If a number N can be represented as P 1 a - P 2 b - P 3 c where P 1- P 2- P 3 are prime factors of N then number of factors of N can be found as a -1 - b -1 - c -1

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Quantitative Aptitude

Factors of a Number

If a number N...

Question

If a number N can be represented as $P_{1}^{a} \times P_{2}^{b} \times P_{3}^{c}$ where $P_{1}, P_{2}, P_{3}$ are prime factors of N then number of factors of N can be found as $(a + 1) \times (b + 1) \times (c + 1)$

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Solution

Number of factors of x is = (2+1)(3+1) = 12

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Similar questions

Q. Let

n = p_{1}^{α_{1}}, p_{2}^{α_{2}}, p_{3}^{α_{3}} . . . p_{k}^{α_{k}}

where (

p_{1}, p_{2}, . . . p_{k}

are primes and

α_{1}, α_{2}, α_{3} . . ., α_{k} ϵ N)

then find the number of ways in which

n

can be expressed as the product of two factors which are prime to one another.

Let

p

be a prime number &

n

be a positive integer, then exponent of prime

p

n!

is denoted by

E_{p} (n!)

& is given by

E_{p} (n!) = [\frac{n}{p}] + [\frac{n}{p^{2}}] + [\frac{n}{p^{3}}] + . . . . . + [\frac{n}{p^{x}}]

where

x

is the largest positive integer such that

p^{x} \leq n < p^{x + 1}

and

[\cdot]

denotes the greatest integer

Again every natural number

N

can be expressed as the product of its prime factors given by

N = P_{1}^{k_{2}} P_{2}^{k_{2}} . . . . P_{r}^{k_{r}}

where

P_{1}, P_{2}, P_{3}, . . . . . . . P_{r}

are prime numbers &

k_{1}

are whole numbers.

The greatest integer

n

for which

77!

is divisible by

3^{n}

Q. A number N when factorised can be written

N = a^{4} \times b^{3} \times c^{7}

. Find the number of perfect squares which are factors of N. (The three prime numbers a, b, c > 2)

Q. If

n

is a natural number such that

n = p_{1}^{a_{1}} \cdot p_{2}^{a_{2}} \cdot p_{3}^{a_{3}} . . . . p_{k}^{a_{k}}

where

p_{1}, p_{2}, p_{3}, . . ., p_{x}

are distinct prime numbers, then show that

log n > k log 2

Q. If

n

is a natural number such that

n = p_{1}^{a_{1}} . p_{2}^{a_{2}} \dots p_{k}^{a_{k}}

and

p_{1}, p_{2}, p_{3}, \dots, p_{k}

are distrinct prime numbers, then

{log}_{e} n \geq

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If a number N can be represented as Pa1×Pb2×Pc3 where P1,P2,P3 are prime factors of N then number of factors of N can be found as (a+1)×(b+1)×(c+1)

Number of factors of x is = (2+1)(3+1) = 12

If a number N can be represented as $P_{1}^{a} \times P_{2}^{b} \times P_{3}^{c}$ where $P_{1}, P_{2}, P_{3}$ are prime factors of N then number of factors of N can be found as $(a + 1) \times (b + 1) \times (c + 1)$