+1(3x+5)= (2x+1)(3x+5) (divided xxx. xx x x 4x x xx 4x 2 - 6x +2x-3 xxxxxxxxxxxxxxxxxx xxxxxxxxxxxx ...
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Long divide (x^4+5x^3+6x^2-x-2) by (x+2) - YouTube


Oct 16, 2013 · What's x^3 + 5x - 7 divided by x - 4? ... Divide 4x^4-2x^3+6x-5 by (x-1) - Duration: ... How to divide x^4 + 3x^3 - x^2 - 4x - 1 by ...

What is the answer to 9x*5-8x*4-7x*3+6x*2+5x+4 divided by 4x ...


What is the answer to 9x*5-8x*4-7x*3+6x*2+5x+4 divided by 4x*2-3x+2? Update Cancel. No ... How do I find the square root of 4x^4+8x^3+8x^2+4x+1? Is [math]x^4+2x^3-7x ...
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3.2 The Factor Theorem and The Remainder Theorem


3.2 The Factor Theorem and The Remainder Theorem 257 ... x2 + 6x+ 7 x 2 x3+4x2 5x 14 2x2 6x2 12x 7x 14 0 ... (of degree at least 1) is divided by x c, ...
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How do you divide (x^3-x^2+4x+4) / (x+2) using polynomial ...


This gives you a difference of: " " " " " " " "x^2 x+2" " |~bar(x^3-x^2+4x+4 ... 3x^2+4x " " " " " "-underline((-3x^2-6x)) ... of #d-2# divided by #d^4-6d^3 ...
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What is #(4x^2-1)/(2x^2-5x-3) * (x^2-6x+9)/(2x^2 ... - Socratic


... you would factor all the polynomials and get: 4x^2-1=(2x-1)(2x+1) x^2-6x+9=(x-3) ... formula: x=(5+-sqrt(25+24))/4=(5+-7)/4 x_1=-1/2; x_2=3 Then 1 ... Socratic ...
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Long Division of Polynomials - analyzemath.com


2x 3 + 4x 2 - 6x Multiply x 2 + 2x - 3 by 2x: ... Vocabulary associated with the long division process 2x 3 + 3x 2 - x + 16 is the dividend x 2 + 2x - 3 is the ...
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Suggested Questions And Answer :


how do you use break apart to divide ?

You need to find the factors of the number you're dividing. If it's an even number you can divide by 2, so do this and note the factors as you find them. Therefore you have so far 2 and another number, we'll call n. Look at n, and if it's even you have another 2, so you would write 2*2 and another n. When you no longer have an even number, or if the first number was odd, you check to see if 3 goes into it. You make a note and divide the number by 3 so you now have ...3 and another n. You may even have 2*2...*3*n. Again you try 3, until you can't divide exactly, then you move on to 5 (numbers ending in 5 or 0 are exactly divisible by 5). You carry on like this noting what factors you've found as you go along. The numbers you pick as divisors will always be prime: in order here are the first few: 2, 3, 5, 7, 11, 13. A prime number is one which can't be divided exactly by a number smaller than it (actually up to its square root). 4 isn't prime because 2 goes into it; 9 isn't prime because 3 goes into it and all primes apart from 2 are odd numbers.  Now's a good time to look at examples. First, 660. It's even so 2 goes into it. So we have 2*330. 330 is also even: 2*2*165. 165 is divisible by 3, so we divide by 3 and we have 2*2*3*55. 55 ends in 5 and it's not divisible by 3, so we divide by 5 and get 11, and now we have 2*2*3*5*11. 11 is prime so we stop. 660=2*2*3*5*11. This breakdown means we can spot the factors other than prime factors. 2*2 means that 4 is a factor; 2*3 means that 6 is a factor; 3*5 means that 15 is a factor; 2*5*11=110 means that 110 is a factor, and so on. So if you needed to divide 660 by 15 you cross out the factors that make 15 (3 and 5) leaving 2*2*11=44. Let's try another example: 2030. 2030=2*1015=2*5*203=2*5*7*29. 29 is prime. How do we know how far to divide? Take 9042=2*4521=2*3*11*137. Is 137 a prime number? We only have to try and divide by numbers up to the square root of 137. What does that mean? The square root is the number which when multiplied by itself is closest to the number we're trying to divide, but smaller than it. 11*11=121; 12*12=144, so we only have to go up to dividing by 11, because 11*11 is less than 137 while 12*12 is greater, and the square of the next prime is 13*13=169, which is bigger than 137; so we only need to go up to primes up to 11, which doesn't divide exactly into 137, so 137 must be a prime number. Now try some numbers yourself!
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Find the remainder when 2^100+3^100+4^100+5^100 is divided by 7.

