41/44 and 1 over 4
1
5-4 = 1 8-5 = 3 = 1*3 17-8 = 9 = 3*3 44-17= 27 = 9*3 let the number be x difference between 44 and x = 27*3 = 81 x = 44 + 81 = 125
Get the last two digits of the number. If that number is divisible by 4 then that number is divisible by 4. Example : 2393844 The last two digits are 44 and since 44/4 = 11 or 44 is divisible by 4 then the number 2,393,844 is divisible by 4.
4 and 4/10
44 = 256
125 (each number is increased by the exponential 3n from the previous: +1, +3, +9, +27, +81)
The sequence continues: 125, 368, 1097, 3284, 9845, 29528, 88577, 265724, 797165, ...
57
41/44 and 1 over 4
Let the numbers be x and y:- If: x+3y = -4 and 3x+y = 44 Then: 3x+9y = -12 and 3x+y = 44 Subtracting 3x+9y = -12 from 3x+y = 44: -8y = 56 => y = -7 By substitution: x = 17 and y = -7 Check: 17+3(-7) = -4 Check: 3(17) -7 = 44 Therefore the two numbers are: 17 and -7
1
Let the numbers be x and y:- If: x+3y = -4 and 3x+y = 44 Solving the above simultaneous equation by elimination method: x = 17 and y = -7 Check: 17+(-21) = -4 Check: 51+(-7) = 44
5 - 4 = 1 = 3^0 8 - 5 = 3 = 3^1 17 - 8 = 9 = 3^2 44-17 = 27 = 3^3 X - 44 = 3^4 = 81 X = 125
5-4 = 1 8-5 = 3 = 1*3 17-8 = 9 = 3*3 44-17= 27 = 9*3 let the number be x difference between 44 and x = 27*3 = 81 x = 44 + 81 = 125
4 < 44 so 4/44 is a proper fraction, not a mixed number.
The answer depends on where the missing number is meant to be.It could be any one of:34, 14, 4, 7, 17, 1914, 10, 4, 7, 17, 1914, 4, 1, 7, 17, 1914, 4, 7, 13, 17, 1914, 4, 7, 17, 25, 1914, 4, 7, 17, 19, 11All of the above are based on polynomials of order 4. There are also infinitely many non-polynomials functions that can generate the sequence of numbers.