S sree340 New member Joined Apr 11, 2008 Messages 2 Apr 11, 2008 #1 210 - 1 is divisible by: 2 to the power of 10 minus 1 a. 31 b. 5 c. 7 d. 9 e. 13 any trick for getting solution
210 - 1 is divisible by: 2 to the power of 10 minus 1 a. 31 b. 5 c. 7 d. 9 e. 13 any trick for getting solution
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Apr 11, 2008 #2 Hello, sree340! \(\displaystyle 2^{10} - 1\,\text{ is divisible by: }\;(a)\;\;31\qquad(b)\;\;5\qquad(c)\;\;7\qquad(d)\;\;9\qquad(e)\;\;13\) Click to expand... \(\displaystyle \text{Factor: }\;2^{10} - 1 \;=\;(2^5-1)(2^5+1) \;=\;(31)(33) \;=\;(31)(3)(11)\)
Hello, sree340! \(\displaystyle 2^{10} - 1\,\text{ is divisible by: }\;(a)\;\;31\qquad(b)\;\;5\qquad(c)\;\;7\qquad(d)\;\;9\qquad(e)\;\;13\) Click to expand... \(\displaystyle \text{Factor: }\;2^{10} - 1 \;=\;(2^5-1)(2^5+1) \;=\;(31)(33) \;=\;(31)(3)(11)\)
