x is a positive integer and has 8 factors while (x+1) has 4 factors.

Question

x is a positive integer and has 8 factors while (x + 1) has 4 factors. If ﻿x < 100﻿ then how many values of x are possible?

A) 0
B) 1
C) 2
D) 3
E) 4

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ManifestDreamMBA · Answer

x has 8 factors
Then it can be expressed as p^7 or p*q^3 or pqr for different prime numbers p, q and r

Looking for p^7 of prime less than 100 : None
Looking for product of primes with cube of a prime less than 100: 8*3 = 24, 8*5 = 40, 8*7 =56, 8*11 = 88, 27*2=54
Looking for product of 3 primes less than 100: 30

x+1 could be 25, 31, 41, 57, 89, 55

x+1 has 4 factors
Then it can be expressed as p^3 or p*q for prime numbers p and q
25 can not be expressed as a cube or product of 2 different primes, out
31 and 41 are prime, out
57 = 3*19
89 is a prime, out
55 = 5*11

We have 2 numbers which satisfy both criteria

notneeded · Answer

can you pls simplify the explanation?

Tishaagarwal13 · Answer

We know x has 8 factors.
So, it can be expressed either as p^7 or p.q^3

Case 1: x = p^7

Taking 2 as an example, 2^7 will be 128.
So, x = 128 and x+1 = 129. 
But this cannot be the answer as it is given that x < 100.

Case 2: x = p.q^3

Let q be 2 and p be 3,5,7

1. x = 3(2^3) = 24. So, x+1 = 25
2. x = 5(2^3) = 40. So, x+1 = 41
3. x = 7(2^3) = 56. So, x+1 = 57
4. x = 11(2^3) = 88. So, x+1 = 89
For 2^3, we cannot take a number beyond this as x will exceed 100.

And, let q be 3 and p be 2
5. x = 2(3^3) = 54. So, x+1 = 55
For 3^3, we cannot take a number beyond this as x will exceed 100.

Now, we know that x+1 should have 4 factors.
From above,
1. 25 has 3 factors - Out
2. 41 has 2 factors - Out
3. 57 has 4 factors (1, 3, 19, 57) - Keep
4. 89 has 2 factors - Out
5. 55 has 4 factors (1, 5, 11, 55) - Keep

So,  the correct answer will be 2 (Option C)

Whoisdmx15 · Answer

Number of factors of a number is given by:  (x+1)*(y+1 ), where the number can be expressed as  a^x*b^y  (where a and b are prime numbers)

For a number to have 8 factors, we ca have  p^7 ,  p^3*q^1

Take  p^7 = 2^7 (More than 100), hence abandon this case

Take  p^3*q^1  = 
 2^3*3 = 24 ,
 2^3*5 = 40 , 
 2^3*7 = 56 ,
 2^3*11= 88 , 
 3^3*2 = 54  (Beyond this any prime number will yield more than 100)

Now checking x+1 needs to have 4 factors,
 25 = 5^2  (Hence 3 factors)
41 = prime hence 2 factors
 57 = 19^1*3^1 = 4 factors
89 = prime hence 2 factors
 55 = 5^1*11^1 = 4  factors

Hence 2 numbers satisfy criterion
 IMO Ans C

haneenk15 · Answer

Hey, how can (pqr)^2 in this case where x has 8 factors? I didn't understand the working of that. (pqr)^1 satisfies the condition, whereas (pqr)^2 has 27 factors.

Secondly, why didn't you work out the 3rd case? Please clarify!

Thanks!

ManifestDreamMBA · Answer

You are right, I just mistyped when I wrote the solution here. Fixed now, thanks!

giuseppemvn · Answer

Very hard. Is there a way to solve this that feels more "rigorous"? there could be at least 10 different x's and the problem becomes very lengthy...

Aarya_San · Answer

This is one of the better explanations. I think you can add pqr as another case (2*3*5) = 30 -> 31 - Out

RaahimAsadShah · Answer

Please can you explain how we came up with case 2 x = p.q*3? Thanks.

Kinshook · Answer

x is a positive integer and has 8 factors while (x + 1) has 4 factors. 
If ﻿x < 100﻿ then how many values of x are possible?

x is of the form:
p^7: None <100
p*q^3: 2×3^3=54; 3*2^3=24; 5*2^3=40; 7*2^3=56; 11*2^3=88;
pqr: 2*3*5=30; 2*3*7=42; 2*3*11=66; 2*3*13=78;

x+1 possibilities:
p^3: None 
pq: 5*11=55; 3*19=57

IMO C

x is a positive integer and has 8 factors while (x+1) has 4 factors.

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x is a positive integer and has 8 factors while (x+1) has 4 factors.

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