Find the smallest number that is divisible by 2, 3, 4, 5, 6 and add 1.
We need the least common multiple of 2, 3, 4, 5, 6.
2 = 2
3 = 3
4 = 2^2
5 = 5
6 = 2 * 3
LCM = product of common and not common prime factors with larger exponent.
LCM = 2^2 * 3 * 5 = 4 * 3 * 5 = 60
To always have a remainder of 1, you need of add 1 to 60.
The number is 61.
Check:
61/2 = 30 remainder 1
61/3 = 20 remainder 1
61/4 = 15 remainder 1
61/5 = 12 remainder 1
61/6 = 10 remainder 1
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I think it best to actually do the problem to determine the correct answer.
Clearing out the fractions, I found that
x(x-2) - (x -1) = 2x -5. This results in x=2 and x=3.
However, 2 is not a solution (but is an extraneous solution), because substituting 2 for x in the 2nd term results in div. by zero.
Thus, only the second answer is correct.
Yes! that would be an isosceles triangle.
13 because absolute is showing the same number but positive