Find two consecutive integers such that the sum of 2 times the first integer and 4 times second integer is 46.
543795
x+1=y\n2x+4y=46\n2x+4(x+1)=46\n2x+4x+4=46\n6x=42\nx=7\ny=8
444950
2P = (2,4)\n4P = (4,sqrt(72))
593943
1) The sum of the positive 3-digit odd integers = 101 + 103 + 105 + ... + 995 + 997 + 999.  Adding the first of these to the last, we get 101 + 999 = 1100.  Adding the second of these to the next to last, we get 103 + 997 = 1100.  This pattern keeps repeating until 549 + 551 = 1100.  The question then becomes, how many of these 1100's do we have?  We have one for 101, 103, 105, ..., 549 (the later numbers share their 1100's with these).  How many numbers are in the set 101, 103, 105, ..., 549?  Well, there are 449 numbers in 101, 102, 103, ..., 549.  Since there is one more odd number than even number, we must have 225 odd numbers and 224 even numbers.  Thus, there are 225 odd numbers in the set 101, 103, 105, ..., 549.  Thus, we have 225 1100's to add together.\n\n225 * 1100 = 247500.\n\n\n\n2) This time we wish to add the numbers of the set 10, 11, 13, 14, ..., 94, 95, 97, 98.  Adding the first to the last we get 10 + 98 = 108.  Adding the second to the second to last we get 11 + 97 = 108.  The pattern keeps repeating until we get 53 + 55 = 108.  How many 108's do we have?  Well, we have one for 10, 11, 13, 14, 16, 17, ..., 52, 53 (again, the later numbers share their 108's with these).  How many numbers are in this set?  In the set 10, 11, 12, 13, ..., 52, 53, there are 44 numbers.  The multiples of three range from 12 to 51--from 3*4 to 3*17.  Thus, there are 14 multiples of three in this range.  This means that there are 30 numbers in the set 10, 11, 13, 14, 16, 17, ..., 52, 53.  Thus, we have 30 108's to add together.\n\n30 * 108 = 3240.\n\n\n\n3) Here, I am assuming that the raise of 10% also applies to the previous year's raise and not just to the "principal" of $18000.  Thus, I am assuming a compound raise instead of 5 simple raises.\n\nDuring the first year, you make $18000.  During the second year, you make 18000 * 1.1 = $19800.  During the third year, you make 19800 * 1.1 = $21780.  During the fourth year, you make 21780 * 1.1 = $23958.  Finally, during the fifth year, you make 23958 * 1.1 = $26353.80.\n\nGood luck.
361589
Sum of all numbers 1 to 100:\nS = n[2a1 + (n - 1)d]/2\n= 100[2(1) + (100 - 1)]/2\n= 100(101)/2\n= 5050\n\nSum of all numbers 3 to 99, multiples of 3:\nS = n[2a1 + (n - 1)d]/2\n= 33[2(3) + (33 - 1)3]/2\n= 33(6 + 96)/2\n= 1683\n\nSum of all numbers 5 to 100, multiples of 5:\nS = n[2a1 + (n - 1)d]/2\n= 20[2(5) + (20 - 1)5]/2\n= 20(10 + 95)/2\n= 1050\n\nSum of all numbers 15 to 90, multiples of 15:\nS = n[2a1 + (n - 1)d]/2\n= 6[2(15) + (6 - 1)15]/2\n= 6(30 + 75)/2\n= 315\n\nSum of integers from 1 to 100 which are not divisible by 3 and 5:\nS = sum(1-100) - sum(3-99) - sum(5-100) + sum(15-90)\n= 5050 - 1683 - 1050 + 315\n= 2632\n\nWe have to add in the sum(15-90) because in removing sum(3-99) and sum(5-100), we're taking out multiples of 15 twice.  Adding sum(15-90) ensures that we only take out 15 once, 30 once, etc.
264056
k=12, assuming you meant 4(k+2) - 2(k-1) = 30, otherwise it won't work out.
