It is not a problem
The answer is 128/(2^(n-1)) if the 1st term is 128. The divisor is found by the realization that these are decreasing powers of two.
The factors of 512 are: 1 2 4 8 16 32 64 128 256 512So: 1·2·256 = 5121·4·128 = 5121·8·64 = 5121·16·32 = 5122·4·64 = 5122·8·32 = 5124·8·16 = 512
A = LW 64 = 64 * 1 or 1 * 64 64= 32 * 2 or 2 * 32 64 = 16 * 4 or 4 * 16
if you want to find out 60 binary so i am doing an easy way 128 64 32 16 8 4 2 1 0 1 0 0 0 0 0 0 we will put 0 in 128,32,16,8,4,2 and 1 because we didn't use that numbers and 1 under 64 because we used that number to make 64 binaries and 128 64 32 16 8 4 2 1 these are called ddn numbers we use these numbers to make any binary number
This sequence is called the doubling sequence.
1, 2, 4, 8, 16, 32, 64, 128, -1, -2, -4, -8, -16, -32, -64, -128
1, 2, 4, 8, 16, 32, 64, 128: 1 x 128, 2 x 64, 4 x 32, 8 x 16
1, 2, 4, 8, 16, 32 1, 2, 4, 8, 16, 32, 64, 128 1, 2, 4, 8, 16, 32, 64, 128, 251, 502, 1004, 2008, 4016, 8032, 16064, 32128
1, 2, 4, 8, 16, 32, 64, 128.
The factors of 128 are: 1, 2, 4, 8, 16, 32, 64, 128.
They are: 1, 2, 4, 8, 16, 32, 64, 128, 256.
1 x 128, 2 x 64, 4 x 32, 8 x 16, 16 x 8, 32 x 4, 64 x 2, 128 x 1 = 128
1, 2, 4, 8, 16, 32, 64, 128.
By: 1 2 4 8 16 32 64 and 128
1, 2, 4, 8, 16, 32, 64, 128.
1, 2, 4, 8, 16, 32, 64, 128.
1 2 4 8 16 32 64 128.