Problem#28: What is the sum of both diagonals in a 1001 by 1001 spiral?

Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:

It can be verified that the sum of the numbers on the diagonals is 101. What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed in the same way?

Solution:

Number present in the diagonals of first spiral is 1

Numbers present in the diagonals of second spiral is 3,5,7 and 9

Numbers present in the diagonals of third spiral is 13,17,21 and 25

Numbers present in the diagonals of Fourth spiral is 31,37,43 and 49

From above observation it is easy to conculde the following,

1. Every spiral has 4 diagonal numbers (except the first 1)

2. Diagonal numbers same spirals are arithmetic sequences

3,5,7 and 9 with 2 as common difference

13,17,21 and 25 as 4 as common difference

3. The value common difference in spiral s is 2(s-1)

Ex: common diff of spiral 3 is 6 and spiral 4 is 8

4. The difference between last number of spiral s-1 and first number of spiral s is also 2(s-1).

Ex: diff between last number and first number of spiral 2 and 3 respectively is 4 ( 13 -9)

diff between last number and first number of spiral 3 and 4 respectively is 6 ( 31 -25)

5. Number of spirals in NxN matrix is N/2 + 1/2.

The following PL/SQL solution is written based on the above findings,

Answer: 669171001

Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:

**21**22 23 24

**25**

20

**7**8**9**1019 6

**1**2 1118

**5**4**3**12**17**16 15 14

**13**

It can be verified that the sum of the numbers on the diagonals is 101. What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed in the same way?

Solution:

Number present in the diagonals of first spiral is 1

Numbers present in the diagonals of second spiral is 3,5,7 and 9

Numbers present in the diagonals of third spiral is 13,17,21 and 25

Numbers present in the diagonals of Fourth spiral is 31,37,43 and 49

From above observation it is easy to conculde the following,

1. Every spiral has 4 diagonal numbers (except the first 1)

2. Diagonal numbers same spirals are arithmetic sequences

3,5,7 and 9 with 2 as common difference

13,17,21 and 25 as 4 as common difference

3. The value common difference in spiral s is 2(s-1)

Ex: common diff of spiral 3 is 6 and spiral 4 is 8

4. The difference between last number of spiral s-1 and first number of spiral s is also 2(s-1).

Ex: diff between last number and first number of spiral 2 and 3 respectively is 4 ( 13 -9)

diff between last number and first number of spiral 3 and 4 respectively is 6 ( 31 -25)

5. Number of spirals in NxN matrix is N/2 + 1/2.

The following PL/SQL solution is written based on the above findings,

DECLARElv_final_value NUMBER (10); lv_iter_value NUMBER; lv_factor NUMBER; BEGIN lv_final_value := 1; lv_iter_value := 1; FOR i IN 2 .. 501 LOOP lv_factor := 2 * (i - 1); lv_final_value := lv_final_value +lv_iter_value * 4 + lv_factor * 10; lv_iter_value := lv_iter_value + lv_factor * 4; END LOOP; DBMS_OUTPUT.PUT_LINE (lv_final_value); END; |

Answer: 669171001