Why Do I Always Make Mistakes in Multi-Digit Multiplication?

[Tyrion101]

Ok, so I know how to do it, that is not the question. The question is, why do I always get them wrong. If there is a problem say 323 * 486 Somewhere in the problem I will mess up, it will always be something different than what I messed up on last time. What am I to do? I am very close to just giving up on the whole idea of doing anything math related anymore.

 

[symbolipoint]

Tyrion101,

Just think of multiplication as sets of segments crossing each other. The result of multiplication is the number of intersections which occur. For multidigit number factors, break each factor into place values. This gives you separate products which can form a sum to evaluate.

By your example, a way I show and explain is that one factor is shown as 300 plus 20 plus 3 segments, and another factor is 400 plus 80 plus 6 segments. The segments of both factors intersect, giving NINE groups of products to be summed.

 

[Tyrion101]

I'm assuming, i multiply 300 x 400, and 20 x 80 etc?

 

[symbolipoint]

I'm assuming, i multiply 300 x 400, and 20 x 80 etc?

Wow! That reply was too fast. I just barely finished sending my first response and you posted what I quoted above. Take the time to read and think; you seem to be on the correct understanding. I could draw the figure, but maybe you could also and it would be more to your benefit.

You see or find the nine partial products to sum?

 

[phinds]

Ok, so I know how to do it, that is not the question. The question is, why do I always get them wrong. If there is a problem say 323 * 486 Somewhere in the problem I will mess up, it will always be something different than what I messed up on last time. What am I to do? I am very close to just giving up on the whole idea of doing anything math related anymore.

The bolded part is HIGHLY doubtful. All you are doing is a sequence of single-digit multiplication, some carries, and then some addition. How many different things can you do wrong? About 3.

The problem has to be that you are careless about the individual steps.

 

[Tyrion101]

Wow! That reply was too fast. I just barely finished sending my first response and you posted what I quoted above. Take the time to read and think; you seem to be on the correct understanding. I could draw the figure, but maybe you could also and it would be more to your benefit.

You see or find the nine partial products to sum?

I think I partially understand, but not about where the 9 products come from.

 

[pwsnafu]

I think I partially understand, but not about where the 9 products come from.

Okay. I don't know how much algebra you have done, but it comes from expanding out brackets. I'll do this slowly. Suppose we have
$(x+y) \times a$
that is, we have two number (one called x and the other called y) add them together and then multiply by a. It turns out this is equal to
$(x+y) \times a = x \times a + y \times a$
Similarly if we have
$(x+y)\times(a+b) = x \times (a+b) + y \times (a+b) = x\times a + x\times b + y \times a + y\times b$
giving four multiplications and three additions. (Remember your order of operations!)

In the case of your problem:
$323 \times 486 = (300+20+3)\times(400+80+6)$
so each bracket has three numbers. Expanding it all out
$300\times400+300\times80+300\times6 + 20\times400+20\times80+20\times6+ 3\times400+3\times80+3\times6$
for a total of nine multiplications.

 

[Tyrion101]

Oh, I was oversimplifying it. I have seen this before, just completely forgot. It will probably take me a bit to digest this. I think

 

[Diffy]

Okay. I don't know how much algebra you have done, but it comes from expanding out brackets. I'll do this slowly. Suppose we have
$(x+y) \times a$
that is, we have two number (one called x and the other called y) add them together and then multiply by a. It turns out this is equal to
$(x+y) \times a = x \times a + y \times a$
Similarly if we have
$(x+y)\times(a+b) = x \times (a+b) + y \times (a+b) = x\times a + x\times b + y \times a + y\times b$
giving four multiplications and three additions. (Remember your order of operations!)

In the case of your problem:
$323 \times 486 = (300+20+3)\times(400+80+6)$
so each bracket has three numbers. Expanding it all out
$300\times400+300\times80+300\times6 + 20\times400+20\times80+20\times6+ 3\times400+3\times80+3\times6$
for a total of nine multiplications.

In this thread, we have someone who claims to have trouble with multiplying two 3 digit numbers.

The solution presented is to do 9 sets of multiplication problems.

There are a few things I would like to get across to the OP.

1) When you do a large multiplication problem take your time and do not rush. When you go fast you open yourself up for error. Eventually, when you start getting less errors, you can start moving faster.

2) Double check your work. Do the whole problem again on a blank sheet of paper where you can't see your work the first time around. Take your time and double check.

3) Write neatly. This can help you so much. It is easy to make mistakes when you have sloppy writing. The digit columns may not line up, or you may misread a 1 for a 7, etc. Take your time and write neatly and orderly.

4) Don't try and re-invent the wheel. Try to master multiplication the way you were taught. Learning a new method might be too much. Take your time and really try to learn the way you were first taught.

5) TAKE YOUR TIME!

6) Practice, Practice, Practice.

 

[jbriggs444]

To the good advice offered by Diffy I would add two possibilities:

Do the multiplication again, but instead of computing 323 * 486, compute 486 * 323. There is less chance that you'll make the exact same mistake twice in a row.


Do the trick of "casting out nines". Add up all the digits in the one factor:

3 + 2 + 3 = 8

So the first factor has a result of "8"

Add up all the digits in the other factor

4 + 8 + 6 = 18; 1 + 8 = 9

So the second factor has a result of "9"

Multiply the results; 8 * 9 = 72

7 + 2 = 9

This tells you that the product should have a casting out nines result of 9.

Now after you compute the product, add up its digits.

156,978; 1 + 5 + 6 + 9 + 7 + 8 = 36 ; 3 + 6 = 9

The result checks!

 

[pwsnafu]

In this thread, we have someone who claims to have trouble with multiplying two 3 digit numbers.

The solution presented is to do 9 sets of multiplication problems.

For humans there is only two choices:

1. Use the long multiplication algorithm, in this case resulting in 9 one-digit multiplications,
2. Get OP to learn his 323-times table.