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if i had 1,3,5,7,11,13,15 how can i get total 30 if i use 5 digits from the group

(1,3,5,7,11,13,15) to get total 30

Research, Knowledge and Information :


Mathematical Puzzles: What is () + () + () = 30 using 1,3,5,7 ...


This allows for the solution 11,3 + 15,7 + 3 = 30. Only use numbers ... Is it possible to make 10000 out of the digits 0, 1, 2, 3, 4, 5, 6, 7 ... (1,3,5,7,9,11,13,15 ...
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Question: "If I had a group of odd numbers 1,3,5,7,9,11,13,15 ...


... 1,3,5,7,9,11,13,15 how can I get total sum ... Question: "if i had a group of odd numbers 1,3,5,7 ... 1,3,5,7,9,11,13,15 how can I get total sum = 30 if I ...
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4 Ways to Add 5 Consecutive Numbers Quickly - wikiHow


How to Add 5 Consecutive Numbers Quickly. Bet someone you can add five consecutive integers faster than they can. ... Ex. 11, 12, 13, 14, 15; 11 x 5 = 55; 55 + 10 = 65;
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Techniques for Adding the Numbers 1 to 100 – BetterExplained


296 Comments on "Techniques for Adding the Numbers 1 to 100" ... /2 = total (5(1+6))/2 = 15. 1 3 5 7 9 = 25 (5 ... 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15.
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Count On


In any day before noon how many times is the sum of the digits of the time equal to 15? ... and 1 so they are 2, 3, 5, 7, 11, 13 ... 30, 31, 30... B) 7, 8, 5, 5, 3, 4 ...
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Math Tricks for All Ages - University of Massachusetts ...


Math Tricks for All Ages ... 1: 15: 14: 4: 12: 6: 7: 9: 8: 10: 11: 5: 13: 3: 2: 16: ... Number everyone in the group from 1 to however many there are. Get a piece of ...
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Types of Numbers - Math Goodies


Types of Numbers, Part I Compiled by ... Primes of two or more digits can only contain the digits 1, 3, 7 because, ... 1,2,3,4,5,7,8,9,10,11,13,14,15,16,17,19,21,22 ...
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Homework Practice and Problem-Solving Practice Workbook


Homework Practice and Problem-Solving ... 2 3 4 5 6 7 8 9 10 079 17 16 15 14 13 12 11 10 09 ... 5. Which two numbers use the digits 3 and 1?
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Suggested Questions And Answer :


