Arithmetic Grid

Algebra Level 4

$\begin{array} { | r | r | r | r | r | r | r | r | } \hline {0} & {XX} & {XX} & {XX} & {XX} & {XX} & {XX} & {XXX} \\ \hline {XX} & {XXX} & {XXX} & {XXX} & {XXX} & {XXX} & {XXX} & {XXX} \\ \hline {XXX} & {XXX} & {XXX} & {XXX} & {XXX} & {XXX} & {XXX} & {XXX} \\ \hline {XXX} & {XXX} & {XXX} & {XXX} & {XXX} & {XXX} & {XXXX} & {XXXX} \\ \hline {XXX} & {XXX} & {XXX} & {XXX} & {XXXX} & {XXXX} & {XXXX} & {XXXX} \\ \hline {XXX} & {XXX} & {XXX} & {XXXX} & {XXXX} & {XXXX} & {XXXX} & {XXXX} \\ \hline {XXX} & {XXX} & {XXXX} & {XXXX} & {XXXX} & {XXXX} & {XXXX} & {XXXX} \\ \hline {XXX} & {XXX} & {A} & {XXXX} & {XXXX} & {XXXX} & {XXXX} & {B} \\ \hline \end{array}$

The numbers in each column form an increasing arithmetic sequence
The numbers in each row form an increasing arithmetic sequence
The number of X's in each spot indicates how many digits are in the number
$A > 1200$
$B > 2900$

Find the value of $A+B$ .

The answer is 4125.

1 solution

Jeremy Galvagni
Apr 25, 2018

Let r1 be the common difference among terms in the first row, r2 be the common difference between terms in the second row, etc.

This solution relies on the insight that r1, r2, ..., r7 must themselves form an arithmetic progression, with common difference d. Similarly, if c1, c2, ..., c7 are the common differences between terms in the columns, respectively, then these are also an arithmetic progression and share the same common difference d.

We can then compute the value of any individual cell as a combination of r1, c1, d, and its location in the grid (given as the ith row and jth column, starting with 0), as follows:

i,j = i c1 + j r1 + i j d

So in particular we are told that:

B = 7c1 + 7r1 + 49d >= 2900

A = 7c1 + 2r1 + 14d >= 1200

Also note that cell (3,5) is only three digits, and thus 3c1 + 5r1 + 15d < 1000.

The first row contains 6 two-digit numbers and then becomes a three-digit number in the final column. This is only possible if r1 = 15 or r1 = 16.

Let r1 = 15. Plugging this in and doing a little simplifying to the above inequalities gives us the following:

[1] c1 + 7d >= 400

[2] c1 + 2d >= 168

[3] c1 + 5d <= 308

Considering [1] and [3], we can see that 2d must be at least 400 - 308 = 92, so d >= 46.

Considering [2] and [3], we can see that 3d can be no more than 308 - 168 = 140, so d <= 46.67.

Therefore, d = 46. Plugging this into the above inequalities shows us that 77.3 <= c1 <= 78.3, so c1 = 78.

With r1, c1, and d we can populate the entire grid and confirm that it conforms to all of the conditions of the puzzle:

0	15	30	45	60	75	90	105
78	139	200	261	322	383	444	505
156	263	370	477	584	691	798	905
234	387	540	693	846	999	1152	1305
312	511	710	909	1108	1307	1506	1705
390	635	880	1125	1370	1615	1860	2105
468	759	1050	1341	1632	1923	2214	2505
546	883	1220	1557	1894	2231	2568	2905

$A=1220, B=2905, A+B=\boxed{4125}$

Arithmetic Grid

The answer is 4125.

1 solution

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