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If p and q are both positive integers, then is p divisible by 9?

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If p and q are both positive integers, then is p divisible by 9?  [#permalink]

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New post 05 Mar 2019, 16:02
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If p and q are both positive integers, then is p divisible by 9?

1) p/10 + q/10 is an integer.

2) p/9 + q/10 is an integer.
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Re: If p and q are both positive integers, then is p divisible by 9?  [#permalink]

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New post 05 Mar 2019, 20:02
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If p and q are both positive integers, then is p divisible by 9?

1) \(\frac{p}{10} + \frac{q}{10}\) is an integer.
Various ways possible..
If p=q=5, \(\frac{p}{10} + \frac{q}{10}=1\) , but p or 5 is not divisible by 9.
If p=9 and q=1, \(\frac{p}{10} + \frac{q}{10}=1\) , but p or 9 is divisible by 9.
Thus, insufficient

2) \(\frac{p}{9} + \frac{q}{10}\) is an integer.
Now the denominator are different in both cases, so let us solve
\(\frac{p}{9} + \frac{q}{10}=x....10p+9q=90x...10p=90x-9q=9(10x-q)\)
Now 10p is surely a multiple of 9, so only way i sthat p is a multiple of 9
Sufficient

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Re: If p and q are both positive integers, then is p divisible by 9?  [#permalink]

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New post 06 Mar 2019, 09:01
selim wrote:
If p and q are both positive integers, then is p divisible by 9?

1) p/10 + q/10 is an integer.

2) p/9 + q/10 is an integer.


IF YOU FIND MY SOLUTION TO BE HELPFUL, PLEASE GIVE ME KUDOS

(1) p/10 + q/10 is an integer

Let p =8, q = 2, This satisfies our constraint and p is not divisible by 9.

Let p =9 and q = 1. This satisfies our constraint and p is divisible by 9 NS

(2) p/9 +q/10 is an integer.

So, our target denominator for the sum of our terms is going to be 90 (we will have to convert p/9 + q/10 to a common denominator of 90). This results in q being multiplied by 9 and p being multiplied by 10 to convert to the LCD of 90.
Then our q term will contribute one of the following values to the numerator of our sum 9(1), 9(2), 9(3)...= 9,18,27,36,45,54,63,72,81,90....
Our p term will be multiplied by 10 when converting to our common denominator. So, our p term will contribute one of the following values to our final sum:
10,20,30,40,50,60,70,80,90,.... Notice, that our sum will only be divisible by 90, if both 10p and 9q are both multiples of 90, therefore p is a multiple of 9 sufficient.
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Re: If p and q are both positive integers, then is p divisible by 9?  [#permalink]

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New post 06 Mar 2019, 09:35
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selim wrote:
If p and q are both positive integers, then is p divisible by 9?

1) p/10 + q/10 is an integer.

2) p/9 + q/10 is an integer.


The property, when any positive integer is divided by 9, is the division leaves a non-terminating decimal except for multiple of 9, i.e.:-

Positive
Integer

1 Divided by 9 = 0.111111111
2 Divided by 9 = 0.222222222
3 Divided by 9 = 0.333333333
4 Divided by 9 = 0.444444444
5 Divided by 9 = 0.555555556
6 Divided by 9 = 0.666666667
7 Divided by 9 = 0.777777778
8 Divided by 9 = 0.888888889
9 Divided by 9 = 1
10 Divided by 9 = 1.111111111
11 Divided by 9 = 1.222222222
12 Divided by 9 = 1.333333333
13 Divided by 9 = 1.444444444
14 Divided by 9 = 1.555555556
15 Divided by 9 = 1.666666667
16 Divided by 9 = 1.777777778
17 Divided by 9 = 1.888888889
18 Divided by 9 = 2
19 Divided by 9 = 2.111111111

Since it is mentioned that both p and q are integers the only possibility is "p" is a multiple of 9 since "q" is also a positive integer.

Correct Ans B - Sufficient!! Straight and simple!!
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Re: If p and q are both positive integers, then is p divisible by 9?   [#permalink] 06 Mar 2019, 09:35
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