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If M and N are integers such that M = N^2 1, is M divisible by 12?

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If M and N are integers such that M = N^2 1, is M divisible by 12? [#permalink] New post   08 Feb 2020, 06:51
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If M and N are integers such that M = \(N^2\) – 1, is M divisible by 12?

(1) N-1 is the square of an even number

(2) N+1 is a two digit number and the sum of its digits is a multiple of 3

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Re: If M and N are integers such that M = N^2 1, is M divisible by 12? [#permalink] New post   08 Feb 2020, 08:03
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Balkrishna wrote:
rheam25 wrote:
If M and N are integers such that M = \(N^2\) – 1, is M divisible by 12?

(1) N-1 is the square of an even number

(2) N+1 is a two digit number and the sum of its digits is a multiple of 3




M = N^2 - 1
M = (N-1) (N+1)

We need to check whether (N-1)(N+1) contains 2^2 and 3.


Statement-1:
N-1 is the square of an even number.
Therefore, (N-1) = (2k)^2 where k is an integer.

So, (N-1) contains 2^2.

Does (N-1) or (N+1) contains 3? Let's check.

If k=3, then (N-1) = 36 and (N+1) = 14. Here, M is divisible by 12.
If k=4, then (N-1) = 64 and (N+1) = 66. Here, M is not divisible by 12.

There may or may not be 3 in (N-1)(N+1).

So, statement 1 is not sufficient alone.


Statement-2:
N+1 is a two digit number and the sum of its digits is a multiple of 3

(N+1) is divisible by 3.

Does (N-1) or (N+1) contains 2^2? Let's check.

If (N+1)=12, then (N-1)=10. Here, M= (N-1)(N+1)= 12*10. M is divisible by 12.
If (N+1)=15, then (N-1)=13. Here, M= (N-1)(N+1)= 15*13. M is not divisible by 12.

There may or may not be 2^2 in (N-1)(N+1).

So, statement 2 is not sufficient alone.


Combining both statements:
N-1 is the square of an even number.
N+1 is a two digit number and the sum of its digits is a multiple of 3

(N-1) contains 2^2 and (N+1) contains 3.

So, M = (N-1)(N+1) contains 2^2 and 3.

Hence, M is divisible by 12.


Therefore, both statements together are sufficient. Answer is C.


hi, Balkrishna
The highlighted part above is divisible by 12.
Can you check one more time ?

I think, Statement 1 is alone sufficient
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Re: If M and N are integers such that M = N^2 1, is M divisible by 12? [#permalink] New post   08 Feb 2020, 16:35
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M = (N-1)(N+1)

Check whether (N-1)(N+1) contains all 2^2 and 3.

1) if e.g. (N-1) =2^2, then (N+1) =2^2+2 =2*(2+1) =2*3.
Thus, (N-1)(N+1) surely contain all 2^2, 3 and another 2 as a bonus.
SUFFICIENT

2) Say (N+1)=3k and (N-1)=3k-2, where k=4,5,6,... Thus, (N-1)(N+1)=3*k*(3k-2).
(N-1)(N+1) is definitely divisible by 3. However, k*(3k-2) is divisible by 4 ONLY IF k is even.
NOT SUFFICIENT

FINAL ANSWER IS (A)
lnm87, lacktutor, eakabuah, Archit3110 try it.

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Re: If M and N are integers such that M = N^2 1, is M divisible by 12? [#permalink] New post   Updated on: 09 Feb 2020, 09:44
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chondro48
Thanks for tagging, it's a good question. Kudos to rheam25...
Here it is basically asking to check if (N-1)(N+1) is divisible by both 4 and 3.
So, (N-1)(N+1) is a pair where N is either even or odd i.e. pairs as (1,3), (3,5), (5,7)... where N is even
OR pairs as (2,4), (4,6), (6,8)... where N is odd

For statement 1
(N-1)(N+1) starts from (4,6), a multiple of 12. As (N-1) would always have 4 present i.e. (N+1) would always have 3 present. Hence M is a multiple of 12.
SUFFICIENT.

For statement 2
Minimum (N+1) is 12 and maximum is 99
So, Minimum (N-1)(N+1) = 10*12 which divisible by 12
Maximum (N-1)(N+1) = 97*99 which is not
INSUFFICIENT.

Answer A.
chondro48 wrote:
M = (N-1)(N+1)

Check whether (N-1)(N+1) contains all 2^2 and 3.

1) if e.g. (N-1) =2^2, then (N+1) =2^2+2 =2*(2+1) =2*3.
Thus, (N-1)(N+1) surely contain all 2^2, 3 and another 2 as a bonus.
SUFFICIENT

2) Say (N+1)=3k and (N-1)=3k-2, where k=4,5,6,... Thus, (N-1)(N+1)=3*k*(3k-2).
(N-1)(N+1) is definitely divisible by 3. However, k*(3k-2) is divisible by 4 ONLY IF k is even.
NOT SUFFICIENT

FINAL ANSWER IS (A)
lnm87, lacktutor, eakabuah, Archit3110 try it.

Posted from my mobile device

Originally posted by unraveled on 09 Feb 2020, 01:10.
Last edited by unraveled on 09 Feb 2020, 09:44, edited 1 time in total.
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Re: If M and N are integers such that M = N^2 1, is M divisible by 12? [#permalink] New post   09 Apr 2020, 08:53
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nvoonna wrote:
For statement 1,

If k=3, then (N-1) = 36 and (N+1) = 14. Here, M is divisible by 12.

