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The ages of three friends are prime numbers. The sum of the ages is

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mrinal2100
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The ages of three friends are prime numbers. The sum of the ages is [#permalink] New post   21 Oct 2010, 11:04
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The ages of three friends are prime numbers. The sum of the ages is less than 51. If the ages are in Arithmetic Progression (AP) and if at least one of the ages is greater than 10, what is the difference between the maximum possible median and minimum possible median of the ages of the three friends?

(A) 0
(B) 1
(C) 13
(D) 6
(E) 8
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D
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Re: The ages of three friends are prime numbers. The sum of the ages is [#permalink] New post   22 Oct 2010, 06:07
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When you want to take 3 numbers in AP, always take them to be a-d, a, a+d (as Gurpreet did above)
Note: On similar lines, 4 numbers in AP should be taken as a-3d, a-d, a+d, a+3d
This is done so that when the numbers are added, we can get rid of d.

If a-d + a + a+d < 51, we get a < 17

To get Minimum Median: Take smallest possible values. We need to take at least one number > 10 so take the greatest number as 11. Can you make an AP of prime numbers where 11 is the greatest number? Sure, 3, 7 and 11.
Note: When looking for prime numbers, consider only those numbers that end with 1/3/7/9

To get Maximum Median, median being 'a' that we assumed above, a <17 so the greatest median possible is 13. Now confirm whether you can you make an AP of primes with 13 as the middle number. I got one- 7, 13, 19. That's enough.

The difference between the maximum and minimum median will be 13 - 7 = 6
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Re: The ages of three friends are prime numbers. The sum of the ages is [#permalink] New post   21 Oct 2010, 12:02
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mrinal2100 wrote:
The ages of three friends are prime numbers. The sum of the ages is less than 51. If the ages are in Arithmetic Progression (AP) and if at least one of the ages is greater than 10, what is the difference between the maximum possible median and minimum possible median of the ages of the three friends?

(A) 0
(B) 1
(C) 13
(D) 6
(E) 8

i got the ans as
Show Spoiler
a


Let the numbers are a-d, a , a+d

sum <51 => a-d+a+a+d < 51 => middle number a <17

prime numbers are 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47

a can be 3,5,7,11,13

if a is 3 -> a-d can be 2 only; a+d not possible thus reject this.
if a is 5, a-d can be 2 and 3; only a-d = 3 satisfies
=> three numbers can be 3,5,7 ->reject this , as it is given at least one is >10

if a = 7 ; a-d can be 3 and a+d can be 11
This satisfies all the conditions hence the median is 7=> min median.

if a = 11, the three element pair is 3,11,19 or 5,11,17
median = 11; we are not sure if its the max median possible

if a = 13, 7,13,19 pair is possible
median = 13; reject 11 as the max.

thus difference = 13-7 = 6
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Re: The ages of three friends are prime numbers. The sum of the ages is [#permalink] New post   Updated on: 21 Oct 2010, 11:51
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Even i do agree with ans: A

Here is my version of explanation:

Let the 3 prime numbers be: x, y, z

Given: x + y + z = 51
Assume: x is smallest prime number among 3 numbers

Since these 3 are in Arithmetic progression:
y = x + d
z = x + 2d
==> x + y + z = x + (x + d) + (x + 2d) = 51
==> x + d = 17

Now, valid combinations are:
S.NoxdyzComments
13141731Valid solution
21161723Valid solution
31701717Valid solution


From above table three valid solutions (1) (2) & (3), The maximum median is: 17, The minimum median is:17
Hence, the difference is zero.

Cheers!
Ravi

Originally posted by nravi549 on 21 Oct 2010, 11:49.
Last edited by nravi549 on 21 Oct 2010, 11:51, edited 1 time in total.
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The ages of three friends are prime numbers. The sum of the ages is [#permalink] New post   05 Nov 2016, 05:40
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mrinal2100 wrote:
The ages of three friends are prime numbers. The sum of the ages is less than 51. If the ages are in Arithmetic Progression (AP) and if at least one of the ages is greater than 10, what is the difference between the maximum possible median and minimum possible median of the ages of the three friends?

