If both multiplicands even, multiplication should also be ... .
555x660 = 111 x 110 x 5 x 6 = 111x110x30 should have double 0's at the end.
The simplified form is 75 + 6xA = 117
A x A = 1 , 11 , 21 , 31 , 41 ... if A is equal to 1 and 9. Looking at the numbers the problem is solved at this point. But have a look at the
2nd "y" for future convenience .
1421 has a digit sum of 8 . The digit sum the product is (2+A)x(4+A) = 8 + 6xA + AxA. This means A should be a multiple of 9.
Multiplication is repeated summation in a sense. By summation we know the digit sum of the summands must be equal to the digit sum of the sum if
no carry is done. In case of carry mod 9 summation (casting out nines) should work because the change then is a multiple of 9 (i.e. carrying a ten from ones' is putting a 1
to tens' which is an effective change of 9 in digits' sum). So casting out nines we may chack our multiplications up to a multiple of 9:
a- mod 9 sum of the digits of one of the multiplicands is put into say right of the cross.
b- mod 9 sum of the digits of the other multiplicand is written into left then.
c- These two are multiplied and mod 9 sum of the digits of the result is written into the top.
d- Mod 9 sum of the digits of the result of the original multiplication is written into the bottom.
Check is OK if the digits in top and and bottom are equal.
Checking "if one multiplicand is a multiple of 5, the result of multiplication is a multiple of 5." is of no further use because any possible error is
a multiple of 9 .
Have a look at ry1 also .
I is certainly ... .
The factors (multiplicands) of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36 . Constant product gives max. sum if facors differ as much as possible.
Subtraction is necessary and it should be done the right way .
Multiplication of the second nu. by 3 once, is enough.
First are multiplications due: 2500x5 =12500 , 88x44 = 352 x 11 = ... .