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shameekv1989
17 Apr 2020, 16:27
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C
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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
A) 0
B) 1
C) 2
D) 3
E) 4
17 Apr 2020, 21:48
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If a number less than 100 has 8 factors then number is of the form x= a^3×b where a and b are the prime numbers.
1. 2^3×3 =24
2. 2^3×5 =40
3. 2^3×7 =56
4. 2^3×11 = 88
5. 3^3×2 =54
Out of this 57 and 55 also has four factors.
There are two values of x possible that is 56 and 54
Option C is the answer
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1. 2^3×3 =24
2. 2^3×5 =40
3. 2^3×7 =56
4. 2^3×11 = 88
5. 3^3×2 =54
Out of this 57 and 55 also has four factors.
There are two values of x possible that is 56 and 54
Option C is the answer
Posted from my mobile device
17 Apr 2020, 22:01
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shameekv1989
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
A) 0
B) 1
C) 2
D) 3
E) 4
OK, this has to do with the number of factors of any positive integer.
x is a positive integer and has 8 factors
Now \(8=(7+1)\), so x can be a^7
\(8=2*4=(1+1)(3+1)\), so x can be ab^3
\(8=2*2*2=(1+1)(1+1)(1+1)\), so x can be abc
(x+1) has 4 factors..
So two possibilities..
\(x+1=p^3....and...x+1=pq\)
Ok let us check if there are any x and x+1 which fit into these categories...
(I) \(x=a^7\)
a=1....but then a is not prime.
a=2, will make x>100
So NO possible values
(II) \(x=ab^3\)
a=3 and b=2....ab^3=3*2^3=24....x+1=25=5^2...Cannot be expressed as p^3 or pq..
a=5 and b=2....ab^3=5*2^3=40....x+1=41...Cannot be expressed as p^3 or pq..
a=7 and b=2....ab^3=7*2^3=56....x+1=57=3*19=pq...YES
a=11 and b=2....ab^3=11*2^3=88....x+1=89...Cannot be expressed as p^3 or pq..
a=13 and b=2....ab^3=13*2^3=104>100
a=2 and b=3....ab^3=2*3^3=54....x+1=55=5*11=pq..YES
a=5 and b=3....ab^3=5*3^3>100
a=2 and b=5....ab^3=2*5^3=250>100...
Beyond this all combinations of a and b will give x>100, as b^3 will become >100.
So NO possible values
(III) \(x=abc\)
a=2, b=3, c=5...x=2*3*5=30....x+1=31...Cannot be expressed as p^3 or pq..
a=2, b=3, c=7...x=2*3*7=42....x+1=43...Cannot be expressed as p^3 or pq..
a=2, b=3, c=11...x=2*3*11=66....x+1=67...Cannot be expressed as p^3 or pq..
a=2, b=3, c=13...x=2*3*13=78....x+1=79...Cannot be expressed as p^3 or pq..
a=2, b=3, c=17...x=2*3*17=102>100..
a=2, b=5, c=7...x=2*5*7=70....x+1=71...Cannot be expressed as p^3 or pq..
a=2, b=5, c=11...x=2*5*11=110>100..
a=3, b=5, c=7...x=3*5*7=105>100.
No more combinations possible such that x <100..
Total two cases..
x=54 and x=56
C
