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If the sum of two whole numbers is < 40, what is the largest possible product of the two numbers?

Also would you please verify how you got the answer. How you set up the equation. Thanks!

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Whole Numbers - Math League

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Sum-product number - Wikipedia

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Suggested Questions And Answer :

If the sum of two whole numbers is < 40, what is the largest possible product of the two numbers?

???????? yer numbers full av holes ????? em be INTEGER x+y=40, then x=y=20 20^2=400 but if gotta be <,  then 19^2=361
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This is the expansion of (ax+by+c)(dx+ey+f) where a to f represent numerical constants (integers), obeying the following arithmetic rules: ad=6, be=-5, cf=-7 (coefficients of x^2, y^2, constant) The only numbers we have for a, b, c, d, e or f must be in the set {1 1 1 2 3 5 6 7}; no other numbers can be used and the arithmetic rules must always apply. This is because the only factors of 5 are 1 and 5; the only factors of 7 are 1 and 7; but the pairs of factors of 6 are 1 and 6 or 2 and 3. I haven't yet considered the signs plus and minus. We also have the sums of two products: ae+bd=13 (coefficient of xy), af+cd=11 (x), ce+bf=36 (y). The largest of these is 36, and there's only one way of using 4 numbers in the set to get 36 using each one once and that is 7*5+1*1. So from this we can find b, c, e and f. The following table shows all 8 combinations for these coefficients: bf+ce Ref b c e f 1 1 -7 -5 1 2 1 -5 -7 1 3 -7 1 1 -5 4 -5 1 1 -7 5 -1 7 5 -1 6 -1 5 7 -1 7 7 -1 -1 5 8 5 -1 -1 7 The remaining two coefficients are a and d and the possible pairs of factors are 1 and 6, or 2 and 3. Here's a table to show all 8 possible combinations: a and d Ref a d 1 1 6 2 6 1 3 -1 -6 4 -6 -1 5 2 3 6 3 2 7 -2 -3 8 -3 -2 The next bit is tedious. We need to try out all the combinations of b, c, e and f against all those of a and d! Yes, that's 64 in all. What we're looking for is the sum of the products ae+bd=13. I'll spare you the details and just show you the results. Ref 1 from the first table with ref 8 from the second gives the required result and also satisfies af+cd=11. So we have the values of a to f: -3, 1, -7, -2, -5, 1, and the factorisation is: (-3x+y-7)(-2x-5y+1). This can also be written: (3x-y+7)(2x+5y-1), which is ref 5 from the first table with ref 6 from the second. Combinations in red have complementary signs to those shown in black.  
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is it possible to use the distributive property to rewrite 85 + 99 as a product of a whole number greater than 1 and the sum of 2 whole numbers?

85+99=184, the factors of which are 2, 4, 8 and 23, so we need two numbers (integers) to add up to one of these factors and the product of the remaining factors will be the multiplier. Let's pick 23 as the sum and 8 as the multiplier. 23 is the sum of many pairs of numbers, so we'll pick 9 and 14. The distributive property is then demonstrable: 8(9+14)=72+112=184.
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how to factorise fully

