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Math Expert V
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For how many ordered pairs of positive integers (x,y) is x + 2y = 100?  [#permalink]

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For how many ordered pairs of positive integers (x,y) is x + 2y = 100?

(A) 33
(B) 49
(C) 50
(D) 99
(E) 100

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For how many ordered pairs of positive integers (x,y) is x + 2y = 100?  [#permalink]

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Bunuel wrote:
For how many ordered pairs of positive integers (x,y) is x + 2y = 100?

(A) 33
(B) 49
(C) 50
(D) 99
(E) 100

pair series will go on from x,y = (2,49), (4,48),(6,47)...... (98,1)
total such pairs 49
IMO B
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Originally posted by Archit3110 on 21 Mar 2019, 04:09.
Last edited by Archit3110 on 21 Mar 2019, 08:26, edited 1 time in total.
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Re: For how many ordered pairs of positive integers (x,y) is x + 2y = 100?  [#permalink]

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Archit3110 wrote:
Bunuel wrote:
For how many ordered pairs of positive integers (x,y) is x + 2y = 100?

(A) 33
(B) 49
(C) 50
(D) 99
(E) 100

pair series will go on from x,y = (100,0 ) to (0,50)
total such pairs = 50
(100,0) to (2,49)
total values of y=49
IMO B

Bro, x and y can't be 0.

x and y are positive integers.
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Re: For how many ordered pairs of positive integers (x,y) is x + 2y = 100?  [#permalink]

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Since x cannot be 0 y must be smaller than 50, therefore there are 49 different possible values of y.
Answer B is correct since pairs are ordered?
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Re: For how many ordered pairs of positive integers (x,y) is x + 2y = 100?  [#permalink]

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I'm supposing the appropriate solution is 16. Since it is x+2y=100 this suggests x is even. so x=2,4, 6, 8, 10, ..., 30, 32. There are 16 requested sets.
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Re: For how many ordered pairs of positive integers (x,y) is x + 2y = 100?  [#permalink]

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jackharrywa wrote:
I'm supposing the appropriate solution is 16. Since it is x+2y=100 this suggests x is even. so x=2,4, 6, 8, 10, ..., 30, 32. There are 16 requested sets.

Why do you stop at 32?

x could also have the value 98 and y=1, by your logic that's 49 possible values of x
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Re: For how many ordered pairs of positive integers (x,y) is x + 2y = 100?  [#permalink]

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Berlin92 wrote:
jackharrywa wrote:
I'm supposing the appropriate solution is 16. Since it is x+2y=100 this suggests x is even. so x=2,4, 6, 8, 10, ..., 30, 32. There are 16 requested sets.

Why do you stop at 32?

x could also have the value 98 and y=1, by your logic that's 49 possible values of x

yes, you are saying absolutely right, it is 49 possible value.
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Re: For how many ordered pairs of positive integers (x,y) is x + 2y = 100?  [#permalink]

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Here is how i am going to solve it...

x + 2y = 100 for positive pair of x and y..

Lets assume that y = 49 then 2y = 98. Then x will be 2. y can't be 50 because then 2y will be 100 and x will become zero. As per problem statement x cant be zero.

now lets reduce the value of y

case 1 - if y = 48 then x becomes 4
case 2 - if y = 47 then x becomes 6
case 3 - if y = 46 then x becomes 8
and so on..

so y can have values from 49 to 1. It will change x values accordingly as well.

correct answer is B.
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For how many ordered pairs of positive integers (x,y) is x + 2y = 100?  [#permalink]

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Hello!

It can be solved with arithmetic progression.

$$nth = a + d(n-1)$$

$$nth = 98$$
$$a = 2$$

Applying the formula $$n = 49$$.

49.

B

Originally posted by jfranciscocuencag on 21 Mar 2019, 18:38.
Last edited by jfranciscocuencag on 25 Mar 2019, 13:05, edited 1 time in total.
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Re: For how many ordered pairs of positive integers (x,y) is x + 2y = 100?  [#permalink]

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Bunuel wrote:
For how many ordered pairs of positive integers (x,y) is x + 2y = 100?

(A) 33
(B) 49
(C) 50
(D) 99
(E) 100

We see that the smallest integer value for y is 1 (and x will then be 98), and the largest value of y is 49 (and x will then be 2). Since y can be any integer from 1 to 49, there are 49 values for y and hence 49 ordered pairs of (x, y).

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For how many ordered pairs of positive integers (x,y) is x + 2y = 100?  [#permalink]

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Bunuel wrote:
For how many ordered pairs of positive integers (x,y) is x + 2y = 100?

(A) 33
(B) 49
(C) 50
(D) 99
(E) 100

x + 2y = 100
x=100-2y
x=2(50-y)
Since x and y have to be positive integers, smallest value of y = 1, and largest value of y=49, hence there are 49 such pairs.

Hope this helps .
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- Stne For how many ordered pairs of positive integers (x,y) is x + 2y = 100?   [#permalink] 29 Mar 2019, 13:03
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