When positive integer n is divided by 3, the remainder is 2

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RBBmba@2014: Legendary Member; Posts: 944; Joined: Wed May 30, 2012 8:21 am; Thanked: 8 times; Followed by:5 members

When positive integer n is divided by 3, the remainder is 2

by RBBmba@2014 » Fri Mar 27, 2015 6:19 am

When positive integer n is divided by 3, the remainder is 2. When n is divided by 7, the remainder is 5. How many values less than 100 can n take?
(A) 0
(B) 2
(C) 3
(D) 4
(E) 5

OA: E

Experts - how can we solve this type of problems in least time ? Can we use LCM concept here for quick solution ?

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Source: Beat The GMAT — Problem Solving | Fri Mar 27, 2015 6:19 am

DavidG@VeritasPrep: Legendary Member; Posts: 2663; Joined: Wed Jan 14, 2015 8:25 am; Location: Boston, MA; Thanked: 1153 times; Followed by:128 members; GMAT Score:770

by DavidG@VeritasPrep » Fri Mar 27, 2015 6:26 am

One approach: brute force. We'll start with the numbers less than 100 that give a remainder of 5 when divided by 7. (We can generate this list by starting with the remainder, and then simply adding the divisor over and over.)

5, 12, 19, 26, 33, 40, 47, 54, 61, 68, 75, 82, 89, 96

Now we just have to find the ones that give a remainder of 2 when divided by 3.

5, 26, 47, 68, 89

Five numbers. Answer is E

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DavidG@VeritasPrep: Legendary Member; Posts: 2663; Joined: Wed Jan 14, 2015 8:25 am; Location: Boston, MA; Thanked: 1153 times; Followed by:128 members; GMAT Score:770

by DavidG@VeritasPrep » Fri Mar 27, 2015 6:30 am

Note also that when we evaluate the numbers in the list, a pattern emerges.

5, 12, 19, 26, 33, 40, 47, 54, 61, 68, 75, 82, 89, 96

5: remainder 2 when divided by 3
12: remainder 0 when divided by 3
19: remainder 1 when divided by 3

26: remainder 2 when divided by 3
33: remainder 0 when divided by 3
40: remainder 1 when divided by 3

And so on. Once you pick up this pattern, you can detect the relevant numbers more quickly.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

by Brent@GMATPrepNow » Fri Mar 27, 2015 6:48 am

One thing to note is that, when listing possible values in a remainder question, many students forget to list the smallest positive value.
For example, when told that, when n is divided by 7, the remainder is 5, the possible values of n are: 5, 12, 19, 26, etc
Many students forget to include 5 in their list. However, 5 divided by 7 equals 0 with remainder 5, so 5 is a valid value of n.

When it comes to remainders, we have a nice rule that says:

If N divided by D, leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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RBBmba@2014: Legendary Member; Posts: 944; Joined: Wed May 30, 2012 8:21 am; Thanked: 8 times; Followed by:5 members

by RBBmba@2014 » Fri Mar 27, 2015 7:54 am

Thanks Dave/Brent for your reply.

I followed exactly the same approach(NOT although by your formula Brent, but it still the same thing) and thought whether there is any better way to solve it faster ?

@ Dave - I didn't get your 2nd method, however!

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DavidG@VeritasPrep: Legendary Member; Posts: 2663; Joined: Wed Jan 14, 2015 8:25 am; Location: Boston, MA; Thanked: 1153 times; Followed by:128 members; GMAT Score:770

by DavidG@VeritasPrep » Fri Mar 27, 2015 9:55 am

It's not really a second method. It's a way to filter through our options a bit more efficiently.

If we don't see a pattern, we'll just grind through our list of numbers that give us a remainder of 5 when divided by 7:
5, 12, 19, 26, 33, 40, 47, 54, 61, 68, 75, 82, 89, 96

Or we can see that the pattern goes like this

5: remainder 2 when divided by 3
12: remainder 0 when divided by 3
19: remainder 1 when divided by 3

26: remainder 2 when divided by 3
33: remainder 0 when divided by 3
40: remainder 1 when divided by 3

For each set of three numbers, starting with 5, 12, 19, only the first term will give us a remainder of 2 when divided by 3. Of 5, 12, 19, only '5' will work. Of 26, 33, 40, only '26 will work, and so forth.