I can offer tips: Look out for constants and coefficients that are multiples of the same number, e.g., if all the coefficients are even, 2 is a factor. If 3 goes into all the coefficients, 3 is a factor. Place the common numerical factor outside brackets, containing the expression, having divided by the common factor. In 2x^2+10x-6 take out 2 to give 2(x^2+5x-3). In 24x^2-60x+36 12 is a common factor so this becomes: 12(2x^2-5x+3). This also applies to equations: 10x+6y=16 becomes 2(5x+3y)=2(8). In the case of an equation, the common factor can be completely removed: 5x+3y=8. Now look for single variable factors. For example, x^2y^3+x^3y^2. Look at x first. We have x^2 and x^3 and they have the common factor x^2, so we get x^2(y^3+xy^3) because x^2 times x is x^3. Now look at y. We have y^2 and y^3 so now the expression becomes x^2y^2(y+x). If the expression had been 2x^2y^3+6x^3y^2, we also have a coefficient with a common factor so we would get: 2x^2y^2(y+3x). If you break down the factors one at a time instead of all at once you won't get confused. After single factors like numbers and single variables we come to binomial factors (two components). These will usually consist of a variable and a constant or another variable, such as x-1, 2x+3, x-2y, etc. The two components are separated by plus or minus. In (2) above we had a binomial component y+x and y+3x. These are factors. It's not as easy to spot them but there are various tricks you can use to help you find them. More often than not you would be asked to factorise a quadratic expression, where the solution would be the product of two binomial factors. Let's start with two such factors and see what happens when we multiply them. Take (x-1) and (x+3). Multiplication gives x(x+3)-(x+3)=x^2+3x-x-3=x^2+2x-3. We can see that the middle term (the x term) is the result of 3-1, where 3 is the number on the second factor and 1 the number in the first factor. The constant term 3 is the result of multiplying the numbers on the factors. Let's pick a quadratic this time and work backwards to its factors; in other words factorise the quadratic. x^2+2x-48. The constant term 48 is the product of the numbers in the factors, and 2 is the difference between those numbers. Now, let's look at a different quadratic: x^2-11x+10. Again the constant 10 is the product of the numbers in the factors, but 11 is the sum of the numbers this time, not the difference. How do we know whether to use the sum or difference? We look at the sign of the constant. We have +10, so the plus tells us to add the numbers (in this case, 10+1). When the sign is minus we use the difference. So in the example of -48 we know that the product is 48 and the difference is 2. The two numbers we need are 6 and 8 because their product is 48 and difference is 2. It couldn't be 12 and 4, for example, because the difference is 8. What about the signs in the factors? We look at the sign of the middle term. If the sign in front of the constant is plus, then the signs in front of the numbers in the factors are either both positive or both negative. If the sign in front of the constant is negative then the sign in the middle term could be plus or minus, but we know that the signs within each factor are going to be different, one will be plus the other minus. The sign in the middle term tells us to use the same sign in front of the larger of the two numbers in the factors. So for x^2+2x-48, the numbers are 6 and 8 and the sign in front of the larger number 8 is the same as +2x, a plus. The factors are (x+8)(x-6). If it had been -2x the factors would have been (x-8)(x+6). Let's look at a more complicated quadratic: 6x^2+5x-21. (You may also see quadratics like 6x^2+5xy-21y^2, which is dealt with in the same way.) The way to approach this type of problem is to look at the factors of the first and last terms. Just take the numbers 6 and 21 and write down their factors as pairs of numbers: 6=(1,6), (2,3) and 21=(1,21), (3,7), (7,3), (21,1). Note that I haven't included (6,1) and (3,2) as pairs of factors for 6. You'll see why in a minute. Now we make a table (see (5) below). In this table the columns A, B, C and D are the factors of 6 (A times C) and 21 (B times D). The table contains all possible arrangements of factors. Column AD is the product of the "outside factors" in columns A and D and column BC is the product of the "inside factors" B and C. The last column depends on the sign in front of the constant 21. The "twiddles" symbol (~) means positive difference if the sign is minus, and the sum if the sign in front of the constant is plus. So in our example we have -21 so twiddles means the difference, not the sum. Therefore in the table we subtract the smaller number in the columns AD and BC from the larger and write the result in the twiddles column. Now we look at the coefficient of the middle term of the quadratic, which is 5 and we look down the twiddles column for 5. We can see it in row 6 of the figures: 2 3 3 7 are the values of A, B, C and D. If the number hadn't been there we've either missed some factors, or there aren't any (the factors may be irrational). We can now write the factors leaving out the operators that join the binomial operands: (2x 3)(3x 7). One of the signs will be plus and the other minus. Which one is which? The sign of the middle term on our example is plus. We look at the AD and BC numbers and associate the sign with the larger product. We are interested in the signs between A and B and C and D. In this case plus associates with AD because 14 is bigger than 9. The plus sign goes in front of the right-hand operand D and the minus sign in front of right-hand operand B. If the sign had been minus (-5x), minus would have gone in front of D and plus in front of B. So that's it: [(Ax-B)(Cx+D)=](2x-3)(3x+7). (The solution to 6x^2+5xy-21y^2 is similar: (2x-3y)(3x+7y).) [In cases where the coefficient of the middle term of the quadratic appears more than once, as in rows 1 and 7, where 15 is in the twiddles column, then it's correct to pick either of them, because it just means that one of the binomial factors can be factorised further as in (1) above.] Quadratic factors A B C D AD BC AD~BC 1 1 6 21 21 6 15 1 3 6 7 7 18 11 1 7 6 3 3 42 39 1 21 6 1 1 126 125 2 1 3 21 42 3 39 2 3 3 7 14 9 5 2 7 3 3 6 21 15 2 21 3 1 2 63 61  
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the sum of two numbers is 10 and their product is more than 21