how many ways are there to add and get the sum of 180

There are an infinite number of ways to get 180 from two numbers, if we count decimals and fractions as well as other real numbers; but if we are limited to positive integers greater than zero and just the sum of two of them, we are limited to x and 180-x. If we also exclude 90+90 because the numbers are the same, then we have 1 to 89 combined with 179 to 91, which is 89 pairs. Moving on to the sum of three different numbers, let's make 1 plus another two different numbers adding up to 179. So we have 2+177, 3+176, ..., 87+92, 88+91, 89+90, which is 88 groups combined with 1. Move on to 2 plus another two different numbers adding up to 178: 3+175, ..., 87+91, 88+90, which is 86 groups. Then we move on to 3 plus 177: 4+173, ..., 86+91, 87+90, 88+89, 85 groups. And so on, with reducing numbers, until we get to 59, 60 and 61. Let's divide the numbers into two groups A and B. In A we start with 1 and in B we put 2 and (180-A-B)=177 as a pair (2,177). Then we put the next pair in group B: (3,176), then (4,175) and keep going till we have used up all the numbers, ending up with (88,90). Then we count how many pairs there are in group B and pair it up with the number in group A, so we start with (1,88) which covers all the combinations of numbers in group B. Now we move to 2 in group A, put all the pairs adding up to 178 in group B, and finally put the count of these pairs with 2 in group A: (2,86). We then move on to 3, and so on, putting in the counts to make up the number pair in group A. When we've finished by putting the last count in group A, which is (59,1), we can forget about group B and look at the pattern in group A. What we see is this: (1,88), (2,86), (3,85), (4,83), (5,82), (6,80), (7,79), ... See how the counts come in pairs with a gap? All the multiples of 3 are missing in the counts sequence (e.g., 87, 84, 81). We find there are 29 pairs and one odd count, 88, which is unpaired. Number the pairs 0 to 28 and refer to the pair number as N. Add the counts in the pairs together so we start with pair 0 as 86+85=171, pair 1 as 165, pair 2 as 159, and so on. The sequence 171, 165, 159, ..., 3 is an arithmetic sequence with a start of 171 and a difference of 6 between each term in the sequence. [Note also that the terms in the series are all multiples of 3: 3*57, 3*55, 3*55, ...] The rule for the Nth term is 171-6N. When N=0 we have the first term 171 and when N=28 the last term is 3. There is one more term at the end which is unpaired made up of the numbers 59, 60 and 61. We can combine this with the unpaired (1,88). We can find the sum of the terms in the series, which will tell us how many ways there are of adding three different integers so that their sum is 180 (like the sum of the angles of a triangle).  To find the sum of the terms of the series we note that there are 29 terms (0 to 28) and they all contain 171, so that's 171*29=4959. We also have to subtract 6(0+1+2+3+...+28)=6*28*29/2=2436. So 4959-2436=2523. [The sum of the series is also 3(57+55+53+...+5+3+1)=2523.] To this we add the "odd couple" 88+1=89 and 2523+89=2612. Add also the 89 which is the number of pairs of integers adding up to 180 we calculated at the beginning. The total so far is 2612+89=2701 ways of adding 2 or 3 positive integers so that their sum is 180. If you want to go further, please feel free to do so!
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area of a region between two curves?