How did 14 come into picture here? Can someone explain me this? rheam25

Its a typo. Please refer other solutions.

Hope you find them helpful.

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Re: If M and N are integers such that M = N^2 1, is M divisible by 12? [#permalink] New post   Updated on: 08 Feb 2020, 08:10
rheam25 wrote:
If M and N are integers such that M = \(N^2\) – 1, is M divisible by 12?

(1) N-1 is the square of an even number

(2) N+1 is a two digit number and the sum of its digits is a multiple of 3




M = N^2 - 1
M = (N-1) (N+1)

We need to check whether (N-1)(N+1) contains 2^2 and 3.


Statement-1:
N-1 is the square of an even number.
Therefore, (N-1) = (2k)^2 where k is an integer.

So, (N-1) contains 2^2.

Does (N-1) or (N+1) contains 3? Let's check.

If k=3, then (N-1) = 36 and (N+1) = 14. Here, M is divisible by 12.
If k=4, then (N-1) = 64 and (N+1) = 66. Here, M is divisible by 12.
If k=5, then (N-1) = 100 and (N+1) = 102. Here, M is divisible by 12.
If k=6, then (N-1) = 144 and (N+1) = 146. Here, M is divisible by 12.

(N-1)(N+1) contains by 2^2 and 3.

So, statement 1 is sufficient alone.


Statement-2:
N+1 is a two digit number and the sum of its digits is a multiple of 3

(N+1) is divisible by 3.

Does (N-1) or (N+1) contains 2^2? Let's check.

If (N+1)=12, then (N-1)=10. Here, M= (N-1)(N+1)= 12*10. M is divisible by 12.
If (N+1)=15, then (N-1)=13. Here, M= (N-1)(N+1)= 15*13. M is not divisible by 12.

There may or may not be 2^2 in (N-1)(N+1).

So, statement 2 is not sufficient alone.



Therefore, Answer is A.

Originally posted by Balkrishna on 08 Feb 2020, 07:22.
Last edited by Balkrishna on 08 Feb 2020, 08:10, edited 2 times in total.
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Re: If M and N are integers such that M = N^2 1, is M divisible by 12? [#permalink] New post   08 Feb 2020, 08:09
lacktutor wrote:
Balkrishna wrote:
rheam25 wrote:
If M and N are integers such that M = \(N^2\) – 1, is M divisible by 12?

(1) N-1 is the square of an even number

(2) N+1 is a two digit number and the sum of its digits is a multiple of 3




M = N^2 - 1
M = (N-1) (N+1)

We need to check whether (N-1)(N+1) contains 2^2 and 3.


Statement-1:
N-1 is the square of an even number.
Therefore, (N-1) = (2k)^2 where k is an integer.

So, (N-1) contains 2^2.

Does (N-1) or (N+1) contains 3? Let's check.

If k=3, then (N-1) = 36 and (N+1) = 14. Here, M is divisible by 12.
If k=4, then (N-1) = 64 and (N+1) = 66. Here, M is not divisible by 12.

There may or may not be 3 in (N-1)(N+1).

So, statement 1 is not sufficient alone.


Statement-2:
N+1 is a two digit number and the sum of its digits is a multiple of 3

(N+1) is divisible by 3.

Does (N-1) or (N+1) contains 2^2? Let's check.

If (N+1)=12, then (N-1)=10. Here, M= (N-1)(N+1)= 12*10. M is divisible by 12.
If (N+1)=15, then (N-1)=13. Here, M= (N-1)(N+1)= 15*13. M is not divisible by 12.

There may or may not be 2^2 in (N-1)(N+1).

So, statement 2 is not sufficient alone.


Combining both statements:
N-1 is the square of an even number.
N+1 is a two digit number and the sum of its digits is a multiple of 3

(N-1) contains 2^2 and (N+1) contains 3.

So, M = (N-1)(N+1) contains 2^2 and 3.

Hence, M is divisible by 12.


Therefore, both statements together are sufficient. Answer is C.


hi, Balkrishna
The highlighted part above is divisible by 12.
Can you check one more time ?

I think, Statement 1 is alone sufficient


lacktutor,

Thank you for highlighting the mistake.

I have modified the solution.
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Re: If M and N are integers such that M = N^2 1, is M divisible by 12? [#permalink] New post   09 Feb 2020, 02:53
thanks for tagging chondro48
heres my take

given
M= ( N+1) * (N-1) check whether its divisible by 3*4
#1
N-1 is the square of an even number
so possible value of N = 5,17,37
we get Yes for all cases hence sufficient
#2
N+1 is a two digit number and the sum of its digits is a multiple of 3
possible value of N = 11 we get yes and N = 14 , where we get no
hence insufficient
IMO A;
rheam25 not sure whether this is really falls under 700 level question bracket... must be 650-700 level ..





rheam25 wrote:
If M and N are integers such that M = \(N^2\) – 1, is M divisible by 12?

(1) N-1 is the square of an even number

(2) N+1 is a two digit number and the sum of its digits is a multiple of 3
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Re: If M and N are integers such that M = N^2 1, is M divisible by 12? [#permalink] New post   09 Apr 2020, 08:26
For statement 1,

If k=3, then (N-1) = 36 and (N+1) = 14. Here, M is divisible by 12.

How did 14 come into picture here? Can someone explain me this? rheam25
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Re: If M and N are integers such that M = N^2 1, is M divisible by 12? [#permalink]
09 Apr 2020, 08:26
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