Can You Simplify This?
Find the smallest integer n greater than 100 such that the fraction 1 0 n + 3 9 n + 5 will be reducible.
The answer is 117.
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3 solutions
Can you please explain you did you get g cd ( 9 n + 5 , n − 2 ) = g cd ( n − 2 , 2 3 ) ?
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Divide 9n+5 by n-2 and 23 is remainder.
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Correct! 23 can also be found by repeated subtraction of n − 2 from 9 n + 5 , which is about the same as finding the remainder.
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@Arulx Z – Thanks for clearing my doubt.
Can't it be 102?
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What n? If yes then no
100 is not a multiple of 23. You can try plugging it into the expression.
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Man , i didn't understand but anyways i'll surely figure it out..
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@Vaibhav Basantani – Feel free to ask any questions. Which part didn't you understand?
1 0 n + 3 9 n + 5 = 1 − 1 0 n + 3 n − 2 = 1 − 1 0 + n − 2 2 3 1 will have 117 smallest n > 100 so that n - 2 is multiple of 23. Answer: 117
Can you explain what your solution has to do with "not an irreducible fraction"?
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If g c d ( 2 3 , n − 2 ) = 1 then there will be a "not an irreducible fraction".Tautological. I think it is self-explanatory.
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I agree that "if g cd ( 9 n + 5 , 1 0 n + 3 ) = 1 then it is an irreducible fraction". I also agree thatn "If g cd ( 2 3 , n − 2 ) = 1 , then it is an irreducible fraction". But how does that relate to the working that you've shown?
What i'm asking, is "Explain how your solution demonstrates that g cd ( 9 n + 5 , 1 0 n + 3 ) = g cd ( 2 3 , n − 2 ) ". Or alternatively, why does " n − 2 2 3 is not irreducible implies that 1 − 1 0 + n − 2 2 3 1 is not irreducible.
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@Calvin Lin – I think this is a property of fractions even continued ones.
Let some k : k ∈ Z , k ∣ 9 n + 5 , 1 0 n + 3
k ∣ 9 n + 5 ⇒ k ∣ 1 0 ( 9 n + 5 ) ⇒ k ∣ 9 0 n + 5 0 .
Also , k ∣ 1 0 n + 3 ⇒ k ∣ 9 ( 1 0 n + 3 ) ⇒ k ∣ 9 0 n + 2 7 .
∴ k ∣ 9 0 n + 5 0 − ( 9 0 n + 2 7 ) ⇒ k ∣ 2 3 ⇒ k = 2 3 .
Therefore if the fraction is not an irreducible fraction then 2 3 divides the numerator and the denominator.
So 2 3 ∣ 9 n + 5 and 2 3 ∣ 1 0 n + 3 ⇒ 2 3 ∣ ( n − 2 ) .
2 3 ∣ ( n − 2 ) ⇒ n = 2 5 , 4 8 , 7 1 , 9 4 , 1 1 7 . It is easy to verify these values of n satisfy the required conditions.
Our answer is 1 1 7 .
Moderator note:
Be careful with your presentation. While I understand what you are trying to say, the statement of k ∣ 1 0 n + 3 = k ∣ 9 ( 1 0 n + 3 ) can be confusing to other people who are trying to understand your solution.
Note that these 2 equations are not equivalent to each other, which is what further confuses the equality sign.
@Calvin Lin Sir can you please suggest the necessary changes in my solution....
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Be explicit about what you mean / what you want to say, and ensure that it is mathematically correct.
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@Calvin Lin Sir I have changed the equality sign with implies ⇒ sign...
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@Ankit Kumar Jain – Good. And because we only have a one directional ⇒ , this tells us that the final answers need not necessarily satisfy the original equation. Hence (for the case of this solution), we still have to verify that the values actually work.
For the fraction to not be in the lowest terms, g cd ( 9 n + 5 , 1 0 n + 3 ) > 1 .
By euclidean algorithm ,
g cd ( 9 n + 5 , 1 0 n + 3 ) = g cd ( 9 n + 5 , n − 2 ) = g cd ( n − 2 , 2 3 )
23 must divide n − 2 if the fraction is not in lowest term. Smallest integer n greater than 100 which satisfies the following property is 2 3 ⋅ 5 + 2 = 1 1 7 .