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26 Mar 2020, 09:07
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If a and b are positive integers, a > b, is \(3^a – 3^b\) divisible by 13?
(I) a – b = 3
(II) a + b = 5
(I) a – b = 3
(II) a + b = 5
26 Mar 2020, 09:37
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sjuniv32
If a and b are positive integers, a > b, is \(3^a – 3^b\) divisible by 13?
(I) a – b = 3
(II) a + b = 5
(I) a – b = 3
(II) a + b = 5
a and b as positive integers and a>b should tell you to further work on the equation.
\(3^a-3^b=3^{a-b+b}-3^b=3^{a-b}*3^b-3^b=3^b(3^{a-b}-1)\), after taking out 3^b as common out of two terms.
(1) a-b=3..
So \(3^b(3^{a-b}-1)=3^b(3^3-1)=3^b(27-1)=3^b*26\)
This is surely divisible by 13 as 13*2=26.
Suff
(2) a+b=5
As a and b are positive, there are only two possibilities
(a) a=4 and n=1...
\(3^a-3^b=3^4-3^1=81-3=78\)...Hence divisible by 13, as 13*6=78.
(b) a=3 and b=2
\(3^a-3^b=3^3-3^2=18\).... No, as 18 is not divisible by 13
Insufficient
A
02 Apr 2020, 03:58
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When I first looked at the question stem, the first set of values I thought of was a=3 and b=0. With these values, 3^3 – 3^0 = 26 which is divisible by 13. BUT, b has to be a positive integer. The above process sets the ball rolling in terms of taking values for a and b.
From the question data, we know that a≠1 and b≠0. We also know that a>b.
From statement I alone, a-b = 3. Let’s try a few values.
If a = 4 and b = 1, a-b = 3. \(3^4\) – \(3^1\) = 81 – 3 = 78, which is divisible by 13.
If a = 5 and b = 2, a-b = 3. \(3^5\) – \(3^2\) = 243 – 9 = 234 which is divisible by 13.
If a = 6 and b= 3, a-b = 3. \(3^6\) – \(3^3\) = 729 – 27 = 702 which is divisible by 13.
We see that for any set of values we take for a and b such that a-b = 3, \(3^a\) – \(3^b\) is always divisible by 3. The mathematical explanation for this has been brilliantly illustrated by chetan2u.
Statement I alone is sufficient to answer the main question with a definite YES. Answer options B, C and E can be eliminated, possible answer options are A or D.
From statement II alone, a+b = 5.
If a=4 and b = 1, a+b=5. \(3^4\) – \(3^1\) = 81 – 3 = 78, which is divisible by 13.
If a = 3 and b = 2, a+b = 5. \(3^3\) – \(3^2\) = 27 – 9 = 18, which is not divisible by 13.
Statement II alone gives us a Yes AND a NO and hence insufficient. Answer option D can be eliminated.
The correct answer option is A.
Hope that helps!
From the question data, we know that a≠1 and b≠0. We also know that a>b.
From statement I alone, a-b = 3. Let’s try a few values.
If a = 4 and b = 1, a-b = 3. \(3^4\) – \(3^1\) = 81 – 3 = 78, which is divisible by 13.
If a = 5 and b = 2, a-b = 3. \(3^5\) – \(3^2\) = 243 – 9 = 234 which is divisible by 13.
If a = 6 and b= 3, a-b = 3. \(3^6\) – \(3^3\) = 729 – 27 = 702 which is divisible by 13.
We see that for any set of values we take for a and b such that a-b = 3, \(3^a\) – \(3^b\) is always divisible by 3. The mathematical explanation for this has been brilliantly illustrated by chetan2u.
Statement I alone is sufficient to answer the main question with a definite YES. Answer options B, C and E can be eliminated, possible answer options are A or D.
From statement II alone, a+b = 5.
If a=4 and b = 1, a+b=5. \(3^4\) – \(3^1\) = 81 – 3 = 78, which is divisible by 13.
If a = 3 and b = 2, a+b = 5. \(3^3\) – \(3^2\) = 27 – 9 = 18, which is not divisible by 13.
Statement II alone gives us a Yes AND a NO and hence insufficient. Answer option D can be eliminated.
The correct answer option is A.
Hope that helps!
