inequalities

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inequalities

by nidhis.1408 » Sat Jul 07, 2012 5:51 am
If x > y, x < 6, and y> -3, what is the largest prime number that could be equal to x+ y?

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by Anurag@Gurome » Sat Jul 07, 2012 5:55 am
nidhis.1408 wrote:If x > y, x < 6, and y> -3, what is the largest prime number that could be equal to x+ y?
-3 < y < x < 6

As both x and y are less than 6, (x + y) must be less than 12.
Hence, the largest prime number that could be equal to (x + y) is the largest prime number number juts less than 12, i.e. 11.

This is possible for say, (x = 5.9 and y = 5.1) or (x = 5.6 and y = 5.4) etc.
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by everything's eventual » Wed Aug 22, 2012 12:52 am
Good Day Sir,

This was my way of solving the problem :

x < 6. So I considered x = 5

-3 < y < x. So y can be -3,-2,-1,0,1,2,3,4

From this max prime value of x + y = 7

To get x + y = 11, x needs to be 6 and y needs to be 5. But this is not possible as x < 6. Y cannot be 6 or more as y< x. So how can we get the sum as 11?

Please advise if there is a mistake in my understanding..

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by everything's eventual » Wed Aug 22, 2012 12:54 am
One error,

I meant y can be -2,-1,0,1,2,3,4...

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by cypherskull » Wed Aug 22, 2012 10:41 pm
From the problem statement, it can be deduced that -

-3< y < x <6.

The greatest value of x can be 5 which means y should then be less than 5.

Replacing the value of x as 5 in (x+y) and testing for the values of y < 5

5 + y (=4) => not prime
5 + y (=3) => not prime
5 + y (=2) => prime (Ans).
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by confuse mind » Wed Aug 22, 2012 11:48 pm
Nowhere given that x and y are integers and thus 11 is the largest possible prime value of x+y