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![If p is a prime number greater than 11 and [m]p^2[/m] has a](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "If p is a prime number greater than 11 and [m]p^2[/m] has a"){kind=icon}
22 Apr 2026, 00:51
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
44% (02:19) correct
56%
(02:34)
wrong
based on 18
sessions
History
Date
Time
Result
Not Attempted Yet
If p is a prime number greater than 11 and \(p^2\) has a units digit of 1, what is the greatest integer that must divide \(p^2 - 1\)?
A) 30
B) 40
C) 96
D) 120
E) 240
A) 30
B) 40
C) 96
D) 120
E) 240
Take an example like 19^2 as p^2.
p^2-1 = 360.
Check with options that greatest factor.
120 is our answer.
For mathematical approach:
120 = 2^3*3*5.
p^2-1 = (p+1)(p-1)
So basically ending with 1 means, the prime number ends in 1 itself or 9.
So it is basically x(x+2) form where the number x either in 8 or 0.
Now any number you take of that form will have atleast 3 2s as factor of its product, along with 3 and 5.
How are we sure it is divisble by 3?
Any multiple of 3 you take it alternates between odd or even for every successive integer multiplier.
Considering that,
When you have a prime number, the number is odd.
Odd number - 3 should give you its multiple. BUT this odd number is not even divisble by 3!
So x-2 or x-1 always is always a multiple of 3 where x is a prime number.
Then either x-2+3 = x+1 is a multiple of 3.
x-1 itself could be multiple of 3 too.
Thus (x+1)(x-1) atleast 1 of them will always has a 3 as its factor when x is a prime number.
So we end up having 8,3,5 as its factors. Thus 120.
p^2-1 = 360.
Check with options that greatest factor.
120 is our answer.
For mathematical approach:
120 = 2^3*3*5.
p^2-1 = (p+1)(p-1)
So basically ending with 1 means, the prime number ends in 1 itself or 9.
So it is basically x(x+2) form where the number x either in 8 or 0.
Now any number you take of that form will have atleast 3 2s as factor of its product, along with 3 and 5.
How are we sure it is divisble by 3?
Any multiple of 3 you take it alternates between odd or even for every successive integer multiplier.
Considering that,
When you have a prime number, the number is odd.
Odd number - 3 should give you its multiple. BUT this odd number is not even divisble by 3!
So x-2 or x-1 always is always a multiple of 3 where x is a prime number.
Then either x-2+3 = x+1 is a multiple of 3.
x-1 itself could be multiple of 3 too.
Thus (x+1)(x-1) atleast 1 of them will always has a 3 as its factor when x is a prime number.
So we end up having 8,3,5 as its factors. Thus 120.
kevincan
22 Apr 2026, 03:04
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Your mathematical approach is better here, as examples can sometimes be deceiving
