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9 solutions
This is a Brilliant ( lol , get it ? ) question on A.P .... You can look at it from a different perspective. We have 10 A.P's with the same common difference 1 , A variable Initial term ( a ) and a variable Final term ( l ) = a + 9
S 1 0 = 5 ( a + a + 9 ) ⟹ 1 0 a + 4 5
we need to find
r = 1 ∑ 1 0 1 0 r + 4 5 = 5 5 0 + 4 5 0 = 1 0 0 0
By symmetry across the center, the average of the entries is 10, and the sum is 1 0 × 1 0 × 1 0 = 1 0 0 0
sir can you please elaborate some more on it
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If we consider any two entries that are symmetric with respect to the center of the grid, then a + b = 2 0 , so that the average comes out to be 10.
Alternatively, you can think of it as forming 50 pairs with a sum of 20 each, in the spirit of Gauss.
This answer is ingenious in its simplicity. We can tell by a glance that the average of each square is 10. So: Sum of sqrs / # of sqrs = 10 ...leading us to... Sum of sqrs = 100 x 10
Let S denote this sum. If we rotate this grid as shown above, then adding their numbers in their respective cells shows that all their terms are 20.
Then 2 S is equal to the sum of all the numbers in the grid below.
⎣ ⎢ ⎢ ⎢ ⎡ 2 0 2 0 ⋮ 2 0 2 0 2 0 ⋮ 2 0 … … ⋱ … 2 0 2 0 ⋮ 2 0 ⎦ ⎥ ⎥ ⎥ ⎤
Hence, since there's a total of ten 20's in each rows and columns, then 2 S = 1 0 × 1 0 × 2 0 ⇒ S = 1 0 0 0 .
Fantastic solution, best yet!
just fabulous!!
Adding the numbers on opposing ends, we have:
S = ( 1 + 1 9 ) + 2 ( 2 + 1 8 ) + 3 ( 3 + 1 7 ) + . . . + 9 ( 9 + 1 1 ) + 1 0 ( 1 0 ) = n = 1 ∑ 9 2 0 n + 1 0 0 = 2 9 ( 1 0 ) ( 2 0 ) + 1 0 0 = 9 0 0 + 1 0 0 = 1 0 0 0
think from any row or column. the pattern is 55+65+75+85...................(just calculate first 2 or 3 terms. you will guess the easy pattern)
so, the summation is 1000.
Opposing numbers from one end of the grid to the opposite (numbers 1 and 100, 2 and 99.... all equal 20. There are 50 pairs of such numbers all equaling 20. 50 x 20 = 1000.
The sum
= 1 + 2 ( 2 ) + 3 ( 3 ) + 4 ( 4 ) + 5 ( 5 ) + 6 ( 6 ) + 7 ( 7 ) + 8 ( 8 ) + 9 ( 9 ) + 1 0 ( 1 0 ) + 1 1 ( 9 ) + 1 2 ( 8 ) + 1 3 ( 7 ) + 1 4 ( 6 ) + 1 5 ( 5 ) + 1 6 ( 4 ) + 1 7 ( 3 ) + 1 8 ( 2 ) + 1 9 = 2 0 + 2 ( 2 + 1 8 ) + 3 ( 3 + 1 7 ) + 4 ( 4 + 1 6 ) + 5 ( 5 + 1 5 ) + 6 ( 6 + 1 4 ) + 7 ( 7 + 1 3 ) + 8 ( 8 + 1 2 ) + 9 ( 9 + 1 1 ) + 1 0 0 = 2 0 ( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 ) + 1 0 0 = 2 0 ( 4 5 ) + 1 0 0 = 9 0 0 + 1 0 0 = 1 0 0 0
The elements of the left to right diagonal line multiplied in number of elements of one row then it is 100*10=1000 Or Number of elements in the row ^3. Then it is 10^3=1000
So I fancied writing some code to do this. I wrote int xzult = 0; for (int j = 0; j<10; j++){ for (int i = 1; i<=10; i++){ xzult = xzult + i + j; System.out.print(i+j + " "); //print the grid } System.out.print("\n"); } System.out.println(xzult); //print the final number }
Top row and rightmost column sum to 190 =19*10.
Remove those numbers and repeat, you get 170 = 17*10.
Keep going until 10 = 10*1.
Summing you get: 10 (1+3+5+...+19) = 10 (100) = 1000 ,
1+2 2+3 3+4 4+5 5+6 6+7 7+8 8+9 9+10 10+11 9+12 8+13 7+14 6+15 5+16 4+17 3+18*2+19=1000
Relevant wiki: Sum of n, n², or n³
The sum of the first column is the tenth triangular number ( 1 + 2 + 3 . . . + 1 0 ), which can be worked out using 2 1 1 0 ( 1 0 + 1 ) = 5 5 . The sum of the second column is equal to 5 5 + 1 0 , since all ten number have been increased by one. The third is equal to 5 5 + 2 0 , and so on up to the tenth column. The sum of all the added on multiples of ten is thus 1 0 + 2 0 . . . + 9 0 , which is equal to 1 0 ( 2 1 9 ( 9 + 1 ) ) = 4 5 0 . The sum of all the original triangle numbers is just 5 5 × 1 0 = 5 5 0 . Therefore the answer is 5 5 0 + 4 5 0 = 1 0 0 0 .