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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.
1) p/10 + q/10 is an integer.
2) p/9 + q/10 is an integer.
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
B
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
B
General Discussion
06 Mar 2019, 09:01
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selim
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.