2^3=8, and 8 divided by 7 leaves a remainder 1. This means that (2^3)^2=2^6 will also have remainder 1, and 2^99 divided by 7 has a remainder of 1 so 2^100 has a remainder of 2*1=2 when divided by 7. 3^6=729 which is 104 rem 1 when divided by 7. So 3^96 also has rem 1 when divided by 7. 3^4=81 which is 11*7 + 4 so 3^100 divided by 7 has rem 4. 4^3=2^6 which has rem 1 when divided by 7; 4^96 also has rem 1 and 4^4=256 which has rem 4 as does 4^100. 5 is 7-2 so that's rem -2; 25 has rem (-2)^2=4; 125 has rem (-2)^3=-8 which has rem -1; (-2)^6 has rem 1. Check: 5^6=15625=7*2232 + 1. So 5^96 gives rem 1 and 5^4 gives rem (-2)^4=16 which is rem 2. (SUMMARY: Each of the numbers 2, 3, 4 and 5 raised to the power of 6 has a remainder of 1 when divided by 7, so since 6 goes into 100 16 times (6*16=96) each of the numbers raised to the power of 96 also has a rem of 1 when divided by 7. Therefore, we only had to take the numbers 16, 81, 256 and 625 (4th powers of 2, 3, 4 and 5) to get to the 100th power, and then we look at the divisibility of 7 into each of these, arriving at 2, 4, 4 and 2 as the remainders, the sum of which gives us the remainder we're looking for when further dividing it by 7.) Now we just add the rems: 2+4+4+2=12 which has a rem of 5 when divided by 7. The whole original expression 2^100+3^100+4^100+5^100  then has rem 5 when divided by 7.
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hpw do you estimate an answer to a problem with a remainder?

The purpose of the estimate is not to get an accurate answer, but just an indication of what the answer should be approximately, so that when the actual answer is determined, you can compare it with the estimate. They should be fairly close. If they are, this gives you confidence in your ability to work out the answer. If they're not, you may have made a mistake in your calculations. With practice, you become used to estimating quickly and that gives you a rough expectation of what the answer should be. So when you're doing a division it's not the remainder you need to be concerned about, but rather a rough idea of what the answer is going to be when calculated thoroughly. For example, if you were asked to divide 490 by 21, you could quickly estimate this by using 500 in place of 490 and 20 in place of 21, because these substitute numbers are close enough, or compatible with the actual numbers. 500 divided by 20 is 25. That can be done quickly in your head. 490/21 is going to be roughly 25. The actual answer is 23 with a remainder of 7. The estimate doesn't try to predict the remainder, but 25 is close enough to 23 to tell you that the real answer is probably correct. If you calculated 490/21 to be something way different from 25, you would suspect something had gone wrong in your calculations. So that's it: the estimate is just a quick fire judgment, before you take the time to do the proper calculation. An estimate is, or should be a quick mental calculation and not a laborious or lengthy exercise. For some numbers there is a way of predicting the remainder without doing any division at all. The obvious one is dividing by 2, because all odd numbers will give remainder 1. Dividing by 4 requires dividing only the last two digits by 4 and noting the remainder. Dividing by 8 requires dividing the last three digits and noting the remainder. Dividing by 5 is just a matter of noting whether the number ends in zero or 5. If it does there's no remainder; if it doesn't the remainder is found by subtracting 5 or zero from the last digit, depending on the size of the digit. If the digit is between 1 and 4 then so is the remainder; if it's between 6 and 9 then the remainder is found by subtracting 5. When dividing by 9, we don't do any division at all: we simply add up the digits making the number and if the result is 10 or more we add the digits of the result and we keep doing this until we end up with a single digit. If this digit is 9 the number is divisible exactly by 9; otherwise the digit is the actual remainder (this method is not surprisingly called the 9's remainder, and it can be used to check addition, subtraction and multiplication with a 90% accuracy level). Division by 11 is slightly more tricky, but consists of, starting with the last digit, add the alternate digits of the number then subtract the sum of the remaining digits. If the result is positive then that is the remainder; if negative then add 11 to find the remainder; for example, to find the remainder of 98765 divided by 11, we add 5+7+9=21 and subtract 6+8=14: 21-14=7, so the remainder is 7. Another example: remainder of dividing 23451 by 11; 1+4+2-(5+3)=-1+11=10, so the remainder is 10. If the result of the alternate subtraction exceeds 11 repeat the process: 190827 gives 7+8+9-(2+0+1)=21 so the next step is 1-2=-1, add 11=10, so 10 is the remainder.
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10 vedic maths rules for class IX

