Brian@VeritasPrep wrote:Hey everyone,

Rahul's solution above is terrific, but let me point out a few strategic things here.

1) When you see that (35^2 - 1) setup, or any version of (x^2 - y^2), you HAVE TO think about the Difference of Squares rule: x^2 - y^2 = (x+y)(x-y). That rule sets up quite a bit for you - namely, it allows you to turn ugly subtraction into multiplication, which should allow you to factor.

2) Pursuant to the above, when you see addition/subtraction of exponents, there's an overwhelming likelihood that you'll need to factor somehow to turn that into multiplication (broken record for those of you reading my posts today, but I can't stress this enough). Difference of Squares is a great tool to have in your arsenal for that.

3) You may find it helpful to break down the resulting divisibility problem (34*36/k) into prime factors:

2*2*2*3*3*17 is divisible by k.

Since the answer choices are:

8 = 2*2*2 (divisible)
9 = 3*3*3 (divisible)
12 = 2*2*3 (divisible)
16 = 2*2*2*2 (we're one 2 short, so this is NOT divisible)
17 = 17 (divisible)

You can attack these systematically. When divisibility is in question, prime factorization helps you to have a system for checking divisibility without needing to treat each division as individual.

Brian@VeritasPrep wrote:Hey everyone,

Rahul's solution above is terrific, but let me point out a few strategic things here.

1) When you see that (35^2 - 1) setup, or any version of (x^2 - y^2), you HAVE TO think about the Difference of Squares rule: x^2 - y^2 = (x+y)(x-y). That rule sets up quite a bit for you - namely, it allows you to turn ugly subtraction into multiplication, which should allow you to factor.

2) Pursuant to the above, when you see addition/subtraction of exponents, there's an overwhelming likelihood that you'll need to factor somehow to turn that into multiplication (broken record for those of you reading my posts today, but I can't stress this enough). Difference of Squares is a great tool to have in your arsenal for that.

3) You may find it helpful to break down the resulting divisibility problem (34*36/k) into prime factors:

2*2*2*3*3*17 is divisible by k.

Since the answer choices are:

8 = 2*2*2 (divisible)
9 = 3*3*3 (divisible)
12 = 2*2*3 (divisible)
16 = 2*2*2*2 (we're one 2 short, so this is NOT divisible)
17 = 17 (divisible)

You can attack these systematically. When divisibility is in question, prime factorization helps you to have a system for checking divisibility without needing to treat each division as individual.

how did u write 9= 3*3*3 above.

ankur.agrawal wrote:

Brian@VeritasPrep wrote:Hey everyone,

Rahul's solution above is terrific, but let me point out a few strategic things here.

1) When you see that (35^2 - 1) setup, or any version of (x^2 - y^2), you HAVE TO think about the Difference of Squares rule: x^2 - y^2 = (x+y)(x-y). That rule sets up quite a bit for you - namely, it allows you to turn ugly subtraction into multiplication, which should allow you to factor.

2) Pursuant to the above, when you see addition/subtraction of exponents, there's an overwhelming likelihood that you'll need to factor somehow to turn that into multiplication (broken record for those of you reading my posts today, but I can't stress this enough). Difference of Squares is a great tool to have in your arsenal for that.

3) You may find it helpful to break down the resulting divisibility problem (34*36/k) into prime factors:

2*2*2*3*3*17 is divisible by k.

Since the answer choices are:

8 = 2*2*2 (divisible)
9 = 3*3*3 (divisible)
12 = 2*2*3 (divisible)
16 = 2*2*2*2 (we're one 2 short, so this is NOT divisible)
17 = 17 (divisible)

You can attack these systematically. When divisibility is in question, prime factorization helps you to have a system for checking divisibility without needing to treat each division as individual.

how did u write 9= 3*3*3 above.

Typo...the 3 key must have stuck! Sorry about that...just fixed it.

klmehta03 wrote:{[(35^2) -1]/k}=integer
integer= (35-1)*(35+1)/k
=1224/k
only when you divide by 16 the divisor is not an integer
so ans D

Brian@VeritasPrep wrote:Hey everyone,

Rahul's solution above is terrific, but let me point out a few strategic things here.

1) When you see that (35^2 - 1) setup, or any version of (x^2 - y^2), you HAVE TO think about the Difference of Squares rule: x^2 - y^2 = (x+y)(x-y). That rule sets up quite a bit for you - namely, it allows you to turn ugly subtraction into multiplication, which should allow you to factor.

2) Pursuant to the above, when you see addition/subtraction of exponents, there's an overwhelming likelihood that you'll need to factor somehow to turn that into multiplication (broken record for those of you reading my posts today, but I can't stress this enough). Difference of Squares is a great tool to have in your arsenal for that.

3) You may find it helpful to break down the resulting divisibility problem (34*36/k) into prime factors:

2*2*2*3*3*17 is divisible by k.

Since the answer choices are:

8 = 2*2*2 (divisible)
9 = 3*3 (divisible)
12 = 2*2*3 (divisible)
16 = 2*2*2*2 (we're one 2 short, so this is NOT divisible)
17 = 17 (divisible)

You can attack these systematically. When divisibility is in question, prime factorization helps you to have a system for checking divisibility without needing to treat each division as individual.

Rahul@gurome wrote:(35^2 - 1)/k = (35 - 1)(35 + 1)/k = (34)(36)/k

We can clearly see that if k = 17, 12, 9 or 8, then (35^2 - 1)/k will give an integer value.
But if k = 16, then 35^2 - 1)/k will not give an integer value.

The correct answer is [spoiler](D)[/spoiler].