860633
WTF? - Your question makes no sense.  I suppose you mean TWO consecutive numbers multiplied together...\n\nBut in that case the answer is impossible:\n\nIf the two consecutive numbers are A and A+1\n\nA x A+1 will be greater than A^2 and less than (A+1)^2  i.e. it will always be between two squares.\n\n:-/\n\n.
277593
In your case use your fingers bye the way 2+2 = 4
9961
0,1,1,2,3\n\nto get the next number add the previous 2 numbers
425151
take the last digit of the number and multiply by the earlier digits to get the next answer\n369*2= 738\n73*8=584\n58*4=232\ntherefore next number will be 23*2=46
714620
47, the next number in the pattern is the sum of the last 2 numbers.
291785
To find the numbers you have to solve these 2 equations simultaneously.\nThe sum of two integers is 16 translates to the equation:\nx + y = 16\nThe sum of their squares is 130 translates to the equation:\nx^2 + y^2 = 130\nSolving for x and y gives you 7 and 9.\nAlways check to see if your answer makes sense.\n7 + 9 = 16\n7^2 + 9^2 = 49 + 81 = 130
1063322
three consecutive even integers are\nx, x+2 and x+4\n\nEquation is\n2((x+2)+(x+4)) = 6x - 12\n\nSolving for x\nx = 12\nSo, 12, 14 and 16
853291
4 is the answer. Here's how:\n4 appears after a break of two numbers, so the number that follows 40 must be 4
316069
The formula for the sum of N consecutive numbers starting with number M is, \nM + (M+1) + (M+2) + ... + (M+N-1)  is:\n\n   (2M+N-1)*N\n   ----------\n       2\nfor 50 consecutive numbers we have N=50 and M=1998\n\nthe sum is Sum=(2*1998+50-1)*50/2=4045*25=101125\n\nfor 25 consecutive numbers N=25, Sum=101125, M=?\n(2M+25-1)*25/2=101125\n(2M+24)=8090\n2M=8090-24=8066\nM=4033\nthe smallest number of the 25 consecutive numbers is 4033
480823
if your array is 5 wide by 46 long, then you just assign an incrementing value to each consecutive cell and you will find that the last cell (5,46) = 5 X 46.\nSee below:\n\n1,2,3,4,5,6,....44,45,46 -cell (1,46) = 1 X 46 = 46\n47,48,49,50,....90,91,92 -cell (2,46) = 2 X 46 = 92\n...\n...\n...228,229,230 - cell (5,46) = 5 X 46 = 230
580109
ans is 70\n(5*2+1) (11*2+2) (24*2+3)\nrule is there r two series 5 11 24 51...\n7 16 34 70\n(7*2+2) (16*2+2) (34*2+2)
756753
8) The sum of consecutive integers starting with 1 and ending in n is: s=n*(n+1)/2\nThe average of that is s/n\nAnd the effect of removing the mystery number (call it x) from the sum is:\n(s-x)/n=(n+1)/2-(x/n)=31.25\nAs the first term is a multiple of 1/2 in order for this to be true x must be 1/4 or 3/4 of n (it must be less than n remember?).\nTesting 1/4:\n(n+1)/2-1/4=31.25\nn+1=63\nn=62\n\nAs 62 isn't a factor of 4 and we need n to be of the form 4*x (in order for x/n to be 1/4) we can't use this n.\n\nTesting 3/4:\n(n+1)/2-3/4=31.25\n(n+1)=64\nn=63\n\nAs n isn't a factor of 4 there is no solution (n needs to be of the form 4*x/3).\n9) (2^1999)*(5^2001)= (5^2)*(2^1999)*(5^1999)= 25*10^1999\nAs all those 0's aren't going to affect the sum we can remove them. Result: 25\nThe sum of 2 and 5 is 7\n10) x*x + 6x - 91 = (x + 13)(x - 7)\nThe working out is mostly experimentation I'm afraid, but at least the result works. :)\n11) Sorry, no idea how to solve that one.\n12) I'm not sure I understand the question but if it's what I think it is then I can't solve it through non-experimental means. Try looking up combinatorics on wikipedia.