2 instead of 5: 34/5 can be calculated by multiplying 34 by 2 instead of dividing by 5. 34*2=68. We just move the decimal point one place to the left: 34/5=6.8. 124/5=24.8 because 124*2=248. Move the decimal point: 248 becomes 24.8. 34*5 is the same as 34/2=17 but we add a zero to make 17 into 170. 73*5 is the same as 73/2=36.5 then move the decimal point one place to the right (or add zero): 36.5 becomes 365=73*5. So we only need to know how to multiply and divide by 2 to divide and multiply by 5. We just move the decimal point. Divisibility by 9 or remainder after dividing by 9. All multiples of 9 contain digits which added together give 9. As we add the digits together, each time the result goes over 9 we add the digits of the result together and use that result and continue in this way up to the last digit. Is 12345 divisible by 9? Add the digits together 1+2+3=6. When we add 4 we get 10, so we add 1 and zero=1 then we add 5 to get 6. The number is not exactly divisible by 9, but the remainder is 6. We can also ignore any 9's in the number. Now try 67959. We can ignore the two 9's. 6+7=13, and 1+3=4; 4+5=9, so 67959 is divisible by 9. Multiplying by 11. Example: 132435*11. We write down the first and last digits 1 ... 5. Now we add the digits in pairs from the left a digit step at a time. So 1+3=4; 3+2=5: 2+4=6; 4+3=7; 3+5=8. Write these new digits between 1 and 5 and we get 1456785=132435*11. But we had no carryovers here. Now try 864753*11. Write down the first and last digits: 8 ... 3. 8+6=14, so we cross out the 8 and replace it with 8+1=9, giving us 94 ... 3. Next pair: 6+4=10. Again we go over 10 so we cross out 4 and make it 5. Now we have 950 ... 3. 4+7=11, so we have 9511 ... 3. 7+5=12, giving us 95122 ... 3; 5+3=8, giving us the final result 9512283.  Divisibility by 11. We add alternate digits and then we add the digits we missed. Subtract one sum from the other and if the result is zero the original number was divisible by 11. Example: 1456785. 1 5 7 5 make up one set of alternate digits and the other set is 4 6 8. 1+5+7=13. We drop the ten and keep 3 in mind to add to 5 to give us 8. Now 4 6 8: 4+6=10, drop the ten and add 0 to 8 to give us 8 (or ignore the zero). 8-8=0 so 11 divides into 1456785. Now 9512283: set 1 is 9 1 2 3 and set 2 is 5 2 8; 9+1=0 (when we drop the ten); 2+3=5; set 1 result is 5; 5+2+8=5 after dropping the ten, and 5-5=0 so 9512283 is divisible by 11. Nines remainder for checking arithmetic. We can check the result of addition, subtraction, multiplication and (carefully) division. Using Method 2 above we can reduce operands to a single digit. Take the following piece of arithmetic: 17*56-19*45+27*84. We'll assume we have carried out this sum and arrived at an answer 2365. We reduce each number to a single digit using Method 2: 8*2-1*9+9*3. 9's have no effect so we can replace 9's by 0's: 8*2 is all that remains. 8*2=16 and 1+6=7. This tells us that the result must reduce to 7 when we apply Method 2: 2+3+6=11; 1+1=2 and 2+5=7. So, although we can't be sure we have the right answer we certainly don't have the wrong answer because we arrived at the number 7 for the operands and the result. For division we simply use the fact that a/b=c+r where c is the quotient and r is the remainder. We can write this as a=b*c+r and then apply Method 2, as long as we have an actual remainder and not a decimal or fraction. Divisibility by 3. This is similar to Method 2. We reduce a number to a single digit. If this digit is 3, 6 or 9 (in other words, divisible by 3) then the whole number is divisible by 3. Divisibility by 6. This is similar to Method 6 but we also need the last digit of the original number to be even (0, 2, 4, 6 or 8). Divisibility by 4. If 4 divides into the last two digits of a number then the whole number is divisible by 4. Using 4 or 2 times 2 instead of 25 for multiplication and division. 469/25=469*4/100=1876/100=18.76. 538*25=538*100/4=134.5*100=13450. We could also double twice: 469*2=938, 938*2=1876, then divide by 100 (shift the decimal point two places to the left). And we can divide by 2 twice: 538/2=269, 269/2=134.5 then multiply by 100 (shift the decimal point two places left or add zeroes). Divisibility by 8. If 8 divides into the last three digits of a number then the whole number is divisible by 8. Using 8 or 2 times 2 times 2 instead of 125 for multiplication and division. Similar to Method 9, using 125=1000/8. Using addition instead of subtraction. 457-178. Complement 178: 821 and add: 457+821=1278, now reduce the thousands digit by 1 and add it to the units: 278+1=279; 457-178=279. Example: 1792-897. First match the length of 897 to 1792 be prefixing a zero: 0897; complement this: 9102. 1792+9102=1894. Reduce the thousands digit by 1 and add to the result: 894+1=895. Example: 14703-2849. 2849 becomes 02849, then complements to 97150. 14703+97150=111853; reduce the ten-thousands digit by 1 and and add to the result: 11854. Squaring numbers ending in 5. Example: 75^2. Start by writing the last two digits, which are always 25. Take the 7 and multiply by 1 more than 7, which is 8, so we get 56. Place this before the 25: 5625 is the square of 75. The square of 25 is ...25, preceded by 2*3=6, so we get 625. All numbers ending in 0 or 5 are exactly divisible by 5 (see also Method 1). All numbers ending in zero are exactly divisible by 10. All numbers ending in 00, 25, 50 or 75 are divisible by 25. Divisibility by 7. Example: is 2401 divisible by 7? Starting from the left with a pair of digits we multiply the first digit by 3 and add the second to it: 24: 3*2+4=10; now we repeat the process because we have 2 digits: 3*1+0=3. We take this single digit and the one following 24, which is a zero: 3*3+0=9. When we get a single digit 7, 8 or 9 we simply subtract 7 from it: in this case we had 9 so 9-7=2 and the single digit is now 2. Finally in this example we bring in the last digit: 3*2+1=7, but 7 is reduced to 0. This tells us the remainder after dividing 2401 by 7 is zero, so 2401 is divisible by 7. Another example: 1378. 3*1+3=6; 3*6=18 before adding the next digit, 7 (we can reduce this to a single digit first): 3*1+8=3*1+1=4; now add the 7: 4+7=4+0=4;  3*4=12; 3*1+2+8=5+1=6, so 6 is the remainder after dividing 1378 by 7.  See also my solution to: http://www.mathhomeworkanswers.org/72132/addition-using-vedic-maths?show=72132#q72132
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Describe two methods for converting a mixed number to a decimal

