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# How many pairs positive integers x and y satisfy 2x + y < 40, if y is

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Math Expert
Joined: 02 Sep 2009
Posts: 65290
How many pairs positive integers x and y satisfy 2x + y < 40, if y is  [#permalink]

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05 May 2020, 02:15
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12
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Difficulty:

95% (hard)

Question Stats:

33% (02:52) correct 67% (02:54) wrong based on 48 sessions

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How many pairs positive integers x and y satisfy 2x + y < 40, if y is a multiple of x?

A. 90
B. 91
C. 92
D. 93
E. 94

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Posts: 8757
Re: How many pairs positive integers x and y satisfy 2x + y < 40, if y is  [#permalink]

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05 May 2020, 04:32
1
1
Bunuel wrote:
How many pairs positive integers x and y satisfy 2x + y < 40, if y is a multiple of x?

A. 90
B. 91
C. 92
D. 93
E. 94

So y = ax where a= 1, 2, 3....

We will now calculate for each value of ax..
1) $$a=1...2x + x < 40....3x<39+1....x<13+\frac{1}{3}.......1\leq{x}\leq{13}....13$$ values
2) $$a=2...2x + 2x < 40....4x<40....x<10.......1\leq{x}<{10}....9$$ values
3) $$a=3...2x + 3x < 40....5x<40....x<8.......1\leq{x}<8....7$$ values
4) $$a=4...2x + 4x < 40....6x<36+4....x<6+\frac{4}{6}.......1\leq{x}\leq{6}....6$$ values
5) $$a=5...2x + 5x < 40....7x<35+5....x<5+\frac{5}{7}.......1\leq{x}\leq{5}....5$$ values
6) Now 40/8 and 40/9 are between 4 and 5, so when a=6 and a=7, we will get x as 1 to 4, so 2*4=8 values
7) Now 40/10 to 40/13 are between 3 and 4, so when a=8 to 11, we will get x as 1 to 3, so 4*3=12 values
8) Now 40/14 and 40/19 are between 2 and 3, so when a=12 to 17, we will get x as 1 and 2, so 6*2=12 values
9) Now 40/20 to 40/39 are between 1 and 2, so when a=18 to 37, we will get x as 1, so 20*1=20 values..

Total = 13+9+7+6+5+8+12+12+20=92

C
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Re: How many pairs positive integers x and y satisfy 2x + y < 40, if y is  [#permalink]

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06 May 2020, 00:39
1
1
Bunuel wrote:
How many pairs positive integers x and y satisfy 2x + y < 40, if y is a multiple of x?

A. 90
B. 91
C. 92
D. 93
E. 94

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2x + y < 40 and y is a multiple of x.

1) When x = 1 => y< 38 => 37 values
2) When x = 2, y < 36 => 17 values (only multiples of 2)
3) When x = 3, y < 34 => 11 values (only multiples of 3)
4) When x = 4, y < 32 => 7 values (only multiples of 4)
5) When x = 5, y < 30 => 5 values (only multiples of 5)
6) When x = 6, y < 28 => 4 values (only multiples of 6)
7) When x = 7, y < 26 => 3 values (only multiples of 7)
8) When x = 8, y < 24 => 2 values (only multiples of 8)
9) When x = 9, y < 22 => 2 values (only multiples of 9)
10) When x = 10, y < 20 => 1 values (only multiples of 10)
11) When x = 11, y < 18 => 1 values (only multiples of 11)
12) When x = 12, y < 16 => 1 values (only multiples of 12)
13) When x = 13, y < 14 => 1 value (only 13)
13) When x = 14, y < 12 => 0 values

Total = 37+17+11+7+5+4+3+2+2+1+1+1+1 = 92

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Re: How many pairs positive integers x and y satisfy 2x + y < 40, if y is  [#permalink]

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06 May 2020, 01:12
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Bunuel wrote:
How many pairs positive integers x and y satisfy 2x + y < 40, if y is a multiple of x?

