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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!

## Research, Knowledge and Information :

### Whole Numbers - Math League

Whole Numbers and Their Basic ... A whole number is divisible by 3 if the sum of all its digits is ... the last two digits form one of the numbers 0, 20, 40, 60, ...

### Finding two numbers given their sum and their product

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### Sum-product number - Wikipedia

1 is a sum-product number in any ... The following table lists the sum-product numbers in bases up to 40 ... the maximum possible value of the sum of digits of n ...

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The sum of two numbers is 12. One number is x. The other number is _____. ... Vocabulary: number, sum, total, solve Review words as needed. Return to this page.

### 2.2 Factoring Whole Numbers - McGraw Hill Education

2.2 Factoring Whole Numbers ... (GCF) of two numbers 4. ... In writing composite numbers as a product of factors, ...

### Types of Numbers - Math Goodies

Types of Numbers, Part I Compiled by ... Rational numbers include the whole numbers (0 ... Every integer greater that 83,159 is expressible by the sum of two abundant ...

## 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

### 6x²+13xy+11x+36y-5y²-7

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.

### 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.

### 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

### Fine two numbers whose sum is 24 and product is as large as possible?

biggest produkt...square x=24/2=12 x^2=144

### 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

### THE SUM OF TWO WHOLE NUMBERS IS 11. THEIR PRODUCT IS 28. FIND THE TWO NUMBERS.

4 and 7 4 + 7 = 11 4 x 7 = 28