Represent the mixed number as N a/b where N is the whole number part and a over b is the fraction. Method 1 Write down N and follow it by a decimal point. Now you need to divide b into a, but because a is smaller than b you need to write a, a decimal point, and as many zeroes as you wish for accuracy. Some decimals terminate (divide exactly) and some recur (a pattern repeats indefinitely). Example: a=3 and b=4 so we have 3/4. Divide b into 3.0000... Treat 3.0 as 30 and divide by 4, putting the answer immediately after the decimal point. So we have .7 and 2 over making 20 with the next zero. Now divide 4 into 20 and write the result after the 7: .75. This is an exact division so we don't need to go any further. Let's say N was 5 so the mixed number is 5 3/4. We've already written N as 5. so now we continue after the decimal point with what we just calculated giving us 5.75. Another example: 7 5/6. We write 7. first. Divide 6 into 5.0000... and we get .8333. This is a recurring decimal. We keep getting the same carryover. Now we attach the whole number part to get 7.83333... Another example: 2 1/16. We write 2. first. But be careful in the next part: 16 divided into 1.0000... 16 doesn't go into 10 so we write .0 as our first number. Then we divide 16 into 100. This is 6 remainder 4. So we have .06. 16 into 40 goes 2 remainder 8. That gives us .062. Finally 16 into 80 is exactly 5 so we have .0625. Attach this to the whole number: 2.0625. Method 2 For N a/b we make the improper fraction b*N+a over b, then divide by b. Example: 2 1/16: 16*2+1=33. So we divide 33.000.... by 16. We end up with 2.0625. Example: 7 5/6: divide 6*7+5=47 by 6. We end up with 7.83333...
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What is its remainder when its divided by 7?

