Problem Solving

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)

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...

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

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).

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.

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...

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!