Let the two numbers be x and y. Their sum is less than 10. Therefore both x and y are single digit numbers in the range [1 .. 9] pairs of numbers adding to 10 1+9   product is 9 2+8   product is 16 3+7   product is 21 4+6   product is 24  (a possibility) 5+5   product is 25  (a possibility) We are not told thast the two numbers must be different numbers. Therefore we have two answers. Answers: 4 and 6, 5 and 5
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how do you solve quadratic equations by factoring

If you know that the quadratic is going to factorise, then you need to look at the coefficient of the x^2 term. The number of x^2 you have can be split into factors. For example, if the coefficient is 6, then we have pairs of factors (1,6),(2,3). Then we look at the constant term and split that into possible pairs of factors. For example, if the constant is 24 then the pairs are (1,24),(2,12),(3,8),(4,6). Then we look at the sign of the constant. If it's + then in the next step we are going to ADD, otherwise we're going to SUBTRACT.  The next bit is like a game of tennis doubles. We'll call the game Quadrattack. Instead of the players being in a tennis court, they're in brackets, where the contents of the brackets consist of a team of two players. There's a pair of brackets, one for each team of two. Here's what the brackets look like as factors for the quadratic: (ax+b)(cx+d). We have two teams a and b versus c and d. The team members each represent a number (which may be printed on their T-shirts). The tables below show all possible "games" (fixtures) for the two teams a and b versus c and d, the second table being the "swapped partner" set where b and d change teams. The tables allow you to solve for a, b, c, d where: AX^2+BX+C=(aX+b)(cX+d) and ac=A and bd=C. The specific example of 6X^2+...+24 is given so ac=6 and bd=24. The tables show all possible values for B. X is any variable (representing x, y, t, etc.). The products ad and bc are calculated by multiplying the numbers on their T-shirts. Only players with the right numbers on their T-shirts can take part in the series of games. Plus or minus (+) in front of C tells you whether to use the Sum column (+) or Diff column (-). On with the game!   Game fixtures Game a b c (=A/a) d (=C/b) ad bc Sum Diff 1 1 1 6 24 24 6 30 18 2 1 2 6 12 12 12 24 0 3 1 3 6 8 8 18 26 10 4 1 4 6 6 6 24 30 18 5 2 1 3 24 48 3 51 45 6 2 2 3 12 24 6 30 18 7 2 3 3 8 16 9 25 7 8 2 4 3 6 12 12 24 0 Swap partners (b and d) Game a b c (=A/a) d (=C/b) ad  bc Sum Diff 9 1 24 6 1 1 144 145 143 10 1 12 6 2 2 72 74 70 11 1 8 6 3 3 48 51 3 12 1 6 6 4 4 36 40 4 13 2 24 3 1 2 72 74 2 14 2 12 3 2 4 36 40 4 15 2 8 3 3 6 24 30 6 16 2 6 3 4 8 18 26 8 In an actual question to solve or factorise a quadratic when you know that it can be factorised into rational real factors, you would set up the tables of fixtures for the games according to the given values of A and C. I've just selected A=6 and C=24 as an example. The repeated values of Sum and Diff effectively tell you that the games are in fact identical, so in the example Games 1, 4 and 6 have the same Sum, and it would not be necessary to play more than one of them if you have +C. Whoever decides the games (devises the quadratics) should pick values of A, B and C so that no two games are the same. In Games 2 and 8 Diff is zero, which means the quadratic contains no X term (B=0). The purpose of these games is to play each one until the Sum (where we have +C) or Diff (where we have -C) matches the given value of B. Both teams receive the prize or points, or the individual players receive points. There's just one more thing to do after the winning game: what is the sign in each bracket? For +C the sign is minus for -B and plus for +B; for -C and -B the sign is minus for the b or d player with bc or ad as the larger number and plus for the other b or d player; for -C and +B the sign is plus for the b or d player with bc or ad as the larger number and minus for the other b or d player. For example, if the quadratic to be solved was 6x^2-10x-24, then Game 3 is the winning game (Diff=10) and the factors are (x-3)(6x+8). Here we have -C (-24) and -B (-10) and bc (3*6) is larger than ad (1*8), so b (3) is minus and d (8) is plus.  
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Fine two numbers whose sum is 24 and product is as large as possible?

biggest produkt...square x=24/2=12 x^2=144
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With using the following #'s 56,27,04,17,93 How do I find my answer of 41?

