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Case 1: 1 x 1=1
Case 2: 11 x 11=121
Case 3: 111 x 111=12321
Case 4: 1111 x 1111=1234321
Case 5: 11111 x 11111=123454321
Case 6: 111111 x 111111=12345654321
Case 7: 1111111 x 1111111=1234567654321
Case 8: 11111111 x 11111111=123456787654321
Case 9: 111111111 x 111111111=12345678987654321

2007-11-23 09:36:24 · 4 answers · asked by love 1 in Science & Mathematics Mathematics

4 answers

Because digits in the product represent the number of ways that two non-negative numbers (which are less than the case number) can add to a certain integer. For example, in case 9.
We start with 16 = (9-1)*2
How many ways can two numbers add to 16 using 8 or less? Only 1: 8+8
For 15, there are 2 ways: 8+7 & 7+8
For 14, there are 3 ways: 8+6, 7+7, 6+8
For 13, 4 ways: 8+5, 7+6, 6+7, 5+8
For 12, 5 ways: 8+4, 7+5, 6+6, 5+7, 4+8
etc.
See the pattern?
It will start going down after we get to half of 16 or 8. (Where there are 9 ways, 8+0, 7+1, 6+2, 5+3, 4+4, 3+5, 2+6, 1+7, 0+8).

If we had used a different case, say case 4, we start with 2*(4-1) = 6
One way to make 6 using 3 or less: 3+3
Two ways to make 5: 2+3 or 3+2
three ways to make 4: 3+1, 2+2, 1+2
four ways to make 3: 3+0, 2+1, 1+2, 0+3
3 ways to make 2: 2+0, 1+1, 0+2
2 ways to make 1: 1+0, 0+1
1 way to make 0: 0+0

This is the case because (as in case 9) we are multiplying:

(1+10+10^2+10^3+ ... + 10^8) * (1+10+10^2+10^3+ ... + 10^8)
= 1 + 2*10^1 + 3*10^2 + ... + 2*10^15 + 10^16

Each term in the first (say, 10^a) will multiply with each term in the second (say 10^b) to make a 10^(a+b) term.
The highest term in the expansion will be 10^16. There is only one way this term could be made, 10^8 * 10^8
There were two ways to generate a 10^15 term. 10^8 * 10^7 and 10^7 * 10^8.
Three ways to get a 10^14 term:
10^8 * 10^6, 10^7 * 10^7, 10^6 * 10^8
etc.
Same pattern, same reason.

2007-11-23 09:49:07 · answer #1 · answered by Scott R 6 · 1 0

The number of 1 is the max number we must to write to

111 --- there ia 3 of 1 so 123 then 21

2007-11-23 09:46:13 · answer #2 · answered by tinhnghichtlmt 3 · 0 0

If you do the long multiplication method you will see that each time you solve one of these cases you will do addition for each term.

I will try to type this out

111
xx1
------
111

then
111
x1x
----
1110

then
111
1xx
-------
11100

add these
+ 111
+ 1110
+ 11100
---------
= 12321

There is the pattern

2007-11-23 09:49:56 · answer #3 · answered by Paul T 2 · 0 0

Nothing much to explain...
For example, for the 111x111 multiplication, you have 111x100 + 111x10 + 111x1

2007-11-23 09:54:03 · answer #4 · answered by cattbarf 7 · 0 0

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