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Updated on: 30 May 2024, 09:44
Originally posted by GMATWhizTeam on 05 Apr 2020, 09:39.
Last edited by Bunuel on 30 May 2024, 09:44, edited 2 times in total.
Last edited by Bunuel on 30 May 2024, 09:44, edited 2 times in total.
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B
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Difficulty:
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49% (01:58) correct
51%
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wrong
based on 172
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If p is a positive integer, what is the remainder when \(p^2 – 1\) is divided by 8?
(1) p is completely divisible by 3
(2) p leaves a remainder when divided by 2.
(1) p is completely divisible by 3
(2) p leaves a remainder when divided by 2.
GMATWhiz Quant Prepathon Lesson 1 Question 2
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05 Apr 2020, 09:59
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p= pos int
(p^2-1)/8 ; remainder=?
(p-1)(p+1)/8
If p= odd, p-1 and p+1 both are even
And since p-1,p,p+1 are consecutive integers, one of p-1 and p+1 must be divisible by 2 and the other must be divisible by 4.
Keeping this is mind lets move to the statements
Statement 1: p is divisible by 3
If p=3, (p-1)(p+1)=8
So remainder=0
But if p=6,(p-1)(p+1)=35
Remainder=3
Insufficient
Statement 2: p is odd
Aa discussed above if p is odd, (p-1)(p+1) will always be divisible by 8
Sufficient
B
Posted from my mobile device
(p^2-1)/8 ; remainder=?
(p-1)(p+1)/8
If p= odd, p-1 and p+1 both are even
And since p-1,p,p+1 are consecutive integers, one of p-1 and p+1 must be divisible by 2 and the other must be divisible by 4.
Keeping this is mind lets move to the statements
Statement 1: p is divisible by 3
If p=3, (p-1)(p+1)=8
So remainder=0
But if p=6,(p-1)(p+1)=35
Remainder=3
Insufficient
Statement 2: p is odd
Aa discussed above if p is odd, (p-1)(p+1) will always be divisible by 8
Sufficient
B
Posted from my mobile device
05 Apr 2020, 10:43
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If p is a positive integer, what is the remainder when p^2–1 is divided by 8?
(1) p is completely divisible by 3
if p=3, remainder is 0 BUT if p= 6, remainder is 3 . insufficient
(2) p leaves a remainder when divided by 2.
so, p = 2k+1 (k ≥0)
now, (p^2–1)/8 =((2k+1)^2-1 )/8 = (4k^2+4K) /8 = 4k(k+1)/8
k(k+1) always divisible by 2 since for any value of k one of these two , one must be divisible by 2
so, 4k(k+1) is divisible by 8 .
so, remainder = 0 . sufficient
correct answer is B
(1) p is completely divisible by 3
if p=3, remainder is 0 BUT if p= 6, remainder is 3 . insufficient
(2) p leaves a remainder when divided by 2.
so, p = 2k+1 (k ≥0)
now, (p^2–1)/8 =((2k+1)^2-1 )/8 = (4k^2+4K) /8 = 4k(k+1)/8
k(k+1) always divisible by 2 since for any value of k one of these two , one must be divisible by 2
so, 4k(k+1) is divisible by 8 .
so, remainder = 0 . sufficient
correct answer is B