The picture below shows the required area between the two curves: y=e^(-x^2) (red) and y=1-cos(x) (blue). The vertical blue line shows the limit of the first quadrant (x=(pi)/2). Each grid square in the picture has an area of 0.04 and the area is roughly 14 complete squares: 14*0.04=0.56 is a rough estimate of the area between the curves. The points where the curves cross is the upper limit of the enclosed area which includes the y axis. I assume that the area beyond that up to x=(pi)/2 is not required because it is not enclosed by the y axis. The intersection is when x=0.941944 approx., that is, when e^(-x^2)=1-cos(x) (x in radians). At this point y=0.411783 approx. These define the upper limit for definite integration. Using S[low,high](...) to denote integration, we integrate to get the area between the two curves: S[0,0.941944]((e^(-x^2)-1+cos(x))dx). As far as I know, there's no solution to the indefinite integral of e^(-x^2), but the above definite integral can be evaluated by approximating dx. If dx=0.01, for example, then we can sum the areas of thin rectangles of width 0.01 over the range. This is tedious but it will yield a reasonable approximation. Let f(x)=e^(-x^2)-1+cos(x). Consider a starting point represented by a rectangle of width 0.01 and height f(0). Its area is 1*0.01=0.01. Now consider a different rectangle with the same width but height f(0.01) and area=0.0099985. The two rectangles "trap" the curve between them, so the true area under the curve between the limits 0 and 0.01 is somewhere between the two rectangular areas. If we take the average of these two areas we get 0.00999925. Then we move to f(0.01) and f(0.02). The two areas this time are 0.01f(0.01) and 0.01f(0.02) and the average area is 0.005(f(0.01)+f(0.02)). A third pair of rectangles will be averaged at 0.005(f(0.02)+f(0.03)). The last pair of rectangles will be narrower than 0.01: the first has a height of f(0.94) and a width of 0.001944, the second has a height of f(0.941944) which approximates to zero (to the nearest 10^-7). So the last average area is 0.000972f(0.94). The total area between the curves and the y axis can be expressed as a series: 0.005(f(0)+f(0.01) + f(0.01)+f(0.02) + ... + f(0.92)+f(0.93) + f(0.93)+f(0.94)) + 0.000972f(0.94). As you can see, f(0.01) to f(0.93) are all repeated, so we have 0.005(f(0)+f(0.94)) + 0.000972f(0.94) + 0.01(f(0.01)+...+f(0.93)). Since f(0)=1 and f(0.94)=).00308040 approx. So the first expression is 0.005*1.00308040=0.0050154020. 0.000972f(0.94)=0.000972*0.00308040=0.000003 approx.  To make matters more manageable, I'll divide the range of values into summed groups: 0.01-0.10, 0.11-0.20, 0.21-0.30, etc., up to 0.81-0.90, then 0.91-0.94. Here are the group results (these figures have a superfluous accuracy and will be rounded off later): 0.01-0.10:  9.946463769  0.11-0.20:  9.630991473 0.21-0.30:  9.036361176 0.31-0.40:  8.183088444 0.41-0.50:  7.104969671 0.51-0.60:  5.842140126 0.61-0.70:  4.437880465 0.71-0.80:  2.935475192 0.81-0.90:  1.375462980 0.91-0.93:  0.104328514 TOTAL: 58.59716181 AREA SUBTOTAL: 0.5859716181 TOTAL AREA: 0.585972+0.005018=0.591 approx. Control summation: 5.415713508. Control area subtotal: 0.5415713508 (based on increments of 0.1 across the range 0.1 to 0.9. We would expect this figure to be approximately the same as the more accurate summation.) It appears that the area between the curves and the y axis is of the order 0.591; this figure is close to the rough estimate given near the beginning of this answer.  
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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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using digits 0-9 only once, what combination of 3 digit numbers add to 1000, one digit will be spare