The easiest way to solve this is to write out for the larger divisors those numbers up to 100 where the dividing conditions are satisfied. For 5 it's 9, 14, 24, 29,...99 and for 4 it's 7, 11, 15, 19,...,99. So for 4 and 5 we only have n, the number we're dividing, = 19, 39, 59, 79 and 99. The only number on this list that divides by 3 with remainder 2 is 59, which, divided by 7 has a remainder of 3. Another way of solving it is to take the numbers 4 and 5 and multiply them together=20. Subtract 1 from this and we get 19. 4 has remainder 3 and 5 has remainder 4 (both requirements of the question) when divided into 19 and this happens for every number separated by 20. Similarly, if we multiply 3, 4 and 5 together we get 60. Subtract 1 to get 59 and we meet all the requirements. This would be repeated every 60 numbers but we only need to go up to 100.
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7/8 (9/10 divided by 5/8) =???? leave as as improper fraction. Need help

1) 7/8 (9/10 divided by 5/8) =???? leave as as improper fraction. Need help start in the parenthesis to divide you switch the top and bottom of the second fraction and multiply 9/10 * 8/5 = 72/50  reduce this: 36/25 now multiply by 7/8 7/8 * 36/25 = 7/4 * 18/25 = 126/100 = 63/50 2). 14a^2/10b^2 divided by 21a^2/15b^2 again flip the second fraction 14a^2/10b^2 * 15b^2/21a^2 rewrite with common together 14a^2/21a^2 * 15b^2/10b^2 = 2/3 * 3/2 = 1 3). 7x^3/3 divided by 14x^2/6 7x^3/3 * 6/14x^3 rewrite 7x^3/14x^3 * 6/3 = 1/2 * 2/1 = 1 4). a^2-2a+1/6 divided by (a^2-1) (a - 1)(a - 1)/6 * 1/(a - 1)(a + 1) = (a - 1) /6(a + 1)
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what is the anwser to this 67.82 divide by 2.8

It's the decimal points that are confusing you, isn't it? There are a number of approaches, but the first thing to do is to move the decimal point in the divisor 2.8. Move it one place to the right to get 28, and move it one place to the right in the dividend to get 678.2. Now you can split 28 into its factors 4 and 7, and divide by one then the other like so:  by 4: 678.2/4=169.55 then by 7: 169.55/7=24.22142857... (Easier to divide by 4 first so that you have a smaller number to divide by 7.) A quick explanation of how you divide the decimal: when dividing 678.2 by 4 you get 169 and you haven't finished because you have 2 to carry and you've just met the decimal point, right? Ignore the decimal point but write the next number in the quotient with a decimal point in front of it and carry on: 169.5. You still have 2 over, so bring down a zero to make it 20 and divide again to get another 5: 169.55. Do the same when you're dividing by 7 and keep bringing down zeroes with each remainder. The number goes on forever, so stop when you have enough decimal places or you get fed up! The other approach is to use long division, bringing in the decimal point as above. Long division will take you longer than splitting into factors because you probably won't know your 28 times table as well as you know your 4 and 7 times tables.
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show work for 4.40 divided by 460

