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Must be a multiple of what number?

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AbeNeedsAnswers Master | Next Rank: 500 Posts Default Avatar
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Must be a multiple of what number?

Post Mon Jul 03, 2017 8:34 pm
If x and y are positive integers such that y is a multiple of 5 and 3x + 4y = 200, then x must be a multiple of which of the following?

A) 3

B) 6

C) 7

D) 8

E) 10

E

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Post Wed Jul 05, 2017 1:00 am
Alternatively, start with

3x + 4y = 200

3x = 200 - 4y

3x = 4 * (50 - y)

From this, we know x is a multiple of 4. We're told in the prompt that it's a multiple of 5. Since it's a multiple of 4 and of 5, it must also be a multiple of 4*5, or 20. That tells us x is also a multiple of ANY factor of 20 and E fits the bill.

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Post Wed Jul 12, 2017 4:26 pm
AbeNeedsAnswers wrote:
If x and y are positive integers such that y is a multiple of 5 and 3x + 4y = 200, then x must be a multiple of which of the following?

A) 3

B) 6

C) 7

D) 8

E) 10
Since y is a multiple of 5, we can let y = 5k and we have:

3x + 4(5k) = 200

3x = 200 - 20k

3x = 20(10 - k)

x = [20(10 - k)]/3

Since x is an integer, (10 - k) must be divisible by 3. Then, x is the product of 20 times some integer [which is the quotient of (10 - k)/3]. Thus, x must be a multiple of 10. For instance, when k = 1, x = 90, when k = 4, x = 40, and when k = 7, k = 20.

Answer: E

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Post Wed Jul 05, 2017 1:00 am
Alternatively, start with

3x + 4y = 200

3x = 200 - 4y

3x = 4 * (50 - y)

From this, we know x is a multiple of 4. We're told in the prompt that it's a multiple of 5. Since it's a multiple of 4 and of 5, it must also be a multiple of 4*5, or 20. That tells us x is also a multiple of ANY factor of 20 and E fits the bill.

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Post Wed Jul 12, 2017 4:26 pm
AbeNeedsAnswers wrote:
If x and y are positive integers such that y is a multiple of 5 and 3x + 4y = 200, then x must be a multiple of which of the following?

A) 3

B) 6

C) 7

D) 8

E) 10
Since y is a multiple of 5, we can let y = 5k and we have:

3x + 4(5k) = 200

3x = 200 - 20k

3x = 20(10 - k)

x = [20(10 - k)]/3

Since x is an integer, (10 - k) must be divisible by 3. Then, x is the product of 20 times some integer [which is the quotient of (10 - k)/3]. Thus, x must be a multiple of 10. For instance, when k = 1, x = 90, when k = 4, x = 40, and when k = 7, k = 20.

Answer: E

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Post Wed Jul 05, 2017 12:58 am
We can write "y is a multiple of 5" as y = 5m, where m is some integer whose value we don't need.

From there:

3x + 4y =>

3x + 4*5m =>

3x + 20m

We're told this equals 200, so

3x + 20m = 200

3x = 200 - 20m

3x = 20 * (10 - m)

So x must be a multiple of 20. 20 is itself a multiple of 10, so answer E fits.

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Post Tue Jul 04, 2017 8:04 am
I would recommend relying on strategy whenever possible; on this problem, using the rule that Mitch pointed out is probably the fastest approach. However, you could also TEST NUMBERS here.

Since y is a multiple of 5, just test any multiple of 5:

y = 5
4y = 20
3x + 4y = 200
3x + 20 = 200
3x = 180
x = 60

Eliminate C and D.

Test another value:
y = 10
4y = 40
3x + 40 = 200
x = 160
x = 160/3 --> not an integer. We have to test another value.

y = 15
4y = 60
3x + 60 = 200
3x = 140
x = 140/3 --> not an integer

y = 20
4y = 80
3x + 80 = 200
3x = 120
x = 40
Eliminate A and B.

The only answer left is E.

This strategy is more cumbersome, but in a pinch it would get you there!

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