You have to have 3 sets of 3 digits each.  The first set totals 10 (for the ones place).  The second and third sets total 9 (for the tens and hundreds place).  The 1 carries over from the total of 10 in the ones place, moving the 9 up to 10 in the tens place and the 9 up to 10 in the hundreds place. Ways to make 10: 019, 028, 037, 046, 127, 136, 145, 235 Ways to make 9:  018, 027, 036, 045, 126, 135, 234 If 10 is 019, 9 leaves 234, not enough ways If 10 is 028, 9 leaves 135, not enough ways If 10 is 037, 9 leaves 126, not enough ways If 10 is 046, 9 leaves 135, not enough ways If 10 is 127, 9 leaves 036, 045, enough ways but 036 and 045 share 0 If 10 is 136, 9 leaves 027, 045, enough ways but 027 and 045 share 0 If 10 is 145, 9 leaves 027, 036, enough ways but 027 and 036 share 0 If 10 is 235, 9 leaves 018, not enough ways. We checked all of the possible combinations of 3 digits totaling 10 and it looks like none of them leave any combinations to make two sets of 3 digits totaling 9 each. Answer:  No solution.
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find all reconstuctions of the sum: ABC+DEF+GHI=2014 if all letters are single digits

There is no solution if A to I each uniquely represent a digit 1 to 9, because the 9's remainder of ABC+DEF+GHI (0) cannot equal the 9's remainder of 2014 (=7). Let me explain. The 9's remainder, or digital root (DR), is obtained by adding the digits of a number, adding the digits of the result, and so on till a single digit results. If the result is 9, the DR is zero and it's the result of dividing the number by 9 and noting the remainder only. E.g., 2014 has a DR of 7. When an arithmetic operation is performed, the DR is preserved in the result. So the DRs of individual numbers in a sum give a result whose DR matches. If we add the numbers 1 to 9 we get 45 with a DR of zero, but 2014 has a DR of 7, so no arrangement can add up to 2014. If 2 is replaced by 0 in the set of available digits, the DR becomes 7 (sum of digits drops to 43, which has a DR of 7). 410+735+869=2014 is just one of many results of applying the following method. Look at the number 2014 and consider its construction. The last digit is the result of adding C, F and I. The result of addition can produce 4, 14 or 24, so a carryover may apply when we add the digits in the tens column, B, E and H. When these are added together, we may have a carryover into the hundreds of 0, 1 or 2. These alternative outcomes can be shown as a tree. The tree: 04 >> 11, >> 19: ............21 >> 18; 14 >> 10, >> 19: ............20 >> 18; 24 >> 09, >> 19: ............19 >> 18. The chevrons separate the units (left), tens (middle) and hundreds (right). The carryover digit is the first digit of a pair. For example, 20 means that 2 is the carryover to the next column. Each pair of digits in the units column is C+F+I; B+E+H in the tens; A+D+G in the hundreds. Accompanying the tree is a table of possible digit summations appearing in the tree. Here's the table: {04 (CFI): 013} {09 (BEH): 018 036 045 135} {10 (BEH): 019 037 046 136 145} {11 (BEH): 038 047 137 056 146} {14 (CFI): 059 149 068 158 167 347 086 176 356} {18 (ADG): 189 369 459 378 468 567} {19 (BEH/ADG): 379 469 478 568} {20 (BEH): 389 479 569 578} {21 (BEH): 489 579 678} {24 (CFI): 789} METHOD: We use trial and error to find suitable digits. Start with units and sum of C, F and I, which can add up to 4, 14 or 24. The table says we can only use 0, 1 and 3 to make 4 with no carryover. The tree says if we go for 04, we must follow with a sum of 11 or 21 in the tens. The table gives all the combinations of digits that sum to 11 or 21. If we go for 11 the tree says we need 19 next so that we get 20 with the carryover to give us the first two digits of 2014. See how it works? Now the fun bit. After picking 013 to start, scan 11 in the table for a trio that doesn't contain 0, 1 or 3. There isn't one, so try 21. We can pick any, because they're all suitable, so try 489. The tree says go for 18 next. Bingo! 567 is there and so we have all the digits: 013489576. We have a result for CFIBEHADG=013489576, so ABCDEFGHI=540781693. There are 27 arrangements of these because we can rotate the units, tens and hundreds independently like the wheels of an arcade jackpot machine. For example: 541+783+690=2014. Every solution leads to 27 arrangements. See how many you can find using the tree and table!
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Factoring by grouping?

Is there a typo here? Maybe it should be 0 = x^3 + 6x^2 -3x - 18 .  This can be solved by grouping like so:   0 = (x^3 + 6x^2) + (-3x - 18 )  This is the grouping step.  I groupred the first two terms, and the last two terms.   0 = x^2(x+6) - 3(x+6)   In this step, I factored out the greatest common factor from each set of parenthesis.   0 =  ( x+6 )(x^2 - 3)  In this step, I factored out the common (x+6) binomial factor from the right side of the equation. (a challenging notion to follow, but it goes something like this:  In  x^2(x+6) - 3(x+6) , think of x^2 as "a", 3 as "b", and (x+6) as "c",  Then x^2(x+6) - 3(x+6) is of the form  ac - bc.  I factored out the "c" to get c(a-b).  Substituting back, using  x^2 as "a", 3 as "b", and (x+6) as "c", get c(a-b) = (x+6)(x^2-3)   Continuing on to solve the equation, if 0 =  ( x+6 )(x^2 - 3) , then 0 = x+6   or    0 = x^2 -3 Solving these equations, get x = -6, x =sqrt(3), x = - sqrt(3)  (remember when taking the square root of each side you must include both the positive and negative results)
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if i had 1,3,5,7,11,13,15 how can i get total 30 if i use 5 digits from the group

