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(35^2 – 1)/k is an integer

by sanju09 » Tue Sep 14, 2010 4:25 am
If k is an integer, and (35^2 - 1)/k is an integer, then k could be each of the following, EXCEPT
(A) 8
(B) 9
(C) 12
(D) 16
(E) 17


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by Rahul@gurome » Tue Sep 14, 2010 4:38 am
(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].
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by Brian@VeritasPrep » Tue Sep 14, 2010 11:30 am
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.
Last edited by Brian@VeritasPrep on Tue Sep 14, 2010 11:59 am, edited 1 time in total.
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by ankur.agrawal » Tue Sep 14, 2010 11:51 am
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.
Thx Brian not only for solving the problems but also for giving us deeper insights.
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by ankur.agrawal » Tue Sep 14, 2010 11:54 am
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.
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by Brian@VeritasPrep » Tue Sep 14, 2010 11:59 am
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.
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by klmehta03 » Mon Sep 20, 2010 7:00 pm
{[(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
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by sanju09 » Mon Sep 20, 2010 8:54 pm
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
very laborious
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by anirudhbhalotia » Mon Dec 13, 2010 9:22 am
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.


Nice! Thanks Brian!

You mentioned Difference of Square Rule, if there is a "+" sign, can we still break it up as (x^2 + y^2)?

In the OG-12 Maths review is this rule mentioned...I don't remember it seeing in the book nor in the Gmat Prep! :-(

Can you share/mention more of such essential concepts which we will definitely need to apply? This will be a HUGE help!

Thanks!
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by fskilnik@GMATH » Mon Dec 13, 2010 10:31 am
Another (more theoretical) approach: (It is what is "behind" Brian´s (3) item, watch out!)

When you are told that a and b are integers such that a/b is an integer (b implicitly non-zero, for sure), that means that b is a divisor of a.

Therefore, from the question stem we know that k must be a divisor of 35^2-1 = (34)(36) = 2^3*3^2*17.

(A) 2^3 is a divisor of the "red thing", for sure.
(B) 3^2 also
(C) 2^2*3 also
(D) 2^4 is NOT, because we have only three 2´s and we "would need" four 2´s.

PUFF!

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by HeintzC2 » Mon Aug 08, 2011 6:17 am
i don't think that we can "clearly" see that if k = 17, 12, 9, or 8, then (35^2-1)/k will give an integer value.

for (34)(36)/k, i think we can clearly see that 17, 12, and 9 all result in integer values (they are factors of either 34 or 36). However, neither 8 nor 16 are factors of 34 or 36, and the math required to multiply 34 by 36 then divide out 8 and 16 would be too time consuming (we could factor a 2 out, leaving 17*72, but that isn't necessarily clear on first glance either).

What we are left with is both options A (8) and D (16). Because 8 is a factor of 16, if the number is divisible by 16, then it must also be divisible by 8, however, the converse is not true (a number divisible by 8 is not necessarily divisible by 16). Therefore, 16 must be the number which results in a non-integer result.
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].
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by Vinoth9132* » Thu Jul 31, 2014 12:45 am
D)16
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