Q) A prime number greater than 3 is divided by 6. Which of the following can be the remainder of the mentioned division?
a) 1
b)2
c)4
d)5
e)both a and d
soln : To solve this problem all you need to know is a prime number greater than 3 can be expressed as 6k+1 or 6k-1. 6k+1 will give remainder 1 and 6k-1 will give 5 as a remainder. Hence E
Factors:
I asked my friend today to list all factors of 100 and this is what he did
1,2,4,5,10,20,25,50,100. This is how an untrained mind calculates factor. One important thing to note in factors is for every single factor you have one more corresponding factor known as factor pair. A trained mind should list factors in the form of factor pair. A factor pair will appear something like this
| 1 | 100 |
| 2 | 50 |
| 4 | 25 |
| 5 | 20 |
| 10 | 10 |
If you multiply a factor by one more factor you should get the number itself. In the above example if you multiply 1 with 100, 2 with 50,…10 with 10 you get 100 in all cases. This is how a trained mind should list factors and should not waste time in finding each and every factor. Just remember that for every factor there is one more factor which if multiplied with will lead to number itself.
Easy way of finding factors:
You guys must be aware of the formula to calculate number of factors of a number. Just to confirm we are learning same maths :D here is the formula. A number N of the form a^p*b^q*C^r will have (p+1)(q+1)(r+1). Hence a number 140 = 2^2*5^1*7^1 will have (2+1)(1+1)(1+1) = 12 factors.
How does this formula work? What if I forget the formula? Did I actually understood the concept behind this?
Well if you are not sure to any of the above questions than read on…
Lets take the same number 140 = 2^2*5^1*7^1. I can represent the factors of 140 in the below presented way
| | 2^0 | 2^1 | 2^2 |
| 5^0 | 1 | 2 | 4 |
| 5^1 | 5 | 10 | 20 |
| 7 | 7 | 14 | 28 |
| 7 | 35 | 70 | 140 |
From the above table we can easily see all the 12 factors. How did I prepared this table?
If the number consists of 3 prime numbers than take the least two prime numbers and represent in the above format. For example: I have represented 2^2 as 2^0 2^1 and 2^2 and take the second least prime number and represent it as the way I did for 5. Now 2^0*5^0=1, 2^1*5^0=2 ….it goes on. So you updated the data for first two prime numbers and you have got two rows of data. Now for the 3rd prime number in this case being 7 just multiply 7 with the values you have obtained in first and 2nd row,
Try Doing the same for the number 150.
If N^2 has got k factors less than N than it will have k factors more than N. Hence the total factors will be 2k+1. We add one coz N will also be a factor of N^2
Example: N^2 = 36. Hence N =6. Number of factors of 36 less than 6 is 1,2,3,4. K=4 hence total number of factors is 2*4+1=9
Question:
If N^2 has got 3 factors less than N than N^4 will have how many factors?
a)7
b)8
c)12
d)13
e)14
Soln: N^2 has got 7 factors hence the number is of the form a^6.
N^4 will be square of a^6 hence a^12. a^12 will have (12+1) 13 factors
Let me know if you have any questions
Thanks,
Quant-Master
Posts
2 things
ReplyDelete1) 150 has 12 factors? I used 2^0,2^1,5^0,5^1,5^2 and 3 in the end to make the table. is that okay?
2) how does a number that has 7 factors take the form a^6?
thanks.
Response to Question 1: that's perfect.
ReplyDeleteResponse to second question: x^y, where x is prime, will have y+1 factors. Hence a^6 will have 7 factors - a^0,a^1,a^2,a^3,a^4,a^5,a^6.
In general a number of the form (a^x)*(b^y)*(c^z)..., where a, b and c are prime, will have (x+1)(y+1)(z+1) factors.
For example 210 = 2^1*3^1*5^1*7^1. Hence 210 will have (1+1)(1+1)(1+1)(1+1) = 16 factors.
Let me know if you have any queries
Thanks,
Quant-Master