3 weeks + 7 days + 11hrs + 13hrs+ 1day =30... or (15-9)+ (13-7) +(11-5) +(9-3)+ (7-1)=30 okay.. :)
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sqrt(3a+10) = sqrt(2a-1) + 2

sqrt(3a+10)=sqrt(2a-1)+2 To remove the radical on the left-hand side of the equation, square both sides of the equation. (~(3a+10))^(2)=(~(2a-1)+2)^(2) Simplify the left-hand side of the equation. 3a+10=(~(2a-1)+2)^(2) Squaring an expression is the same as multiplying the expression by itself 2 times. 3a+10=(~(2a-1)+2)(~(2a-1)+2) Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials.  First, multiply the first two terms in each binomial group.  Next, multiply the outer terms in each group, followed by the inner terms.  Finally, multiply the last two terms in each group. 3a+10=(~(2a-1)*~(2a-1)+~(2a-1)*2+2*~(2a-1)+2*2) Simplify the FOIL expression by multiplying and combining all like terms. 3a+10=(~(2a-1)^(2)+4~(2a-1)+4) Remove the parentheses around the expression ~(2a-1)^(2)+4~(2a-1)+4. 3a+10=~(2a-1)^(2)+4~(2a-1)+4 Raising a square root to the square power results in the expression inside the root. 3a+10=(2a-1)+4~(2a-1)+4 Add 4 to -1 to get 3. 3a+10=2a+3+4~(2a-1) Since a is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation. 2a+3+4~(2a-1)=3a+10 Move all terms not containing ~(2a-1) to the right-hand side of the equation. 4~(2a-1)=-2a-3+3a+10 Simplify the right-hand side of the equation. 4~(2a-1)=a+7 Divide each term in the equation by 4. (4~(2a-1))/(4)=(a)/(4)+(7)/(4) Simplify the left-hand side of the equation by canceling the common terms. ~(2a-1)=(a)/(4)+(7)/(4) To remove the radical on the left-hand side of the equation, square both sides of the equation. (~(2a-1))^(2)=((a)/(4)+(7)/(4))^(2) Simplify the left-hand side of the equation. 2a-1=((a)/(4)+(7)/(4))^(2) Combine the numerators of all expressions that have common denominators. 2a-1=((a+7)/(4))^(2) Expand the exponent of 2 to the inside factor (a+7). 2a-1=((a+7)^(2))/((4)^(2)) Expand the exponent 2 to 4. 2a-1=((a+7)^(2))/(4^(2)) Simplify the exponents of 4^(2). 2a-1=((a+7)^(2))/(16) Multiply each term in the equation by 16. 2a*16-1*16=((a+7)^(2))/(16)*16 Simplify the left-hand side of the equation by multiplying out all the terms. 32a-16=((a+7)^(2))/(16)*16 Simplify the right-hand side of the equation by simplifying each term. 32a-16=(a+7)^(2) Since (a+7)^(2) contains the variable to solve for, move it to the left-hand side of the equation by subtracting (a+7)^(2) from both sides. 32a-16-(a+7)^(2)=0 Squaring an expression is the same as multiplying the expression by itself 2 times. 32a-16-((a+7)(a+7))=0 Multiply -1 by each term inside the parentheses. 32a-16-a^(2)-14a-49=0 Since 32a and -14a are like terms, add -14a to 32a to get 18a. 18a-16-a^(2)-49=0 Subtract 49 from -16 to get -65. 18a-65-a^(2)=0 Move all terms not containing a to the right-hand side of the equation. -a^(2)+18a-65=0 Multiply each term in the equation by -1. a^(2)-18a+65=0 For a polynomial of the form x^(2)+bx+c, find two factors of c (65) that add up to b (-18).  In this problem -5*-13=65 and -5-13=-18, so insert -5 as the right hand term of one factor and -13 as the right-hand term of the other factor. (a-5)(a-13)=0 Set each of the factors of the left-hand side of the equation equal to 0. a-5=0_a-13=0 Since -5 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 5 to both sides. a=5_a-13=0 Set each of the factors of the left-hand side of the equation equal to 0. a=5_a-13=0 Since -13 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 13 to both sides. a=5_a=13 The complete solution is the set of the individual solutions. a=5,13
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How many numbers can be made with 3 beads and 3 columns, and then 4 beads and four columns