Let A, B, C, D, E represent the five numbers and OP1 to OP4 represent 4 binary operations: add, subtract, multiply, divide. Although the letters represent the given numbers, we don't know which unique number each represents. Then we can write OP4(OP3(OP2(OP1(A,B),C),D),E)=41 where OPn(x,y) represents the binary operation OPn between two operands x and y. We can also write: OP3(OP1(A,B),OP2(C,D))=OP4(41,E) as an alternative equation.  We are looking for a solution to either equation where 04...93 can be substituted for the algebraic letters and the four operators can be substituted for OPn. Because the operation is binary, requiring two operands, and there is an odd number of operands, OP4(41,E) must apply in either equation. So we only have to work out whether to use OP3(OP2(OP1(A,B),C),D)=OP4(41,E) or OP3(OP1(A,B),OP2(C,D))=OP4(41,E). Also we know that add and subtract have equal priority and multiply and divide have equal priority but a higher priority than add and subtract. 41 ia a prime number so we know that OP4 excludes division. If OP4 is subtraction, we will work with the absolute difference |41-E| rather than the actual difference. For addition and multiplication the order of the operands doesn't matter. This gives us a table (OP4 TABLE) of all possibilities,    04 17 27 56 93 + 45 58 68 97 134 - 37 24 14 11 52 * 164 697 1107 2296 3813 This table shows all possible results for the expression on the left-hand side of either of the two equations. The table below shows OPn(x,y): In this table the cells contain (R+C,|R-C|,RC,R/C or C/R) where R=row header and C=column header. The X cells are redundant. Now consider first: OP3(OP1(A,B),OP2(C,D)). To work out all possible results we have to take each value in each cell and apply operations between it and each value in all other cells not having the same row or column. For example, if we take 21 out of row headed 17, we are using (R,C)=(17,4), so we are restricted to operations involving (56,27), (93,27) and (93,56). We can only use the numbers in these cells, but the improper fractions can be inverted if necessary. If the result of the operation matches a number in the only cell in the OP4 TABLE with the remaining R and C values then we have solved the problem. As it happens, the sum of 21 from (17,4) and 37 from (93,56) is 58 which is in (+,17) of the OP4 TABLE. Unfortunately, the column number 17 is the same as the row number in (17,4). To qualify as a solution it would have to be in the 27 column of the OP4 TABLE. But let's see if the arithmetic is correct: (17+4)+(93-56)=21+37=58=41+17; so 41=(17+4)+(93-56)-17. Yes the arithmetic is correct, but 17 has been used twice and we haven't used 27. So the arithmetic actually simplifies to 41=4+93-56, because 17 cancels out and, in fact, we only used 3 numbers out of the 5. Still, you get the idea. (4*17)+(56-27)=68+29=97=41+56; 41=4*17+56-27-56 is another failed example because 93 is missing and 56 is duplicated, so this simplifies to 41=4*17-27, thereby omitting 56 and 93. Similarly, (27+4)+(93-27)=31+66=97=41+56; so 41=(27+4)+(93-27)-56=4+93-56, omitting 17 and 27. The question doesn't make it clear whether all 5 numbers have to be used just once to produce 41, or whether 41 has to be made up of only the given numbers, but not necessarily all of them. In the latter case, it's clear we have found some solutions. The method I used was to create a table made up of all possible OPn(x,y) as the row and column headers. Then I eliminated all cells where the (x,y) components of the row and column contained an element in common. Each uneliminated cell contained (sum,difference,product,quotient) values. When this table had been completed, I looked for occurrences of the numbers in the cells of the OP4 TABLE, and I highlighted them. The final task was to see if there were any highlighted cells such that the numbers that matched cells in the OP4 TABLE were in a column, the header of which made up the set of five given numbers. It happened that there were none, so OP3(OP1(A,B),OP2(C,D))=OP4(41,E) could not be satisfied. No arrangement of A, B, C, D or E with OP1 to OP4 could be found. More to follow...
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how to solve it need help