5, 12, 19, 26, 33, 40, 47, 54, 61, 68, 75, 82, 89, 96
(We've got five bolded terms total)

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RBBmba@2014: Legendary Member; Posts: 944; Joined: Wed May 30, 2012 8:21 am; Thanked: 8 times; Followed by:5 members

by RBBmba@2014 » Fri Mar 27, 2015 10:10 am

DavidG@VeritasPrep wrote:It's not really a second method. It's a way to filter through our options a bit more efficiently.

If we don't see a pattern, we'll just grind through our list of numbers that give us a remainder of 5 when divided by 7:
5, 12, 19, 26, 33, 40, 47, 54, 61, 68, 75, 82, 89, 96

Or we can see that the pattern goes like this

5: remainder 2 when divided by 3
12: remainder 0 when divided by 3
19: remainder 1 when divided by 3

26: remainder 2 when divided by 3
33: remainder 0 when divided by 3
40: remainder 1 when divided by 3

For each set of three numbers, starting with 5, 12, 19, only the first term will give us a remainder of 2 when divided by 3. Of 5, 12, 19, only '5' will work. Of 26, 33, 40, only '26 will work, and so forth.

5, 12, 19, 26, 33, 40, 47, 54, 61, 68, 75, 82, 89, 96
(We've got five bolded terms total)

Aha! now got your Sir.

Yes, I also went through this pattern but somewhat different way. As I see the gap between any two of these intended numbers is 21, which is LCM of 3 & 7. This is also correct approach, I think. Right ?

5, 26,47,68,89. Each is bigger than immediately preceding one by 21 (LCM of 3 & 7).

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

remainder

by GMATGuruNY » Fri Mar 27, 2015 10:34 am

A quick lesson on remainders:

When x is divided by 5, the remainder is 3.
In other words, x is 3 more than a multiple of 5:
x = 5a + 3.

When x is divided by 7, the remainder is 4.
In other words, x is 4 more than a multiple of 7:
x = 7b + 4.

Combined, the statements above imply that when x is divided by 35 -- the LOWEST COMMON MULTIPLE OF 5 AND 7 -- there will be a constant remainder R.
Put another way, x is R more than a multiple of 35:
x = 35c + R.

To determine the value of R:
Make a list of values that satisfy the first statement:
When x is divided by 5, the remainder is 3.
x = 5a + 3 = 3, 8, 13, 18...
Make a list of values that satisfy the second statement:
When x is divided by 7, the remainder is 4.
x = 7b + 4 = 4, 11, 18...
The value of R is the SMALLEST VALUE COMMON TO BOTH LISTS:
R = 18.

Putting it all together:
x = 35c + 18.

Another example:
When x is divided by 3, the remainder is 1.
x = 3a + 1 = 1, 4, 7, 10, 13...
When x is divided by 11, the remainder is 2.
x = 11b + 2 = 2, 13...

Thus, when x is divided by 33 -- the LCM of 3 and 11 -- the remainder will be 13 (the smallest value common to both lists).
x = 33c + 13 = 13, 46, 79...

Onto the problem at hand:

RBBmba@2014 wrote:When positive integer n is divided by 3, the remainder is 2. When n is divided by 7, the remainder is 5. How many values less than 100 can n take?
(A) 0
(B) 2
(C) 3
(D) 4
(E) 5

When n is divided by 3, the remainder is 2.
n = 3a + 2 = 2, 5, 8....
When n is divided by 7, the remainder is 5.
n = 7b + 5 = 5...

Thus, when n is divided by 21 -- the LCM of 3 and 7 -- the remainder will be 5 (the smallest value common to both lists).
n = 21c + 5 = 5, 26, 47, 68, 89.
Total number of options for n = 5.

The correct answer is E.

Last edited by GMATGuruNY on Sat Dec 30, 2017 10:41 pm, edited 1 time in total.

