dina98
Should not it be D? The first statement may work even if X is 5+7. If X = 12, 36 can be a factor of 15!. Am I wrong?
15! = 15*.... * 12 * .. 3 * 2 * 1.
No. Statement 1 is not sufficient. Look below.
if x = 12, then \(3^{12}\)should be a factor of 15!
For maximum power of \(A^n\)in M! , we use the formula: M/A + M/\(A^2\) + M/\(A^3\) .... M/\(A^n\) where \(A^n\)< M
Thus, maximum p9ower of 3 in 15! = 15/3+ 15/3^2 = 5+1 = 6. So in other words, 3,\(3^2\),\(3^3\), \(3^4\), \(3^5\), \(3^6\) are the only powers of 3 that will give you as a factor of 15!. Anything greater than \(3^6\) (\(3^7\) or \(3^{12}\)) will not give you an integer when you calculate 15!/\(3^{12}\).
I believe you are confusing '^' symbol with the symbol '*'.
"^" is 'raised to the power of' and NOT the multiplication symbol. We write multiplication symbol as either 'x' or '*'
3^12 = \(3^{12}\)= 3*3*3*3*3*3*3*3*3*3*3*3 \(\neq\) 36
3*12 = 36
Hope this helps.