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# The number of positive integer valued pairs (x, y),

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CEO
Joined: 15 Aug 2003
Posts: 3469
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The number of positive integer valued pairs (x, y), [#permalink]  17 Sep 2003, 03:23
00:00

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(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 1 sessions
The number of positive integer valued pairs (x, y), satisfying 4x-17y = 1 and x < 1000 is:

1. 59
2. 57
3. 55
4. 58

my solution took a lot of time..can anyone provide a faster one?
My Solution :

x= (1+ 17y)/4

1+17y is an integer divisible by 4...now check for divisibility

y=1 , x not an integer
y=2 , x not an integer
y=3, x integer ==> 1st pair
y=4 , x not an integer
y=5, x not an integer
y=6 , nope
y=7 , yes
ok...now lets check for a pattern....
y=11, yes
y=15 , yes
y=19, yes

Its a progression with first term =3 and constant increment by 4
Ok..but whats the maximum of y..

we know x<1000
1+17y<4000
17y<3999
y<240...

We get y =234

Nth term of a series = a + (n-1)*d
a= first term
n= total number of terms in series
d= constant increment

so 235 = 3 + (n-1) 4
232/4 = n-1
n=59

thanks
Senior Manager
Joined: 21 Aug 2003
Posts: 258
Location: Bangalore
Followers: 1

Kudos [?]: 4 [0], given: 0

There is shorter way:
rearranging the given equation i get:
x = (y+1)/4 + 4y
Now we are looking for integer values of x,
Therefore, (y+1)/4 must be an integer.
At x=1000, y=235.2
So what i am looking for is number of multiples of 4 between 0 & 235.
(Because 1 less than mutilple of 4 will make (y+1)/4 integer)

Now u can calculate that there are 58 mutiples of 4 between 0 & 235.
does my answer matches official ans.
-Vicks
CEO
Joined: 15 Aug 2003
Posts: 3469
Followers: 61

Kudos [?]: 701 [0], given: 781

Vicky wrote:
There is shorter way:
rearranging the given equation i get:
x = (y+1)/4 + 4y
Now we are looking for integer values of x,
Therefore, (y+1)/4 must be an integer.
At x=1000, y=235.2
So what i am looking for is number of multiples of 4 between 0 & 235.
(Because 1 less than mutilple of 4 will make (y+1)/4 integer)

Now u can calculate that there are 58 mutiples of 4 between 0 & 235.
does my answer matches official ans.
-Vicks

Great Approach! Getting y be itself is a big leap towards the solution.
Just a bit off..but thats enough to get it wrong unfortunately.

Heres what i found out:

you are looking for y+1 /4
So you are looking for multiples of between 0 and 236.2

Thats 59.

Thanks for the solution
Praetorian
Senior Manager
Joined: 21 Aug 2003
Posts: 258
Location: Bangalore
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thanks for the correction. Really bad that after solving that much one does a silly mistake.
-vicks
Intern
Joined: 28 Aug 2003
Posts: 36
Location: USA
Followers: 0

Kudos [?]: 0 [0], given: 0

Vicky wrote:
thanks for the correction. Really bad that after solving that much one does a silly mistake.
-vicks

Good question! Vicks and Praet, thanks for the discussion.
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