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-Dirty-Wizard- t1_itimwug wrote

51x52=Y

^F (50x50)+ ^O (50x2) + ^I (1x50) + ^L (1x2) = Y

Solve the equations in the parentheses first. Then add them together separately.

This is the FOIL method. First, outside, inner, last. You break the equation down into smaller equations that are easier to handle and then add them all together. That process it’s self is called factoring. You can look up videos on factoring to help you understand the concept of FOILing better.

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Moskau50 t1_itimxds wrote

You're trying to solve 51 x 52 = ?

You can break it up into pieces, which it seems like you're trying to do.

51 = 50 + 1 and 52 = 50 + 2

So now you have (50 + 1) x (50 + 2)

There is the distributive property of multiplication, which says that you can "split" the multiplication of a sum into the sum of the products. In other words:

3 x 7
3 x (2 + 5)
(3 x 2) + (3 x 5)
6 + 15 = 21

So you can do the same thing here by distributing each term in the equation (50 + 1) x (50 + 2), using FOIL:

First term of each: 50 x 50

Outside terms: 50 x 2

Inside terms: 1 x 50

Last terms of each: 1 x 2

Now you have (50 x 50) + (50 x 2) + (1 x 50) + (1 x 2). Do the individual multiplications, and you get 2500 + 100 + 50 + 2 = 2652

2

Plinklypl0nk OP t1_itinz88 wrote

Thank you so much I can multiply two digit numbers now :)))) does this also work for three digit numbers.

1

mafiaknight t1_itin3ug wrote

Do you want a breakdown or the math worked out here?
Set one over the other for ease of keeping track. Then multiply the first number of the second term by the two numbers of the first term. Then do the same with the second number of the second term. Write the second set under the first and offset by one place. Now add them together.

51
52
—
255
0102
—
2652

Checking with my calculator: 51x52=2652

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Plinklypl0nk OP t1_itin9vo wrote

I’m still confused I’m sorry

2

Em_Adespoton t1_itinw69 wrote

Check the FOIL examples then.

Carrying is due to the number base you’re working in… in this case you’re using base 10.

Remember: at the end of the day you’re saying you’ve got 51 groups each with 52 elements and you want to add them all up. The rest is just different ways of arranging those elements into groups that are easier to work with.

2

mafiaknight t1_itiqhzt wrote

(This is MUCH harder to explain over text...way easier to show you...I’ll do my best)

This is a way to break down large number multiplication into single digit multiplication.

The first term we have is 51. So I decided to put that on top.
The second term we have is 52. So I put that under it.
That looks like this:
51
52
—

Now I can multiply the individual numbers together, but I have to make sure I write them down in the right order.

So, first we multiply the top term by the first digit of the bottom term (a 5)^((from 52 into both the 5 and 1 from 51)).

Ok, what’s 5x5?
25

What’s 5x1?
5

So the first two get written on the same line right?
That gives us 25 in the 10s spot and 5 in the 1s spot. (The 2 carrying over to the 100s spot automatically). (So we have 250 + 5 to break it ALL the way down) or more simply: 255.

Now what’s 2x5?
10

And 2x1?
2

When we move to the next digit in the second term we also need to move to a new line. But remember that the next digit (the 2) is one place to the right of the previous digit. So we need to move our answers one digit to the right as well.
Now we take the 10 in the 10s place (letting it’s 1 fall in the hundreds place.) and we set the 2 in the 1s place. (To break it down again we get 100 + 2) or more simply 102.

Now we have 255 and 102 right? And 255 goes down on the first line, with 102 on the second, but one place to the right.

That looks like this:
255
_ 102
Right?

So the 2 drops down and adds with the nothing for a 2.
2- - -
Then the 5 drops down and adds with the 1 for 6.
2 6 - -
Then the next 5 drops and adds with the 0 for 5.
2 6 5 -
We’ve run out of numbers on the first line, but have one more on the next. We’ll drop a 0 (to fill that place) and add it to the 2 we have left for 2.
2 6 5 2
And that’s our answer.