a) The table below shows the probability distribution, where P is probability: RAINFALL P Profit (RM) Heavy 0.3 28000 Moderate 0.35 55000 Little 0.2 15000 None 0.15 7000 The table below shows cumulative probability from heavy to no rainfall: RAINFALL P Profit (RM) Heavy 0.3 8400 Heavy/moderate 0.65 27659 Heavy/moderate/little 0.85 30650 All 1 31700 Explanation of profit: e.g., Heavy/moderate/little 0.3*28000+0.35*55000+0.2*15000=8400+19250+3000=30650 Cumulative probability from no to heavy rainfall: RAINFALL P Profit (RM) None 0.15 1050 None/little 0.35 4050 None/little/moderate 0.7 23300 All 1 31700 b) Both tables give expected profit as RM31700. The figure is obtained by multiplying the probability of each possibility by the individual profit then adding all these products together. c) The average (mean) of the figures is 31700/4=7925. The dataset is 8400, 19250, 3000, 1050 To obtain the SD we work out the difference of each of these from the mean and square it: 225625, 128255625, 24255625, 47265625 then add these together: 200002500, divide by 4: 50000625 and take the square root: 7071.11 approx. d) The SD gives an effective margin of error on the expected profit: RM7925+7071 which puts the range of the profit between RM854 and RM14996. This does seem too much of a range to be true, so I guess I've made a wrong assumption. To make a better estimate, we need to look at all the rainfall possibilities in more detail. Consider how the meteorologist got his figures. Let's say he measured rainfall over a period of 365 days and discovered the following: No rainfall for 54.75 days Little rainfall for 73 days Moderate rainfall for 127.75 days Heavy rainfall for 109.5 days These periods of time would give the probabilities quoted. Based on these, and using a prorata value for the farmer's profit, we can calculate his profit for these periods: 54.75/365*7000+73/365*15000+127.75/365*55000+109.5/365*28000=31700, as expected from calculations earlier. However, the cumulative probabilities are wrong, because, apart from when the cumulative probability is 1, the time periods for other values needs to be adjusted prorata for the whole year. No rainfall for the whole year produces a profit of RM7000; but little or no rainfall requires adjustment. The time period is 54.75+73=127.75 which is 0.35 year. So the profit has to be extrapolated based on the factor 1/0.35. So we have (54.75/365*7000+73/365*15000)/0.35=RM11571.43. For rainfall ranging from none to moderate, the time period is 0.7 year. The adjustment or extrapolation factor is 1/0.7. This gives us 23300/0.7=RM33285.71. And, of course, all types of rainfall give us RM31700. Working from heavy to no rainfall: RAINFALL P Profit (RM) Heavy 0.3 28000 Heavy/moderate 0.65 42538.46 Heavy/moderate/little 0.85 36058.82 All 1 31700 And from no to heavy rainfall: RAINFALL P Profit (RM) None 0.15 7000 None/little 0.35 11571.43 None/little/moderate 0.7 33285.71 All 1 31700 In both the last two tables we have datasets for which we can calculate mean and SD for 7 rainfall patterns: No rainfall at all; little or no rain; all but heavy rain; a mixture of all types; heavy to moderate; some daily rain; only heavy rain. Using figures from the tables we arrive at a mean of RM27164.92 and a SD of RM12094.61, which gives a range for the profit of RM15070.31 to RM39259.53. (SD is square root of VARIANCE. Both mean and variance are calculated by dividing the sums of the relevant columns by 7, the number of rainfall types.) RAINFALL Profit Profit-mean (Profit-mean)^2 None 7000 -20164.92 406624000 Little or none 11571.43 -15593.49 243156930 No heavy rain 33285.71 6120.79 37464070  Mixed 31700 4535.08 20566951  Heavy or moderate 42538.46  15373.54  236345732 Some daily rain 36058.82  8893.9  79101457  Only heavy rain 28000 835.08  697359  MEAN: 27164.92  VARIANCE: 146279500  
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4 and 7 4 + 7 = 11 4 x 7 = 28
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