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x is a positive integer less than 500. When x is divided by

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by GMATGuruNY » Wed Mar 15, 2017 5:02 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:
ziyuenlau wrote:x is a positive integer less than 500. When x is divided by 7, the remainder is 1; when x is divided by 3, the remainder is 2. How many x are there?

A. 21
B. 22
C. 23
D. 24
E. 25
When x is divided by 7, the remainder is 1.
x = 7a + 1 = 1, 8...
When x is divided by 3, the remainder is 2.
x = 3b + 2 = 2, 5, 8...

Thus, when x is divided by 21 -- the LCM of 7 and 3 -- the remainder will be 8 (the smallest value common to both lists).
x = 21c + 8.

Since x must be less than 500, we get:
21c + 8 < 500
21c < 492
c < 23.4.

Since c can be any integer between 0 and 23, inclusive -- for a total of 24 values -- there are 24 options for x.

The correct answer is D.
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by [email protected] » Wed Mar 15, 2017 9:46 am
Hi ziyuenlau,

If you're comfortable doing the basic arithmetic behind this question, then you can define the patterns involved with a bit of 'brute force' math.

To start, we need to find a POSITIVE integer (X) that when divided by 7 has a remainder of 1 AND when divided by 3 has a remainder of 2.

Let's start with X=8.... 8/7 = 1r1, 8/3 = 2r2, so X=8 IS an option

How about X=15... 15/7 = 2r1, but 15/3 = 5r0 - NOT an option

X = 22... 22/7 = 3r1, but 22/3 = 7r1 - NOT an option

X = 29... 29/7 = 4r1, 29/3 = 9r2 - IS an option

Notice how by adding 21 to X, we end up with another number that 'fits' what we're looking for. THAT is the pattern here. Now we just need to determine how many of those positive integers are less than 500. You can do that algebraically or list them out (putting them in sets of 5 helps with the organization)....

8, 29, 50, 71, 92

There's a shortcut here... since every 5 terms will include 5 'increases' of 21 (and that would be a total of 105), we can just list every 5th term for the next few rows...

- - - - 197
- - - - 302
- - - - 407

We're getting close to 500 now, so I'll list out the last few terms:

428, 449, 470, 491

4 'sets' of 5 + 4 = 24 total possible values for X

Final Answer: D

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by Matt@VeritasPrep » Wed Mar 15, 2017 4:59 pm
One approach: look for a pattern, then run with it.

Let's start by making lists of each type of number.

7 with r1: 1, 8, 15, 22, 29, 36, 43, ...
3 with r2: 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, ...

Notice the numbers on each list? 8, 29, ...

Further, notice that every THIRD number on the first list will fit the pattern, since the second list is considered when divided by 3. So our pattern is every number of form 21x + 8, where x is any nonnegative integer.

From there, we only need to go the closest such number to 500. 24*21 gets us above 500, so the the largest number is 21*23 + 8. Since we started with 21*0 + 8, we're counting from 21*0 to 21*23, for a total of 24 numbers.
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by Matt@VeritasPrep » Wed Mar 15, 2017 5:00 pm
If you're interested in the broader math behind this, check out any good resource on the Chinese Remainder Theorem. The wiki is a little dense, but a Google search will turn up lots of other resources, some of which you might like more than others.
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by Scott@TargetTestPrep » Fri Dec 15, 2017 9:25 am
hazelnut01 wrote:x is a positive integer less than 500. When x is divided by 7, the remainder is 1; when x is divided by 3, the remainder is 2. How many x are there?

A. 21
B. 22
C. 23
D. 24
E. 25
We are given that when x is divided by 7, the remainder is 1. Thus, x can be any integer that is 1 more than a multiple of 7:

1, 8, 15, 22, 29, ...

We are also given that when x is divided by 3, the remainder is 2. Thus, x can be any integer that is 2 more than a multiple of 3:

2, 5, 8, 11, 14, 17, 20, 23, 26, 29, ...

The first integer that both sets have in common is 8 and the next is 29. The difference is 21, and so that is the common difference we can use to determine all values that x can take on. The next few numbers in this evenly spaced set will be 29 + 21 = 50 and 50 + 21 = 71, etc.

We have an evenly spaced set of numbers. We see that the largest number under 500 that works for x is 491. Now, let's determine the number of values of x:

(largest integer in the set - smallest number in the set)/21 + 1

(491 - 8)/21 + 1 = 483/21 + 1 = 23 + 1 = 24

Answer: D

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