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RBBmba@2014: Legendary Member; Posts: 944; Joined: Wed May 30, 2012 8:21 am; Thanked: 8 times; Followed by:5 members

by RBBmba@2014 » Sat Mar 28, 2015 12:20 am

GMATGuruNY wrote:A quick lesson on remainders:

Combined, the statements above imply that when x is divided by 35 -- the LOWEST COMMON MULTIPLE OF 5 AND 7 -- there will be a constant remainder R.
Put another way, x is R more than a multiple of 35:
x = 35c + R.

To determine the value of R:
Make a list of values that satisfy the first statement:
When x is divided by 5, the remainder is 3.
x = 5a + 3 = 3, 8, 13, 18...
Make a list of values that satisfy the second statement:
When x is divided by 7, the remainder is 4.
x = 7b + 4 = 4, 11, 18...
The value of R is the SMALLEST VALUE COMMON TO BOTH LISTS:
R = 18.

Putting it all together:
x = 35c + 18.
Onto the problem at hand:

RBBmba@2014 wrote:When positive integer n is divided by 3, the remainder is 2. When n is divided by 7, the remainder is 5. How many values less than 100 can n take?
(A) 0
(B) 2
(C) 3
(D) 4
(E) 5

Thus, when n is divided by 21 -- the LCM of 3 and 7 -- the remainder will be 5 (the smallest value common to both lists).

Hi Mitch - thanks for your reply.

A quick clarification - would you please clarify the above parts in 'GREEN' that how we can determine value of R(remainder) based on Mathematical logic other than considering the pattern ?

Look forward to your reply.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Sat Mar 28, 2015 6:36 am

RBBmba@2014 wrote:Hi Mitch - thanks for your reply.

A quick clarification - would you please clarify the above parts in 'GREEN' that how we can determine value of R(remainder) based on Mathematical logic other than considering the pattern ?

Look forward to your reply.

Statement 1: When x is divided by 5, the remainder is 3.
x = 5a + 3 = 3, 8, 13, 18...
Statement 2: When x is divided by 7, the remainder is 4.
x = 7b + 4 = 4, 11, 18...
In each case, the SMALLEST valid value is equal to the REMAINDER.

The same reasoning holds true when the statements are combined to yield x = 35c + R.
The SMALLEST value that satisfies both statements -- 18 -- must be equal to the REMAINDER:
x = 35c + 18 = 18, 53, 88...

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RBBmba@2014: Legendary Member; Posts: 944; Joined: Wed May 30, 2012 8:21 am; Thanked: 8 times; Followed by:5 members

by RBBmba@2014 » Sat Mar 28, 2015 10:53 am

GMATGuruNY wrote:
RBBmba@2014 wrote:Hi Mitch - thanks for your reply.

A quick clarification - would you please clarify the above parts in 'GREEN' that how we can determine value of R(remainder) based on Mathematical logic other than considering the pattern ?

Look forward to your reply.
Statement 1: When x is divided by 5, the remainder is 3.
x = 5a + 3 = 3, 8, 13, 18...
Statement 2: When x is divided by 7, the remainder is 4.
x = 7b + 4 = 4, 11, 18...
In each case, the SMALLEST valid value is equal to the REMAINDER.

The same reasoning holds true when the statements are combined to yield x = 35c + R.
The SMALLEST value that satisfies both statements -- 18 -- must be equal to the REMAINDER:
x = 35c + 18 = 18, 53, 88...

Thanks Mitch for clarifying. So, this worked for the combined statements because 35 is LCM of 5 & 7. Right ?

If it's not LCM of the given numbers (re 5 & 7), then will this logic/approach work ? I guess, NOT.

Correct me please if wrong!

Quote

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Mon Mar 30, 2015 1:47 pm

RBBmba@2014 wrote:
Thanks Mitch for clarifying. So, this worked for the combined statements because 35 is LCM of 5 & 7. Right ?

If it's not LCM of the given numbers (re 5 & 7), then will this logic/approach work ? I guess, NOT.

Correct me please if wrong!

Correct.
if the first statement discusses division by 5, and the second statement discusses division by 7, then combining the two statements implies division by the LCM of 5 and 7 (35).