(A) 0
(B) 1
(C) 13
(D) 6
(E) 8


Other approach for the sake of diversity.

To maximize the sum (get it close to 51) prime numbers in progression must be \(>3\), and every prime number which is \(>3\) can be expressed as \(6k+1\) or \(6k-1\).

Let’s use this to find maximum median directly.

We will construct arithmetic progression in which every next prime has the same difference with previous one.

\(6k+1+6(k+1)+1+6(k+2)+1<51\)

from where our \(k<1,7\) The only integer value \(k\) can take is \(1\). (In case \((6k-1)\) we stiil have \(k<2\))

We have following max sum: \(6*1+1+6*2+1+6*3+1=7+13+19\)
Our Max median is \(13\).

To find Min median we’ll take maximum value in AP as \(11\), next smaller prime is \(7\). And to keep same common difference we take \(3\) as a third element of AP. So smaller median is \(7\).

Difference is \(13-7=6\)
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Re: The ages of three friends are prime numbers. The sum of the ages is [#permalink] New post   23 Mar 2017, 11:46
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mrinal2100 wrote:
The ages of three friends are prime numbers. The sum of the ages is less than 51. If the ages are in Arithmetic Progression (AP) and if at least one of the ages is greater than 10, what is the difference between the maximum possible median and minimum possible median of the ages of the three friends?

(A) 0
(B) 1
(C) 13
(D) 6
(E) 8



i used answer choices...

find the lowest set..
ie { 3,7,11}
median=7
now add answer choices to 7 ..result will be prime again
7+0=7 ..No (we could not find such set with max. values)
7+1=8 ..not prime
7+13=20..not prime
7+6=13...Yes ,keep it
7+8=15..not prime

Ans D
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Re: The ages of three friends are prime numbers. The sum of the ages is [#permalink] New post   21 Oct 2010, 11:29
The largest possible set of primes that fulfills this is 11, 13 and 17, since they have a common difference of 2 and they are prime numbers. So the largest possible median has to be 13.

Let's look at the smallest possible set of primes.

2, 3, 5 - Not in AP
3, 5, 7 - Not in AP
5, 7, 11 - AP!
7, 11, 13 - AP with common difference -2. 13 - 2 = 11, 11 - 4 = 7. But the median here is bigger than the median in the previous case. Hence we can rule this out.

The smallest possible set of primes that fulfills this is 5, 7 and 11. So this case, the median is 7 and hence the difference is 13 - 7 = 6

Is the answer D?
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Re: The ages of three friends are prime numbers. The sum of the ages is [#permalink] New post   21 Oct 2010, 12:03
whiplash2411 wrote:
The largest possible set of primes that fulfills this is 11, 13 and 17, since they have a common difference of 2 and they are prime numbers. So the largest possible median has to be 13.

Let's look at the smallest possible set of primes.

2, 3, 5 - Not in AP
3, 5, 7 - Not in AP
5, 7, 11 - AP!
7, 11, 13 - AP with common difference -2. 13 - 2 = 11, 11 - 4 = 7. But the median here is bigger than the median in the previous case. Hence we can rule this out.

The smallest possible set of primes that fulfills this is 5, 7 and 11. So this case, the median is 7 and hence the difference is 13 - 7 = 6

Is the answer D?


how the bolded parts are in AP?? check again.
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Re: The ages of three friends are prime numbers. The sum of the ages is [#permalink] New post   21 Oct 2010, 12:05
nravi549 wrote:
Even i do agree with ans: A

Here is my version of explanation:

Let the 3 prime numbers be: x, y, z

Given: x + y + z = 51
Assume: x is smallest prime number among 3 numbers

Since these 3 are in Arithmetic progression:
y = x + d
z = x + 2d
==> x + y + z = x + (x + d) + (x + 2d) = 51
==> x + d = 17

Now, valid combinations are:
S.NoxdyzComments
13141731Valid solution
21161723Valid solution
31701717Valid solution


From above table three valid solutions (1) (2) & (3), The maximum median is: 17, The minimum median is:17
Hence, the difference is zero.

