Jagranjosh.com has come up with important concept i.e. power of 2 to help the aspirants in attempting all the questions speedily. Here, we are providing basic concepts to make the calculation faster than doing from long and traditional ones. Practice and thorough learning of some values can help the candidates in cracking Quantitative Aptitude/Numerical Ability Section in Bank Exam, SSC Exam, Railway and Other Exam.
A useful property for the powers of 2
Powers of 2 are very helpful in calculations. Candidates should memorise powers of 2 upto 12 so that it can be used in the questions.
2^{0} |
1 |
2^{1} |
2 |
2^{2} |
4 |
2^{3} |
8 |
2^{4} |
16 |
2^{5} |
32 |
2^{6} |
64 |
2^{7} |
128 |
2^{8} |
256 |
2^{9} |
512 |
2^{10} |
1024 |
2^{11} |
2048 |
2^{12} |
4096 |
- The sum of powers of 2 from 0 to any number n will be equal to 2^{n+1} – 1.
If a number is written from 1 to N as a sum of one or more of the integers of a given set of integers, then it can easily be done from powers of 2. The set of integers used by us comprise of all the powers of 2 starting form 1 (i.e. 2^{0}) to the largest power of 2 less than or equal to N.
For example: If you want to build all the integers upto 255, the numbers 1, 2, 4, 8, 16, 32, 64, 128 are sufficient as 255=1+2+4+8+16+32+64+128.
- Differently, if we have one weight each of 1, 2, 4, 8, 16, 32, 64 and 128 kg, then all the items would be measured from 1 kg to 255 kg using one or more of the given weights (the weights used only in one pan of the weighing scales).
Example: How much minimum number of weights are required to weigh all possible weights upto 512 Kg (Putting all the weights only in one side of pan)
Solution: 512=2^{9}. Minimum Number of weights required=9+1=10. The weights will be 1,2, 4, 8, 16, 32, 64,128, 256 kg
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