Daniel's quadruple differences

Wrapping our head around this one is a trick, but with careful reading, the meaning can be determined.

For each ordered quadruple, $(a, b, c, d)$ ,

$k = (a-b)+(a-c)+(a-d)+(b-c)+(b-d)+(c-d)$ $= 3a+b-c-3d = 3(a-d) + (b-c)$

Since the difference between the highest and lowest values, $(a-d)$ , must be at least 3 but not more than 9 and the difference between the middle two terms, $(b-c)$ , must be at least 1 but not more than 7, we can now consider the short list of values that k can be.

• When $a-d = 3$ => $b-c = 1$ => $k = 10$ . This can happen in 7 ways... (4, 3, 2, 1) through (10, 9, 8, 7)

• $a-d = 4$ => $b-c = 1, 2$ => $k = 13, 14$ . This can happen in 12 and 6 ways respectively.

• $a-d = 5$ => $b-c = 1, 2, 3$ => $k = 16, 17, 18$ . This can happen in 15, 10 and 5 ways respectively.

• $a-d = 6$ => $b-c = 1, 2, 3, 4$ => $k = 19, 20, 21, 22$ . This can happen in 16, 12, 8 and 4 ways respectively.

• $a-d = 7$ => $b-c = 1, 2, 3, 4, 5$ => $k = 22, 23, 24, 25, 26$ . This can happen in 15, 12, 9, 6 and 3 ways respectively.

• $a-d = 8$ => $b-c = 1, 2, 3, 4, 5, 6$ => $k = 25, 26, 27, 28, 29, 30$ . This can happen in 12, 10, 8, 6, 4 and 2 ways respectively.

• $a-d = 9$ => $b-c = 1, 2, 3, 4, 5, 6, 7$ => $k = 28, 29, 30, 31, 32, 33, 34$ . This can happen in 7, 6, 5, 4, 3, 2 and 1 ways respectively.

Thus $N(10)=7, N(13)=12, N(14)=6,..., N(34)=1$ . However, the highest value of $N(k)$ occurs when $k=22$ . $N(22)=19$ so the solution is 22 .

The 4 ordered quadruples where $a-d=6$ and $b-c=4$ thus $k=22$ are

(7, 6, 2, 1), (8, 7, 3, 2), (9, 8, 4, 3), (10, 9, 5, 4)

and the 15 ordered quadruples where $a-d = 7$ and $b-c = 1$ thus $k=22$

(8, 3, 2, 1), (8, 4, 3, 1), (8, 5, 4, 1), (8, 6, 5, 1), (8, 7, 6, 1), (9, 4, 3, 2), (9, 5, 4, 2), (9, 6, 5, 2), (9, 7, 8, 2), (9, 8, 7, 2), (10, 5, 4, 3), (10, 6, 5, 3), (10, 7, 6, 3), (10, 8, 7, 3), (10, 9, 8, 3)

Since $d<c<b<a$ , six pairwise positive differences are $a-b,a-c,a-d,b-c,b-d,c-d$ . Their sum is $3(a-d)+(b-c)$ . Hence the for the given $k$ we should have

$3(a-d)+(b-c)=k$

Now, since $a<11$ , maximum possible value of $a$ is $10$ . Using similar logic about lower bound and differences we can see that maximum possible value of $d$ is $7$ , minimum possible value of $d$ is $1$ , maximum possible value of $b$ is $9$ and so on. (Its quite easy to get all of these).

This gives us possible values for $x=a-d$ and $y=b-c$ as follows:

$x=(3,4,5,6,7,8,9)$ and $y=(1,2,3,4,5,6,7)$ .

Also observe that, we must have $(a-d)>b-c+1$ . This gives:

$y<(x-1)$ .

Thus, for a given value of $x$ , only certain values of $y$ are allowed. These allowed $(x,y)$ pairs and corresponding $k$ values are listed below ( $k=3x+y$ ):

$x=3 \Rightarrow y=1 \Rightarrow k=10$

$x=4 \Rightarrow y=1,2 \Rightarrow k=13,14$

$x=5 \Rightarrow y=1,2,3 \Rightarrow k=16,17,18$

$x=6 \Rightarrow y=1,2,3,4 \Rightarrow k=19,20,21,22$

$x=7 \Rightarrow y=1,2,3,4,5 \Rightarrow k=22,23,24,25,26$

$x=8 \Rightarrow y=1,2,3,4,5,6 \Rightarrow k=25,26,27,28,29,30$

$x=9 \Rightarrow y=1,2,3,4,5,6,7 \Rightarrow k=28,29,30,31,32,33,34$

Now we want to find out number of ordered quadruples for each pair $(x,y)$ . To do this, observe that for given $x$ , a can take values from $x+1$ to $10$ . Hence there are $(10-x)$ choices for $a$ . Then for a given $a$ , we have $d=a-x$ . Also then $b$ can take values from $(a-1)$ till $c$ becomes $d+1 = a-x+1$ .

Hence, $c$ varies from $a-x+1$ to $a-1-y$ . Combining these, we get $a-1-y-(a-x+1)+1=x-y-1$ choices for $c$ and then for a given $c$ , $d$ is determined.

Thus, in total for a pair $(x,y)$ , there are $(10-x)*(x-y-1)$ quadruples. We can see that this value is greatest when $y=1$ . But from above list, we see that, some $k$ values are repeated. Using this formula for those $k$ values for which either $y=1$ or $k$ is repeated or both, we see that for $k=22$ , we get $N(k)=19$ and there is no other $k$ which gives $N(k)$ equal to or greater than $19$ . Hence the required answer is $\boxed{22}$ .

k=(a-b)+(a-c)+(a-d)+(b-c)+(b-d)+(c-d)=3(a-d)+(b-c)........(1)

from the condition given above

a E [4,10],b E [3,9], c E [2,8] ,d E [1,7] ................(2)

thus from (2) min(a-d)=3,min(b-c)=1 and max(a-d)=9,max(b-c)=7....................(3)

thus from (1) and (3) k E[10,34]

also from (3) min((a-d)-(b-c))=2 and max((a-d)-(b-c))=8.................................(4)

now from 1,k=3p+q where p=a-d and q=b-c

hence from(3) and (4) p E [3,9],q E [1,7] and p-q E [2,8]

Case 1: q=1 i.e k=3p+1

it will cover the numbers 10,13,16,19,22,25,28 as p E [3,9]

N(10)=7 1,N(13)=6 2,N(16)=5 3,N(19)=4 4.......

note that N(K)=m*(8-m) where m is integer greater than 1

N(K)=7,12,15,16,15,12,7 respectively for above k

proceeding in the above way we'll find the value of N(K) for 3p+2,3p+3,3p+4,3p+5,3p+6,3p+7

to save the time rather than proceeding the above way,we'll find the value of k such that we'll get

the value of N(K) from multiple types i.e 3p+1,3p+4,3p+7 for different p maybe the same

such numbers are 22,25,26.28.29 and 30 if we keep in mind (2),(3) and (4)

i.e 22=3(7)+(1)=3(6)+4,25=3(8)+1=3(7)+4,26=3(8)+2=3(7)+5,28=3(9)+1=3(8)+4

note that above numbers are written in the form 3p+q

N(22)=3 5+4 1=19,N(25)=3 4+2 3=18,N(26)=2 5+3 1=13,

so the answer is 22.....