GATE2014-2-49

Question

The number of distinct positive integral factors of $2014$ is _____________

Comments

https://gateoverflow.in/2015/gate2014-2-49

check this

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@Prudhvi Phanindra

please take care of category of questions

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$N = \text{Any Natural Number}$

$N = a^{p}\times b^{q}\times c^{r}$
Where, $a,b$ and $c$ are prime numbers. And $p,q$ and $r$ are natural numbers.
$n = $ Number of factors (divisors)

$ n = (p+1)(q+1)(r+1)$

$S_{n} = $ Sum of factors (divisors)

$S_{n} = \dfrac{(a^{p+1}-1)(b^{q+1}-1(c^{r+1}-1)}{(a-1)(b-1)(c-1)}$

$P_{n} = $ Product of factors (divisors)

$P_{n} = (N)^{\frac{n}{2}},\:\:$ Where $n = $ Number of factors

Reference

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@Lakshman Patel RJIT

can you give the source of those formulas...i want to know how sum and product of factors can be calculated

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@chirudeepnamini

Click on the above comment reference.

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@Lakshman Patel RJIT thankyou:)

Answer 1

First do prime factorization of 2014 - 2¹ x 19¹ x 53¹

Now to get a factor of 2014, we can choose any combination of the prime factors including 0. i.e; 2⁰ and 2¹ are possible and similarly for other prime factors also, there are 2 possibilities. So, total number of positive integral factors

$= 2 \times 2 \times 2 = 8$

(When all the powers of prime factors are 0, we get 1 and when all the powers are maximum, we get the given number.)

Comments

What is done after prime factorization? I did not understand, Is there a difference between prime factors and integral factors?

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Any integral factor will be a multiple of prime factors. So, if we decompose a number into a multiple of prime factors, number of integral factors will be the number of ways in which we can multiply the prime factors.

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How to think in exam 1007= 19*53 ?

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For numbers like 1007, where the factors are not clear, the following strategy should be employed.

Suppose we want to find the factors of 1007, now we know 1007 can be written as x*y (x, y != 1 or 1007) if 1007 has some prime factor. It is not very hard to see that both x and y cannot be greater than $\left \lfloor \sqrt{1007} \right \rfloor$ because if both are greater than $\left \lfloor \sqrt{1007} \right \rfloor$ then x*y > 1007 but x*y = 1007.

So one of them has to be less than $\left \lfloor \sqrt{1007} \right \rfloor$ . Now if we can find which is this no. less than $\left \lfloor \sqrt{1007} \right \rfloor$, we can determine x*y. Here $\left \lfloor \sqrt{1007} \right \rfloor$ = 31. So, we check for divisibility with 31,29,23,19. We find 1007 is divisible by 19 and = 19*53. We know 53 is prime. So 1007 = 19*53.

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Now someone can think is it even possible to factorize 1007 right? So to be sure about this Prime number implies it is of the form 6n +1 (converse need not be true.). As it is not of the form 6n+1 proceed with method given by @kumar Ashish.

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Great observation.. reading more about it, every prime other than 2 and 3 are of the form 6n+1 or 6n-1

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For deducing factors of 1007, arbitrary see multiples of 20 with 40 or 50 like this and confirm the range. Like what i did before seeing the proper solution - I multiplied 20*50  - its 1000 so the numbers must be around it. Now see what can be multiplied to get 7 as last digit (coz if it was some even number like 1008 or 1400 you would not have had any difficulty)  so obviously its either 1 and 7 or 3 and 9 Now with 3 and 9 you can't form 13*19 its way less , so just 53*19.

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But $1007=6*168 - 1 (6n-1)$ and is not a prime,so dont use the $6n+1$ or $6n-1$ method to judge primality.It is better to divide by factors $<\sqrt{1007}$.

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if power of 2 was like 2²  then we could have three combinations as 2^0 2^1 and 2^2. Right?

Answer 2

2014 = 2 x 19 x 53

total  positive integral factor=(2x2x2)=8

Answer 3

the factors are 1, 2, 19, 38, 53, 106, 1007 and 2014. So the total makes it 8 number of integral factors.

Comments

Integral factor is the no way how can you multiply prime factor @Regina here 107 is not true it will 1007 please correct it. @Arjun sir explained it in very good way please check this.

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@ Regina Phalange

is there any general formula for this .. ?

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@rishu_darkshadow

No. It is simple prime factorization

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okzz..

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good

Answer 4

answer - 8