A. 90
B. 91
C. 92
D. 93
E. 94

Project PS Butler

Are You Up For the Challenge: 700 Level Questions

It's much harder and lengthy than expected questioning GMAT so I can only call it a fine practice to do just to clear concepts

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Re: How many pairs positive integers x and y satisfy 2x + y < 40, if y is  [#permalink]

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09 May 2020, 08:48
Bunuel - Please can we have we see your method of cracking this. This is one of the questions that might take more time than usual, hence approach needs to be crisp.
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Re: How many pairs positive integers x and y satisfy 2x + y < 40, if y is  [#permalink]

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10 May 2020, 01:08
Can we solve this graphically?
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Re: How many pairs positive integers x and y satisfy 2x + y < 40, if y is  [#permalink]

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13 May 2020, 09:55
Bunuel wrote:
How many pairs positive integers x and y satisfy 2x + y < 40, if y is a multiple of x?

A. 90
B. 91
C. 92
D. 93
E. 94

Project PS Butler

Are You Up For the Challenge: 700 Level Questions

Is there a shorter solution Bunuel?
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Re: How many pairs positive integers x and y satisfy 2x + y < 40, if y is  [#permalink]

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16 May 2020, 08:24
1
Bunuel wrote:
How many pairs positive integers x and y satisfy 2x + y < 40, if y is a multiple of x?

A. 90
B. 91
C. 92
D. 93
E. 94

Since x and y are positive integers and y is a multiple of x, then y ≥ x.

If x = 1, we have:

2(1) + y < 40

y < 38

So y can be any integer from 1 to 37, inclusive. Therefore, there are 37 pairs of positive integers x and y (or solutions) when x = 1.

If x = 2, we have:

2(2) + y < 40

y < 36

So y can be any even integer from 2 to 34, inclusive. Therefore, there are 17 solutions when x = 2.

If x = 3, we have:

2(3) + y < 40

y < 34

So y can be any multiple of 3 from 3 to 33, inclusive. Therefore, there are 11 solutions when x = 3.

If x = 4, we have:

2(4) + y < 40

y < 32

So y can be any multiples of 4 from 4 to 28, inclusive. Therefore, there are 7 solutions when x = 4.

If x = 5, we have:

2(5) + y < 40

y < 30

So y can be any multiples of 5 from 5 to 25, inclusive. Therefore, there are 5 solutions when x = 5.

If x = 6, we have:

2(6) + y < 40

y < 28

So y can be any multiples of 6 from 6 to 24, inclusive. Therefore, there are 4 solutions when x = 6.

If x = 7, we have:

2(7) + y < 40

y < 26

So y can be any multiples of 7 from 7 to 21, inclusive. Therefore, there are 3 solutions when x = 7.

If x = 8, we have:

2(8) + y < 40

y < 24

So y can be any multiples of 8 from 8 to 16, inclusive. Therefore, there are 2 solutions when x = 8.

If x = 9, we have:

2(9) + y < 40

y < 22

So y can be any multiples of 9 from 9 to 18, inclusive. Therefore, there are 2 solutions when x = 9.

If x = 10, we have:

2(10) + y < 40

y < 20

So y can only be 10. Therefore, there is only 1 solution when x = 10.

Similarly, when x = 11, 12 or 13, there is 1 solution for each of these x-values. If x ≥ 14, then there are no solutions. (For example, if x = 14, 2(14) + y < 40 → y < 12. However, recall that y ≥ x since y is a multiple of x.)

Therefore, the total number of solutions is:

37 + 17 + 11 + 7 + 5 + 4 + 3 + 2 x 2 + 1 x 4 = 54 + 18 + 12 + 4 + 4 = 72 + 20 = 92

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Re: How many pairs positive integers x and y satisfy 2x + y < 40, if y is   [#permalink] 16 May 2020, 08:24