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If p, p + 2, p + 6 and p + 8 are all prime numbers greater than 100, t 24 Oct 2021, 10:59
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25% (02:45) correct 75% (02:32) wrong based on 81 sessions

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If p, p + 2, p + 6 and p + 8 are all prime numbers greater than 100, then (p + 8)\(^2\) −(p + 6)\(^2\) + (p + 2)\(^2\) −p\(^2 \)must be divisible by which of the following?

I) 8
II) 48
III) 120

A) none

B) I only

C) I and II

D) I and III

E) I, II and III
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D
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If p, p + 2, p + 6 and p + 8 are all prime numbers greater than 100, t 24 Oct 2021, 20:32
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nislam
If p, p + 2, p + 6 and p + 8 are all prime numbers greater than 100, then (p + 8)\(^2\) −(p + 6)\(^2\) + (p + 2)\(^2\) −p\(^2 \)must be divisible by which of the following?

I) 8
II) 48
III) 120

A) none

B) I only

C) I and II

D) I and III

E) I, II and III
Show more


Are we supposed to know prime numbers greater than 100 in GMAT…..May not be.

Let us solve it other wise.
(p + 8)\(^2\) −(p + 6)\(^2\) + (p + 2)\(^2\) −p\(^2 \)
\((p+8-(p+6))(p+8+p+6)+(p+2-p)(p+2+p)=2(2p+14)+2(2p+2)\)
\(4p+28+4p+4=8p+32=8(p+4)\)

So the expression is a multiple of 8 and p+4
Now, if p, p+2, p+6 and p+8 are prime numbers, the numbers have to be xy1, xy3, xy7 and xy9.
Thus, p+4 will be xy5 or in other words p+4 is a multiple of 5.

Also three consecutive odd numbers will surely have a multiple of 3.
Since p, p+2, p+6 and p+8 are prime numbers, they are NOT multiples of 3. This means p+4 is a multiple of 3.

Hence, 8(p+4) is surely a multiple of 8, 5 and 3 or 8*5*3=120.
Thus 8 and 120 are correct.

What about 48?
p+4 is odd, so 8(p+4) is a multiple of 8 but not if 16, while 48=16*3.
Hence, 48 is not a divisor.

D
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If p, p + 2, p + 6 and p + 8 are all prime numbers greater than 100, t 24 Oct 2021, 20:57
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chetan2u

So the expression is a multiple of 9 and p+4
Now, if p, p+2, p+6 and p+8 are prime numbers, the numbers have to be xy1, xy3, xy7 and xy9.
Thus, p+4 will be xy5 or in other words, p+4 is a multiple of 5.

Also three consecutive odd numbers will surely have a multiple of 3.
Since p, p+2, p+6 and p+8 are prime numbers, they are NOT multiples of 3. This means p+4 is a multiple of 3.
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Sir, I didn't quite understand this part. How 8(p+4) is a multiple of 9 and it's shortened down to a multiple of 5&3 too?

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If p, p + 2, p + 6 and p + 8 are all prime numbers greater than 100, t 24 Oct 2021, 22:11
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yashikaaggarwal
chetan2u

So the expression is a multiple of 9 and p+4
Now, if p, p+2, p+6 and p+8 are prime numbers, the numbers have to be xy1, xy3, xy7 and xy9.
Thus, p+4 will be xy5 or in other words, p+4 is a multiple of 5.

Also three consecutive odd numbers will surely have a multiple of 3.
Since p, p+2, p+6 and p+8 are prime numbers, they are NOT multiples of 3. This means p+4 is a multiple of 3.
Sir, I didn't quite understand this part. How 8(p+4) is a multiple of 9 and it's shortened down to a multiple of 5&3 too?

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Yashika, 8(p+4) is a multiple of 8 and p+4. Now p+4 is also a multiple of at least 3 and 5. So 8(p+4) is a multiple of 8, 5 and 3.

That was a typo. 9 was meant to be 8. Edited.
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If p, p + 2, p + 6 and p + 8 are all prime numbers greater than 100, t 25 Oct 2021, 01:33
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Tough question!!