Consider the different combinations of 3 beads: 3 together, 3 all separate, and a bead and a pair of beads. Now the columns. With 3 columns we can choose which column to put the 3 beads together in; that gives us 3 different numbers (003, 030, 300). Separate all the beads, one per column, that's another number (111). That leaves us with a single bead and a pair, i.e., the digits 0, 1 and 2. There are 6 ways of permuting three different objects. Add this to the other 4 and we get 10, so there are 10 numbers possible using three beads and three columns. With 4 beads, all kept together and 4 columns, there are 4 numbers. Separate the beads and put one in each column and we get one more number. That leaves us with two pairs, and a single bead and 3 together. The numbers made of two pairs consist of the digits 0, 0, 2 and 2. If this had been four different digits we would have 24 different permutations, but we only have two different digits, so we need to reduce the permutations by a factor of 2 twice, so 24/4=6. Finally we have the digits 0, 0, 1 and 3. If this had been just 0, 1 and 3, i.e., three different digits it would be similar to the 3 bead problem and there would be 6 ways of permuting these digits. The extra 0 can be at the beginning or end of each of these 6 permutations making 12 altogether. Another way of looking at it is to reduce the 24 permutations of 4 different objects by a factor of 2 to compensate for two digits (0) being the same. So to summarise by adding together all these in order we have 4 (all beads together) + 1 (all beads separate) + 6 (two pairs) + 12 (single bead and 3 beads together) = 23 different numbers.
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seven times a two digit number

Short answer:  The two digit number is 36. Long answer: 7 * x1x2 = 4 * x2x1 "If the difference between the number is 3. . ." Last digits: x2: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 7*x2:  0, 7, 4, 1, 8, 5, 2, 9, 6, 1 x1:  0, 1, 2, 3, 4, 5, 6, 7, 8, 9 4*x1:  0, 4, 8, 2, 6, 0, 4, 8, 2, 6 The only way this works is when these last digits for 7*x2 and 4*x1 are the same.  That means the 7*x2 line can only be: 7*x2:  0, 4, 8, 2, 6 Which means the possible values for x2 are: x2:  0, 2, 4, 6, 8 Now let's look at x1.  Right now the possible values for x1 are: x1:  0, 1, 2, 3, 4, 5, 6, 7, 8, 9 But we have to end up with a choice from x1 and x2 having a difference of 3 (odd number).  There is no way to get an odd number by subtracting an even number from an even number.  That means x1 has to be odd.  Our possible values for x1 are now: x1:  1, 3, 5, 7, 9 And our possible values for x2 are: x2:  0, 2, 4, 6, 8 The possible combinations for x1x2 and x2x1 are: 10, 01 12, 21 14, 41 16, 61 18, 81 30, 03 32, 23 34, 43 36, 63 38, 83 50, 05 52, 25 54, 45 56, 65 58, 85 70, 07 72, 27 74, 47 76, 67 78, 87 90, 09 92, 29 94, 49 96, 69 98, 89 We want 7*x1x2 = 4*x2x1, so we can do 7 * the first column and 4 * the second column: 70, 4 84, 84 98, 164 112, 244 126, 324 210, 12 224, 92 238, 172 252, 252 266, 332 350, 20 364, 100 378, 180 392, 260 406, 340 490, 28 504, 108 518, 188 532, 268 546, 348 630, 36 644, 116 658, 196 672, 276 686, 356 But since we want 7*x1x2 to equal 4*x2x1, that list reduces to: 84, 84 252, 252 The corresponding x1 and x2 values are: 12, 21 36, 63 But the difference between x1 and x2 is 3, so we can't use x1 = 1, x2=2.  We have to use x1 = 3, x2 = 6. Answer:  The two digit number is 36. Check:  7 * 36 = 4 * 63 252 = 252 good. 6 - 3 = 3 good.
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