Solve using proofs/justification

Hi, I have a question that needs to be solved using "clear, concise and accurate mathematical justification/proofs." I am having some difficulty on understanding where to start and how to solve the problem. Any help will be appreciated. Below is the question:

1) You are given four digits (all strictly positive, i.e. greater than 0)

arranged in a square with a distinct digit from 1 to 9 in each quarter. For example:

5 6

3 2

You make this arrangement into 4 integers by reading horizontally and vertically

and then adding these numbers up, to get the sum S. So, in the above example,

S = 56 + 32 + 53 + 62 = 203

In how many diﬀerent ways can you ﬁll in such a square so that S = 200?

Re: Solve using proofs/justification

Quote:

Originally Posted by

**NFS1** Hi, I have a question that needs to be solved using "clear, concise and accurate mathematical justification/proofs." I am having some difficulty on understanding where to start and how to solve the problem. Any help will be appreciated. Below is the question:

1) You are given four digits (all strictly positive, i.e. greater than 0)

arranged in a square with a distinct digit from 1 to 9 in each quarter. For example:

5 6

3 2

You make this arrangement into 4 integers by reading horizontally and vertically

and then adding these numbers up, to get the sum S. So, in the above example,

S = 56 + 32 + 53 + 62 = 203

In how many different ways can you fill in such a square so that S = 200?

I won't do the question for you, but here's how you might start. Suppose that the four distinct digits are , arranged in the form . The four numbers to be added are , , and Adding them up, we get the equation For the left side to be an even number, must be even, say . Then the equation becomes

For each possible value of from 1 to 8 ( cannot be 9) there is exactly one possible pair of values for k and d. So you can make a table, like this:

. . . . . . . . . .

(I have just filled in the first row and column). For each column in the table, you then have to see how many ways you can choose and so that (remembering each time that the four digits must all be distinct). Having done that, you can just count up the total number of solutions.

Re: Solve using proofs/justification

Hello, NFS1!

We have the array: .

The two-digit numbers are: .

Their sum is 200: .

. .

. .

We have: .

is a two-digit number

. . and must be divisible by 11.

There are only 8 cases to consider.

. . . 2 ways

. . . 4 ways

. . . 2 ways

. . . 6 ways

. . . 2 ways

. . . 4 ways

. . . 2 ways

. . . 0 ways

Hence, there are: .

Re: Solve using proofs/justification

Quote:

Originally Posted by

**Soroban**

No they can't.

Re: Solve using proofs/justification

both Opalg's and Soroban's method both give that a=8 is not possible. although their reasoning in finding b+c differ, both give the same possible pairs (b,c).

so, given a in {1,2,3,4,5,6,7,8}, if 10a+d+11(b+c)/2 = 100, there can be only one possible choice for (b+c)/2 and d.

for example, if a = 3, then 70 - d is divisible by 11. since d < 10 (in fact, it's actually < 9), (b+c)/2 has to be 6, so b+c = 12, and from

70 - d = 66, we get d = 4.

if we choose a = 4, we do NOT get d = 3. if a = 4, then 60 - d is divisible by 11, so (b+c)/2 = 5, hence b+c = 10. from 60 - d = 55, we get d = 5.

so each pair (a,d) is uniquely defined, we need only count them once.

Re: Solve using proofs/justification

Quote:

Originally Posted by

**Deveno** both Opalg's and Soroban's method both give that a=8 is not possible. although their reasoning in finding b+c differ, both give the same possible pairs (b,c).

so, given a in {1,2,3,4,5,6,7,8}, if 10a+d+11(b+c)/2 = 100, there can be only one possible choice for (b+c)/2 and d.

for example, if a = 3, then 70 - d is divisible by 11. since d < 10 (in fact, it's actually < 9), (b+c)/2 has to be 6, so b+c = 12, and from

70 - d = 66, we get d = 4.

if we choose a = 4, we do NOT get d = 3. if a = 4, then 60 - d is divisible by 11, so (b+c)/2 = 5, hence b+c = 10. from 60 - d = 55, we get d = 5.

so each pair (a,d) is uniquely defined, we need only count them once.

so they are only 22 pairs??

Re: Solve using proofs/justification

I also have a second part to the problem that i could also do with some help with:

B) In the above example S = 203. Clearly

83 <= S <= 357: (1) (less than or equal to)

(if you allow repetitions then 44 <= S <= 396) For how many values of S satisfying

the inequality (1) can a square not be constructed. For example, it is not possible

to construct a square with S = 4

Re: Solve using proofs/justification

Quote:

Originally Posted by

**NFS1** I also have a second part to the problem that i could also do with some help with:

B) In the above example S = 203. Clearly

(if you allow repetitions then 44 <= S <= 396) For how many values of S satisfying

the inequality (1) can a square not be constructed. For example, it is not possible

to construct a square with S = 4

I don't see any easy way to do this. You have to go back to the equation and check (for each value of S) whether there is a solution in which a, b, c and d are distinct digits. The smallest possible value of S is 83 (obtained when a=1, b=2, c=3, d=4) and the largest possible value is 357 (a=9, b=8, c=7, d=6). The only values of S for which no solution is possible are going to occur when S is fairly close to those extreme values. When S is well inside the interval [83,357] (for example, when S=200, as in the original problem) there is enough flexibility in the choice of a,b,c,d to ensure that there will always be at least one solution.

Notice also that if (a,b,c,d) is a solution for S, then (10–a,10–b,10–c,10–d) will be a solution for 440–S. So the values of S for which there is no solution will come in pairs adding up to 440. I think that there are 22 such values, namely

. . . . . . . . . .

Re: Solve using proofs/justification

Quote:

Originally Posted by

**Opalg** I don't see any easy way to do this. You have to go back to the equation

and check (for each value of S) whether there is a solution in which a, b, c and d are distinct digits. The smallest possible value of S is 83 (obtained when a=1, b=2, c=3, d=4) and the largest possible value is 357 (a=9, b=8, c=7, d=6). The only values of S for which no solution is possible are going to occur when S is fairly close to those extreme values. When S is well inside the interval [83,357] (for example, when S=200, as in the original problem) there is enough flexibility in the choice of a,b,c,d to ensure that there will always be at least one solution.

Notice also that if (a,b,c,d) is a solution for S, then (10–a,10–b,10–c,10–d) will be a solution for 440–S. So the values of S for which there is no solution will come in pairs adding up to 440. I think that there are 22 such values, namely

. . . . . . . . . .

Thanks for the reply, but how did you reach the 22 values using "clear, concise and accurate mathematical justification/proofs." Sorry for being a pain

Re: Solve using proofs/justification

Quote:

Originally Posted by

**NFS1** how did you reach the 22 values using "clear, concise and accurate mathematical justification/proofs."

I didn't. That suggested solution is just a mixture of trial and error, and guesswork. I was hoping that someone else would confirm or correct it, and hopefully indicate a more convincing method.