(p + 8)^2 −(p + 6)^2 + (p + 2)^2 −p^2 = 8p + 32
= 8(p+4)

Let the numbers be

p (P), p+1 (NP), p+2(P), p+3(NP), p+4(NP), p+5(NP), p+6(P), p+7(NP), p+8(P)

where P - Prime, NP - Non Prime

From the above we can safely say that
p+1, p+3, p+5, p+7 are even numbers (alternate even numbers) and
p+4 is an odd number (non-prime and non-even number)

Now p+4 is divisible by 3 because
1. We have a multiple of 3 every 3 digits and p+1 (even), p+2 (Prime) and p+3 (even) are not multiple of 3

Similarly p+4 is also divisible by 5 because
1. We have a multiple of 5 every 5 digits and p(Prime) p+1 (even), p+2 (Prime), p+3 (even) are consecutive numbers and also not multiple of 5. Hence p-1 and p+4 have to be multiples of 5,

From these statement we can see that p+4 is a multiple of both 5 and 3, effectively a multiple of 15.

Hence 8(p+4) would be a multiple of 120.

For 48 we have to test if p + 4 is divisible by 6. However, we know that p+4 is an odd number and if it has to be a multiple of 6, it needs to have 2 as a factor. Since p+4 is an odd number, 2 cannot be a factor, therefore 48 cannot be a divisor for 8(p+4)

Hence, Option D should be the answer.
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If p, p + 2, p + 6 and p + 8 are all prime numbers greater than 100, t 24 Jan 2022, 11:37
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sanjitscorps18
Tough question!!

(p + 8)^2 −(p + 6)^2 + (p + 2)^2 −p^2 = 8p + 32
= 8(p+4)

Let the numbers be

p (P), p+1 (NP), p+2(P), p+3(NP), p+4(NP), p+5(NP), p+6(P), p+7(NP), p+8(P)

where P - Prime, NP - Non Prime

From the above we can safely say that
p+1, p+3, p+5, p+7 are even numbers (alternate even numbers) and
p+4 is an odd number (non-prime and non-even number)

Now p+4 is divisible by 3 because
1. We have a multiple of 3 every 3 digits and p+1 (even), p+2 (Prime) and p+3 (even) are not multiple of 3

Similarly p+4 is also divisible by 5 because
1. We have a multiple of 5 every 5 digits and p(Prime) p+1 (even), p+2 (Prime), p+3 (even) are consecutive numbers and also not multiple of 5. Hence p-1 and p+4 have to be multiples of 5,

From these statement we can see that p+4 is a multiple of both 5 and 3, effectively a multiple of 15.

Hence 8(p+4) would be a multiple of 120.

For 48 we have to test if p + 4 is divisible by 6. However, we know that p+4 is an odd number and if it has to be a multiple of 6, it needs to have 2 as a factor. Since p+4 is an odd number, 2 cannot be a factor, therefore 48 cannot be a divisor for 8(p+4)

Hence, Option D should be the answer.
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I understood why p+4 has to be a multiple of 3. This is because p+1 has to be a multiple of 3. But I am still not clear why p+4 should be a multiple of even 5. Any 5 consecutive integers will throw up one multiple of 5. Amongst p, p+1, p+2, p+3 and p+4: p and p+2 will definitely not be multiples of 5. Can you explain for the remaining numbers?
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If p, p + 2, p + 6 and p + 8 are all prime numbers greater than 100, t 27 Jan 2022, 08:15
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took me a while but this is how i solved it. p, P+2, p+6, p+8, kinda look like 1, 3, 7, and 9, constraint they p, P+2, p+6, p+8 >100 then i tested (101,103, 107, 109) 201not prime move on 303 not prime no need to test there

i stopped there, 101,103, 107, 109 moved on with these 109^2-107^2+103^2-101^2 = 840 which is = 2^3*3*5*7

i) 8 which is 2^3 divides above
ii) 48 which is 2^4*3, does not divide
iii) 120 which is 2^3*3*5 divides above

which means on i and iii true

D is correct
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If p, p + 2, p + 6 and p + 8 are all prime numbers greater than 100, t 02 May 2026, 05:44
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but for 103 the 8(p+4) is not divisible by 120 , the question specifies must be so...?
nislam
If p, p + 2, p + 6 and p + 8 are all prime numbers greater than 100, then (p + 8)\(^2\) −(p + 6)\(^2\) + (p + 2)\(^2\) −p\(^2 \)must be divisible by which of the following?

I) 8
II) 48
III) 120

A) none

B) I only

C) I and II

D) I and III

E) I, II and III
Show more
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