Latest Elitmus Aptitude Question SOLUTION: you are given number N.give 2*N has 28 factor and 3*N has 30 factor.calculate how many factor will be in 6*N..?

35 factors.
N= 2^5 *3^3
2N= 2^6 * 3^3 .. so factors are 7*4=28
3N= 2^5 *3^4... factors are 6*5=30
6N = 2^6*3^4.. factors are 7*5=35

HI Manav.. You are considering only one case as there can be many cases because the Maximum limit of N is not given in question.For such type of questions the Maximum Limit should be given.

It is told in question that 2*N has 28 factors.
28 = 7 * 4 or 14*2 or 7 * 2 * 2
Manav is considering the case of (7 * 4) only.

For this type of questions always start like
N = 2^p * 3^q * 5^r * ......so on(beacuse 2,3,5,7,11,13,17.. are prime factors)

since the maximum limit is not given in question so it is not possible to give a single answer...

For the question how 7*4 = 28 is coming in MANAV's Solution---
Lets for an example u need to represent 48 in prime factorization--
u will write - 48 = 2^4 * 3^1
Now if u want to know how many factors does 48 has.. we need to do [(4+1) * (1+1)]= 10 factors total

IN General N = 2^p * 3^q * 5^r * 7^s.....so on
Total no of factors are = (p+1) * (q+1) * (r+1) * (s+1)......

Thanks....

same as MANAV.this is right answer.

please explain how it become factor 30 or 28,please help me out!

i understand the solution

Hey Manav can u plz help me to get ur solutions .. i ma understing ... plz help me yr

Firstly we know that any no. N can be expressed as N=(p1^a).(p2^b).(p3^c)......
Here p1,p2,p3....are prime numbers, and a, b, c and its respective powers.

The no. of factors of N is given by (a+1)(b+1)(c+1).....
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Given, 2N has 28 factors and 3N has 30 factors.

Therefore for 2N----->(a+1)(b+1)(c+1)..... =28
all possible ways to do that is

1) 28=(1+1)(1+1)(6+1) ,and if so it would mean 2N=(p1^1).(p2^1).(p3^6)
or
2) 28=(1+1)(13+1) would correspond to 2N=(p1^1).(p2^13)
or
3) 28=(3+1)(6+1) would correspond to 2N=(p1^3).(p2^6)

now since 2N is a multiple of 2, hence any one of p1, p2 or p3 has to be 2(prime no.).This is obviously true for all three cases.
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Knowing this we can guess 2N to be the following,

form case 1) 2N=(2^1).(p2^1).(p3^6) or 2N=(p1^1).(p2^1).(2^6)
=>N=(p2^1).(p3^6) =>N=(p1^1).(p2^1).(2^5)

no.of factors=(1+1)(6+1)=14 (1+1)(1+1)(5+1)=24
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form case 2) 2N=(2^1).(p2^13) or 2N=(p1^1).(2^13)
=>N=(p2^13) =>N=(p1^1).(2^12)

no.of factors=(13+1)=14 (1+1)(12+1)=26
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form case 3) 2N=(2^3).(p2^6) or 2N=(p1^3).(2^6)
=>N=(2^2).(p2^6) =>N=(p1^3).(2^5)

no.of factors=(2+1)(6+1)=21 (3+1)(5+1)=24
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similarly for 3N----->(a+1)(b+1)(c+1)..... =30
all possible ways to do that is

1) 30=(1+1)(2+1)(4+1) ,and if so it would mean 2N=(p1^1).(p2^2).(p3^4)
or
2) 30=(4+1)(5+1) would correspond to 2N=(p1^4).(p2^5)
or
3) 30=(2+1)(9+1) would correspond to 2N=(p1^2).(p2^9)
or
4) 30=(1+1)(14+1) would correspond to 2N=(p1^1).(p2^14)

now since 3N is a multiple of 3, hence any one of p1, p2 or p3 has to be 3(prime no.).This is obviously true for all four cases.
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Knowing this we can guess 3N to be the following,

form case 1) 2N=(3^1).(p2^2).(p3^4) or 2N=(p1^1).(3^2).(p3^4) or 2N=(p1^1).(p2^2).(3^4)
=>N=(p2^2).(p3^4) =>N=(p1^1).(3^1).(p3^4) =>N=(p1^1).(p2^2).(3^3)

no.of factors=(2+1)(4+1)=15 (1+1)(1+1)(4+1)=20 (1+1)(2+1)(3+1)=24
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form case 2) 2N=(3^4).(p2^5) or 2N=(p1^4).(3^5)
=>N=(3^3).(p2^5) =>N=(p1^4).(3^4)

no.of factors=(3+1)(5+1)=24 (4+1)(4+1)=25
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form case 3) 2N=(3^2).(p2^9) or 2N=(p1^2).(3^9)
=>N=(3^1).(p2^6) =>N=(p1^2).(3^8)

no.of factors=(1+1)(9+1)=20 (2+1)(8+1)=27
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form case 4) 2N=(3^1).(p2^14) or 2N=(p1^1).(3^14)
=>N=(p2^14) =>N=(p1^1).(3^13)

no.of factors=(14+1)=15 (1+1)(13+1)=28
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from all the above cases we see, N=(p1^3).(2^5) whose sum of factors is 24 and,
N=(3^3).(p2^5) whose sum of factors is also 24

hence, N=(3^3).(2^5)
6N=(3^4).(2^6)
sum of factors of 6N is (4+1)(6+1)=5*7=35.