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19 May 2022, 08:30
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E
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Difficulty:
45%
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Question Stats:
64% (01:25) correct
36%
(01:38)
wrong
based on 713
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If n and p are different positive prime numbers, which of the integers n^4, p^3, and np has (have) exactly 4 positive divisors?
(A) n^4 only
(B) p^3 only
(C) np only
(D) n^4 and np
(E) p^3 and np
(A) n^4 only
(B) p^3 only
(C) np only
(D) n^4 and np
(E) p^3 and np
23 Aug 2023, 21:23
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If n and p are different positive prime numbers, which of the integers n^4, p^3, and np has (have) exactly 4 positive divisors?
No. of factors of n^4 = (4+1)=5
No. of factors of p^3 = (3+1)=4
No. of factors of np = (1+1)(1+1)=4
Hence E
No. of factors of n^4 = (4+1)=5
No. of factors of p^3 = (3+1)=4
No. of factors of np = (1+1)(1+1)=4
Hence E
General Discussion
19 May 2022, 08:44
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IMO E
Lets assume n is 2 and p is 3
now 2^4= 2*2*2*2*1
List of divisors = 1,2,4,8 and 16
Count = 5
p^3 = 3^3 = 3*3*3 =
List of divisors = 1,3,9, and 27
Count = 4
np = 2*3 =6
List of divisors= 1,2,3, and 6
Count=4
Lets assume n is 2 and p is 3
now 2^4= 2*2*2*2*1
List of divisors = 1,2,4,8 and 16
Count = 5
p^3 = 3^3 = 3*3*3 =
List of divisors = 1,3,9, and 27
Count = 4
np = 2*3 =6
List of divisors= 1,2,3, and 6
Count=4
06 Jul 2026, 23:15
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n and p = different prime num
n^4, p^3 and np, which of has exact 4 divisors.
for n^4, its not possible no matter what value we try. because it will have always 5 divisors.
for examp, 2^4 =26 (4+1=5) so it will have 5 divisors. 1,2,4,8,16. we can try any num and it will give us always 5 divisors.
for p^3, it will always has 4 divisors. because (3+1 = 4).
for example, 3^3 = 27. divisors = 1,3,9,27. any num will get to this same 4 divisors.
for np, n^a * p^b = (a+1) (b+1). here a and b will be 1 for each.
(1+1)(1+1) = 4
so np will always has 4 divisors.
so choice E
n^4, p^3 and np, which of has exact 4 divisors.
for n^4, its not possible no matter what value we try. because it will have always 5 divisors.
for examp, 2^4 =26 (4+1=5) so it will have 5 divisors. 1,2,4,8,16. we can try any num and it will give us always 5 divisors.
for p^3, it will always has 4 divisors. because (3+1 = 4).
for example, 3^3 = 27. divisors = 1,3,9,27. any num will get to this same 4 divisors.
for np, n^a * p^b = (a+1) (b+1). here a and b will be 1 for each.
(1+1)(1+1) = 4
so np will always has 4 divisors.
so choice E
Bunuel