4.40 divided by 460. Since 460=10*46, we can simplify the problem by dividing 4.40 by 10 first. To do this we move the decimal point one place to the left: 0.440. We can also divide this result by 2 because 46=2*23. 0.440 divided by 2=0.220. All that remains is to divide 0.220 by 23. We use long division as if there were no decimal point. 23 won't divide into 2 or 22 so in the answer we need two zeroes after the dec pt. 23 divides into 220 9 times (9*23=207) leaving a remainder 13, which we combine with another zero to make 130. 9 is placed in the answer making 0.009. 23 goes into 130 5 times so 5 goes next to 9 in the answer. And so on, calculating the remainder and bringing down zeroes indefinitely for as many dec places as necessary. To 3 significant figures the answer is 0.00957, close to 0.01 which you quoted.
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64239 is divided by a certain number.while dividing the number 175,114 and,213 appears as three successive remainders.find the divisor ?

It's not clear what "successive remainders" means. There is only one remainder after a division: (dividend)=(quotient)*(divisor)+(remainder). For example, if 175 is the remainder after division then 286 divides into 64239 224 times, remainder 175. The largest remainder is 213, implying that the divisor is bigger than 213. It's clear that the divisor cannot be divided again by the same divisor to yield a remainder in excess of the divisor. In the example 286 will not divide into 224, so 224 would be a second remainder. If the divisor has to be used three times to divide successively into each quotient to obtain each remainder then the divisor cannot exceed 40 (the approximate cube root of 64239) and therefore the remainders cannot exceed 40. See explanations in square brackets below. The only other interpretation I can find to the question is that in the process of long division, we have the successive remainders as we proceed with the division. But that doesn't seem to work either. [The numbers a, b and c are integers and b=cx+213; a=bx+114; 64239=ax+175, making ax=64064. Combining these equations we get: a=64064/x=bx+114, so b=64064/x^2-114/x=cx+213, and c=64064/x^3-114/x^2-213/x. Because 64064=ax, we see that 64064 must be divisible by x. 64064 factorises: 2^6*7*11*13. Because the largest remainder is 213, x>213, because x must be bigger than any remainder it creates. But the first term 64064/x^3 can only yield an integer value if x^3<64064, that is, x<40, which contradicts the requirement x>213.] [If the successive remainders appear in the long division, and the divisor has 3 digits as expected, and three remainders suggest that the other factor also has 3 digits, then if 175 is the first remainder it must be the result of dividing the divisor into 642. Therefore the multiple of the divisor =642-175=467. Since 467 is prime then the first digit of the other factor must be 1. 64239 divides by 467 137 times with 260 as the remainder. This is not one of the three remainders, so 175 cannot be the first remainder. Try 114: 642-114=528=2^4*3*11. We need a factor >213. 264, 352. This time we appear to have partial success, because 176 (less than 213, but half of 352) goes into 64239 364 times with a remainder of 175! So we have two of the three required remainders. Unfortunately the third remainder is 87, so 114 cannot be the first remainder. Finally, try 213: 642-213=429=3*11*13. The only factor bigger than 213 is 429 itself, and the final remainder after dividing 64239 by 429 is 318, not included in the three remainders given. So the assumption that the remainders arise during long division is also false.] 64064=2^6*7*11*13. Leaving aside the powers of 2, and just looking at combinations of 7, 11 and 13, we get 7*11=77; 7*13=91; 11*13=143, and 7*11*13=1001. By progressively doubling these products we can discover factors bigger than 213: 308=4*77; 364=4*91; 286=2*143, etc. We can also take 7, 11 and 13 and multiply them by powers of 2: 224=32*7; 352=32*11; 416=32*13. We can write the factors of 64064 in pairs, starting with a factor a little bigger than 213: (224,286), (308,208), (352,182), (364,176), (416,154), (448,143), (616,104), (704,91), (1001,64), (1232,52), (2002,32), (2464,26), (4004,16), (4928,13), (8008,8), (16016,4), (32032,2). These are (a,x) or (x,a) pairs. If we take 175 as the remainder, and ignore the other two, then we have a choice of factors. The 3-digit factors are (224,286), (308,208), (352,182), (364,176), (416,154), (448,143), (616,104). The only pair in which each factor is greater than 213 is (224,286).
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