Addition is one dimensional whreas multiplication is multidimensional: 1 mm. x 1 mm. is 1 mm.^{2} , 1mm. x 1 mm. x 1 mm. is
1 mm.^{3} etc.
First we have to find the numbers, the middle one is 90/5 = 18 .
Four times the number should be 112 .
The difference is about 200 means the greater may have 3 at tens'. And of course 1 at ones'. The smallest of these numbers is 301.
Ax100+Bx10+C -Cx100-Bx10-A = 99x(A-C) gives that A-C =2 . If we get C = 0 we have contradiction. Taking C = 1 gives A=3 . The smallest of such numbers is
301 and its digits reversed form is 103 which satisfy the restriction.
The rectangle may be 5 or 6 .
That the difference is a multiple of 99 we already know. Candidates of difference are 0 , 99, 198, 297, 396, 495, 594, 693, 792 and 891.
If the difference is 0, there are 9 possibilities:111, 222, 333, ..., 999. 444 satisfies the sum.
Odd differences does not give a natural number when the summation condition is also considered. Difference 198 gives 543 and 345. Difference
396 gives 642 and 246. Difference 594 gives 741 and 147. Here we have to remember that, among the numbers with a definite sum, the nearer the numbers the greater the product
as in the case of the area of a square.
The first multiplicand is 1000x(100-1)=1000 x 99 . The product should be around 100 000 x 100 000 i.e. 10 billions.
1000x1000(100-1)x(100+1) = ? as (100-1)x(100+1) = 100^{2} - 1^{2} = 9999.
Being questioned is the product of the sum of integers from 1 to 10 and from 1 to 15 .
This is the feedback!