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RenB
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If n is a positive two-digit integer, how many different values of n a 10 Jan 2024, 19:05
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If n is a positive two-digit integer, how many different values of n allow n^3 - n to be a multiple of 12?

A. 70
B. 67
C. 63
D. 60
E. 57
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If n is a positive two-digit integer, how many different values of n a 10 Jan 2024, 19:13
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RenB
If n is a positive two-digit integer, how many different values of n allow n^3 - n to be a multiple of 12?
A. 70
B. 67
C. 63
D. 60
E. 57
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n^3-n= n(n^2-1)= n(n-1)(n-2)-> 3 consecutive integers
3 consecutive integers are divisible by 3!=6. Thus 3 is a factor. For the product to be divisible by 12, it should also be divisible by 4.
Considering 2 cases for the value of n:
n is odd->
Then n+1 and n-1 will be divisible by 2, thus the product will be divisible by 4.
Half the numbers between 10 and 99 [90 numbers: (99-10)+1] are odd -> 45 numbers

n is even->
Then n-1 and n+1 will be odd. The product will be divisible by 4 only when n is divisible by 4. The number of numbers divible by 4 between 10 and 99 is 22

Add 45 and 22= 67 numbers
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If n is a positive two-digit integer, how many different values of n a 25 May 2024, 00:00
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Total two-digit integers are 90.

n^3-n= n(n^2-1)= (n-1)*(n)*(n+1) -> 3 consecutive integers

For n=10 product of 3 consecutive integer is not multiple of 12,
For n=14 product is not multiple of 12.

These number will form arithmetic progression starting from 10 and ending in 98. 
Series will be 10,14,18,....90,94,98

Number of terms
98= 10+(n-1) *4
n=23

There are 23 two-digits number which are not multiple of 12.

Total multiple of 12 which are two - digit integers are 90 - 23 = 67­
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If n is a positive two-digit integer, how many different values of n a 07 Jul 2024, 08:49
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RenB
n^3-n= n(n^2-1)= n(n-1)(n-2)-> 3 consecutive integers
3 consecutive integers are divisible by 3!=6. Thus 3 is a factor. For the product to be divisible by 12, it should also be divisible by 4.
Considering 2 cases for the value of n:
n is odd->
Then n+1 and n-1 will be divisible by 2, thus the product will be divisible by 4.
Half the numbers between 10 and 99 [90 numbers: (99-10)+1] are odd -> 45 numbers

n is even->
Then n-1 and n+1 will be odd. The product will be divisible by 4 only when n is divisible by 4. The number of numbers divible by 4 between 10 and 99 is 22

Add 45 and 22= 67 numbers
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In the first case did you assume that since n is odd so it must be a multiple of 3?­
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If n is a positive two-digit integer, how many different values of n a 22 Aug 2024, 00:34
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If n is a positive two-digit integer, how many different values of n allow n^3 - n to be a multiple of 12?

A. 70
B. 67
C. 63
D. 60
E. 57
Show more
­
\((n^3 - n) = (n-1)n(n+1)\)

Look at the first 12 two digit integers. The pattern will be repeated in each of the next 12 integers.

10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21

To be divisible by 12, we need to check divisibility by 3 and 4.
If n = 10, 9*10*11 will not be divisible by 12.
If n = 11, 10*11*12 will be.
If n =12, it will be.
If n = 13, it will be.
If n = 14, 13*14*15 will not be divisible by 12.
...
We find that of the 12 numbers, 9 are divisible by 12. In every 4 numbers, the first number is not a possible value of n but other 3 are. If you are able to identify this pattern early on, you don't even need to check for all 12 numbers.

Hence 3 out of every 4 numbers will be divisible by 12.

From 10 to 99, we have 90 numbers which gives us 3* 90/4 = 3 * 22.5
The 22 complete cycles of 4 will give us 66 values of n and of the leftover 2 numbers, one will be an acceptable value of n
Hence 67 values of n.

Answer (B)

RenB - https://gmatclub.com/forum/if-an-intege ... l#p3444072
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If n is a positive two-digit integer, how many different values of n a 27 Sep 2025, 07:36
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If n is a positive two-digit integer, how many different values of n allow n^3 - n to be a multiple of 12?

n^3 - n = (n-1)n(n+1)

(n-1)n(n+1) is a multiple of 3.

If n is odd, (n-1)n(n+1) is a multiple of 2*4 = 8 & 3, i.e. a multiple of 24 or 12 also
n = {11,13,,....99}: Number of such possible values of n = (99-11)/2 + 1 = 45

If n is a multiple of 4; (n-1)n(n+1) is a multiple of 12.
n = {12, 16,...., 96}: Number of such possible values of n = (96-12)/4 + 1 = 22

Number of values of n that allow n^3 - n to be a multiple of 12 = 45+22 = 67

IMO B
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