Cheers!
Ravi


Read closely. The sum of ages are less than 51, not equal to 51.
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Re: The ages of three friends are prime numbers. The sum of the ages is [#permalink] New post   21 Oct 2010, 12:07
Oops..mis read it.
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Re: The ages of three friends are prime numbers. The sum of the ages is [#permalink] New post   21 Oct 2010, 12:09
The sum of the ages is less than 51. Not 51.

And AP doesn't necessarily mean the common difference is positive. It can be negative also. If you consider the series 13, 11 and 7, we have common difference of -2. It's a descending Arithmetic Progression.
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Re: The ages of three friends are prime numbers. The sum of the ages is [#permalink] New post   21 Oct 2010, 12:10
But, either way. Once we go beyond the first AP I mentioned, the median becomes larger and larger, so we can disregard the medians that are in between the largest and smallest. But this brings me to a question. When it says it's LESSER, is 51 included?
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Re: The ages of three friends are prime numbers. The sum of the ages is [#permalink] New post   21 Oct 2010, 12:10
whiplash2411 wrote:
The sum of the ages is less than 51. Not 51.

And AP doesn't necessarily mean the common difference is positive. It can be negative also. If you consider the series 13, 11 and 7, we have common difference of -2. It's a descending Arithmetic Progression.


does 11+ (-2) = 7? :P

moreover it does not matter whether the d is -ve or +ve. one of the number will always be greater than the median and the other smaller.
We are only concerned with the MEDIAN.
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Re: The ages of three friends are prime numbers. The sum of the ages is [#permalink] New post   21 Oct 2010, 12:20
I'm sleep deprived. Ignore that part of my answer.
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Re: The ages of three friends are prime numbers. The sum of the ages is [#permalink] New post   18 Jun 2016, 07:37
VeritasPrepKarishma wrote:
When you want to take 3 numbers in AP, always take them to be a-d, a, a+d (as Gurpreet did above)
Note: On similar lines, 4 numbers in AP should be taken as a-3d, a-d, a+d, a+3d
This is done so that when the numbers are added, we can get rid of d.

If a-d + a + a+d < 51, we get a < 17

To get Minimum Median: Take smallest possible values. We need to take at least one number > 10 so take the greatest number as 11. Can you make an AP of prime numbers where 11 is the greatest number? Sure, 3, 7 and 11.
Note: When looking for prime numbers, consider only those numbers that end with 1/3/7/9

To get Maximum Median, median being 'a' that we assumed above, a <17 so the greatest median possible is 13. Now confirm whether you can you make an AP of primes with 13 as the middle number. I got one- 7, 13, 19. That's enough.

The difference between the maximum and minimum median will be 13 - 7 = 6


Your approach is good. But the marked portion does not seem to be correct.
I think it should be:a-2d, a-d, a, a+d
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The ages of three friends are prime numbers. The sum of the ages is [#permalink] New post   23 Mar 2017, 13:36
mrinal2100 wrote:
The ages of three friends are prime numbers. The sum of the ages is less than 51. If the ages are in Arithmetic Progression (AP) and if at least one of the ages is greater than 10, what is the difference between the maximum possible median and minimum possible median of the ages of the three friends?

(A) 0
(B) 1
(C) 13
(D) 6
(E) 8


lowest possible AP=3,7,11
highest possible AP=7,13,19 or 3,13,23
13-7=6
D
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Re: The ages of three friends are prime numbers. The sum of the ages is [#permalink] New post   12 Sep 2019, 11:27
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mrinal2100 wrote:
The ages of three friends are prime numbers. The sum of the ages is less than 51. If the ages are in Arithmetic Progression (AP) and if at least one of the ages is greater than 10, what is the difference between the maximum possible median and minimum possible median of the ages of the three friends?

(A) 0
(B) 1
(C) 13
(D) 6
(E) 8


The set with the minimum possible median is {3, 7, 11}, and one of the sets with the maximum possible median is {3, 13, 23}. (Note that {7, 13, 19} is another one.) Therefore, the difference is 13 - 7 = 6.

(Note: The fact that the sum of the ages is less than 51 and the ages are in an arithmetic progression means the median age must be less than 51/3 = 17.)

Answer: D
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Re: The ages of three friends are prime numbers. The sum of the ages is [#permalink] New post   12 Nov 2022, 22:20
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