So, to write it out again:
51
x52
——
2 5 5
+ 1 0 2
————
2 6 5 2

This works for any size number multiplied with any other number of any size.

0

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1

Target880 t1_itip3fr wrote

51 x 52 = ( 50 + 1 ) x (50 +2)

(a + b) x (c +d) = a x (c + d) + b x (c + d) = a x c + a x d + b x c + b x d

It is this expansion you need to use. Think of it each variable in the first parent multiplied by each variable in the second and add them together

( 50 + 1 ) x (50 +2) = 50 x (50 + 2) + 1 x ( 50 + 2) = 50 x 50 + 50 x 2 + 1 x 50 + 1 x 2 = 50 x 50 + 3 x 50 + 2 = 2500 + 150 + 2 = 2652

​

I would not ever get to 50 x 50 if tried to do it by hand or in my head 50 x 52 + 1 x 52 would be my start. 1 x 52 is 52 so we just add it at the end

Then try to get to a multiple of 10. 50 is 100 /2 so 50 x 52 = (100 * 52) /2 = 5200/2. You can directly get it is 2600 but you can split it too.

Ignore the 00 at the end. 52 / 2 = 26 or 50 / 2 + 2/2 = 25+1 =26. We cannot just add back the 00 at the end and get 2600

The result is no 2600 +51= 2651.

Tying to get to a multiple of 10 other another number that will be easy is a good idea 99 x 52 is directly hard but 100 x 52 - 1 x 52 is not. You get 5200 - 52 and you can get 5148 from that

1

EvenSpoonier t1_itiqw8j wrote

When you multiply, the first thing you do is take one number, and you think of it as a set of a certain size. It doesn't actually matter which of the two numbers you use, as long as you're consistent about it. So we'll say it's the first number, to keep things consistent.

So we have 51 objects. We'll think of this as a sit of little blocks, one inch on a side. They could instead be 1 centimeter, 1 mile, or whatever; what matters is each of these little blocks means 1. And we have 51 of them. Let's put them in a line. There are 51 blocks in the line, so the line is 51 long.

The next thing you do is you make a number of sets, the same size as that first one. The number of sets you make is equal to the second number (or really, whatever number you didn't use in the first step).

  • If the second number is 0, then you don't actually have any sets. Throw away the first one, and your answer is zero. This is always true, no matter what the first number was: if you don't have any sets, you don't have anything.

  • If the second number is 1, then you already have your one set. You made it in that last step. So that's your answer. 51 times 1 is 51. This is, once again, always true no matter what the other number is. (If it's zero times one you have a set that doesn't have anything in it, so you still have nothing.

  • If the last number is anything else, you're going to make more sets, same as the first. Remember that that first set counts. So let's say the second number is 2, you have two of your lines of blocks, each of length 51.

Let's point all of these lines in the same direction, but then stack them on top of each other. So if the lines are horizontal, we stack them vertically, and vice versa. This turns our line into a rectangle. For our 51 * 2 problem, it's 51 long and 2 high. For 51 * 52, it's 51 long and 52 high. This is the key to understanding multiplication: what you're really doing is making rectangles.

Now, the answer to our multiplication goes back to the blocks: how many do you have, in total? Each block, remember, is 1. You could just count the blocks, but those numbers can get pretty big pretty quickly: even for 51 * 2 there are 102 blocks total, and for 51 * 52 there are actually thousands of blocks (2652 of them, to be precise). You really don't want to count all those blocks yourself, if you can avoid it: counting will work, but it takes a long time. And that's where all this stuff where we fiddle with these numbers comes into play. It's all just shortcuts to avoid having to count out those blocks. But in the end, it's all about making rectangles. And that goes in reverse: every rectangle you make is a multiplication problem.

Does this help?

1