PS Factor/Factorial without finding list of values

Expert replies

wayofjungle: Senior | Next Rank: 100 Posts; Posts: 77; Joined: Fri Mar 11, 2011 8:55 am; Thanked: 12 times

by wayofjungle » Mon Mar 28, 2011 11:30 am

If j and k are positive integers, j – 2 is divisible by 4 and k – 5 is divisible by 4, all of the following could be the value of j – k EXCEPT:

43
33
21
13
5

Do you understand how to solve this without coming up with lists of values?

[spoiler]This is an EXCEPT question, so you're looking for the choice that cannot be the value of j – k. Since we know j – 2 and k – 5 are both divisible by 4, we can come up with values of j = 2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50... and k = 5, 9, 13, 17, 21, 25, 29, 33... j – k = 14 – 9 = 5, or 26 – 13 = 13, or 34 – 13 = 21 or 46 – 13 = 33. The only answer that you can't eliminate is A, so the answer is A. [/spoiler]

Source: Beat The GMAT — Quantitative Reasoning | Mon Mar 28, 2011 11:30 am

VivianKerr: GMAT Instructor; Posts: 1035; Joined: Fri Dec 17, 2010 11:13 amFollowed by: 365 members; Location: Los Angeles, CA; Thanked: 474 times

by VivianKerr » Tue Mar 29, 2011 7:53 am

I did a similar approach, and as far as I know there's no shortcut:

(j - 2) / 4

(k - 5) / 4

Let's say j = 6 and k = 9. The difference will be a multiple of 3. ELIMINATE B and C.

Now let's keep k = 9 and just keep adding 4 to j, seeing how the difference changes.

If j = 14 and k = 9, the difference is 5. ELIMINATE E.

If j = 22 and k = 9, the difference is 13. ELIMINATE D.

I think it's easier to keep k the same at 9, and just continue to increase j, looking for the answer choices among the differences. It's basically Backsolving.

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kevincanspain: GMAT Instructor; Posts: 613; Joined: Thu Mar 22, 2007 6:17 amFollowed by: 64 members; Location: madrid; Thanked: 171 times; GMAT Score:790

by kevincanspain » Tue Mar 29, 2011 8:07 am

wayofjungle wrote:If j and k are positive integers, j – 2 is divisible by 4 and k – 5 is divisible by 4, all of the following could be the value of j – k EXCEPT:

43
33
21
13
5

Do you understand how to solve this without coming up with lists of values?

[spoiler]This is an EXCEPT question, so you're looking for the choice that cannot be the value of j – k. Since we know j – 2 and k – 5 are both divisible by 4, we can come up with values of j = 2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50... and k = 5, 9, 13, 17, 21, 25, 29, 33... j – k = 14 – 9 = 5, or 26 – 13 = 13, or 34 – 13 = 21 or 46 – 13 = 33. The only answer that you can't eliminate is A, so the answer is A. [/spoiler]

You could reason that since j - 2 and k - 5 are both multiples of 4, so is their difference, j - k + 3. Thus j - k is 3 less than a multiple of 4. Since 43 is 1 less than a multiple of 4, it is the exception we are looking for.

Kevin Armstrong
GMAT Instructor
Gmatclasses
Madrid

manpsingh87: Master | Next Rank: 500 Posts; Posts: 436; Joined: Tue Feb 08, 2011 3:07 amFollowed by: 6 members; Thanked: 72 times

by manpsingh87 » Tue Mar 29, 2011 8:21 am

wayofjungle wrote:If j and k are positive integers, j – 2 is divisible by 4 and k – 5 is divisible by 4, all of the following could be the value of j – k EXCEPT:

43
33
21
13
5

Do you understand how to solve this without coming up with lists of values?

[spoiler]This is an EXCEPT question, so you're looking for the choice that cannot be the value of j – k. Since we know j – 2 and k – 5 are both divisible by 4, we can come up with values of j = 2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50... and k = 5, 9, 13, 17, 21, 25, 29, 33... j – k = 14 – 9 = 5, or 26 – 13 = 13, or 34 – 13 = 21 or 46 – 13 = 33. The only answer that you can't eliminate is A, so the answer is A. [/spoiler]

as j-2 is divisible by 4 therefore, j-2 can be written as j-2=4m ; j=4m+2; where m is an integer,

similarly k-5=4n; k=4n+5; where n is an integer;

now j-k= 4(m-n) - 3;
as both m and n are integers, there their difference would also be an integer, i.e. m-n would yield an integral value,
therefore j-k= 4t-3; where t is an integer;

for t=9 j-k=33; for t=6 j-k=21; for t=4 j-k=13; for t=2 j-k =5

i.e. all values except A are possible hence A

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