Questions

[H15]

1) 213 x 411 or 204 x 312
2) The reciprocal of 2-5i is a+bi. Find a.
3) If x+y=-1 and xy= -2 then x^3+y^3=

[RussiaPSJA]

I'll attempt to help out with my limited Number Sense knowledge :p

1) 213 x 411 or 204 x 312

Alright, if you know how to do foil to multiply 2 by 2 digit numbers, then you utilize the same concept here. Most people would break the numbers up into a 2 digit and 1 digit number, but personally I'm bad at multiplying big numbers, so I like to stick to multiplying 1 digit numbers. Personally, I find my method to be very quick, efficient and a lot more accurate (personally) than the other method.

Pretend we have 2 3 digit numbers, abc and def. We're going to work this from the back forward. Note that a, b, and c are all digits of the first 3 digit number and d, e, and f are all digits of the other 3 digit number.

abc x def:

The last digit is going to be c * f.

For the next digit you're going to cross multiply with bc and ef, so the next digit will be b * f + c * e (notice how you cross-multiply and then you add them).

This next digit is the most complicated digit, because now we're going to go a step further and cross-multiply through the whole number. We're going to do a * f + b * e + c * d, if you notice the pattern is easy to catch on to.

The next digit is going to be what we did earlier, where we cross multiplied bc and ef, but this time we're going to cross multiply ab and de. So this digit will be a * e + b * d.

The final digit is simple. You simply multiply the first 2 digits (a & d), so it's a * d.

In that whole mess you're going to have to account for carry-overs. So let's do what we just did with the two examples you wanted the answer to.

The first one: 213 x 411

The last digit is 3 * 1 (our c and f from earlier), so it's 3.
The next digit is 3 * 1 + 1 * 1 (our b * f + c * e), which is 4.
The next digit is 2 * 1 + 1 * 1 + 3 * 4 (our a * f + b * e + c * d), which is 15. Notice that it is 2 digits. Now we'll be carrying over. Write down the 5 and keep the 1 in mind.
Our next digit is going to be 2 * 1 + 4 * 1 + 1 (the carry over). Notice this is our a * e + b * d step. With the carry over this becomes 7.
Now, we finish it off by doing 2 *4, which is 8.

The final answer is 87,543.

Now, the second one: 204 x 312

The last digit: 2 * 4 = 8
The second to last digit: 4 * 1 + 2 * 0 = 4
The middle digit: 2 * 2 + 1 * 0 + 4 * 3 = 16 [Notice the carry over necessary. Write the 6 and keep the 1 in mind]
The second digit: 2 * 1 + 0 *3 + 1 (carry over) = 3
The first digit: 2 * 3 = 6

Your answer: 63,648

2) The reciprocal of 2-5i is a+bi. Find a.

Alright, so the reciprocal basically means 1/(2-5i), so to eliminate this dilemma we have we must multiply by the conjugate of the the base which is 2 + 5i. So, when we do 1/(2-5i) * (2 + 5i)/(2 + 5i) we obtain (2 + 5i)/(29). Now, we want a, correct? A would be 2/29.

From what we could conclude, the shortcut for this problem is two steps; the denominator and the numerator.

The denominator is just going to be the two numbers squared and added: 2^2 + 5^2 = 29.
The numerator is just going to be the conjugate of what is given, so you change the sign of the i coefficient.

From this we could conclude that if you are given this question and they want b, you'll be able to get it by squaring and adding them for the denominator and then switching the sign of the i coefficient. So b would be 5/29. Some advice, from what I know the tests get harder as the year progresses, so as the year progresses you'll be required to simplify your answer (I think :p).

3) If x+y=-1 and xy= -2 then x^3+y^3=

Alright, this scenario is something that you should memorize. When given those, you cube x + y, and subtract 3 * (x + y) * (x*y). Note, if it said x^3 - y^3 you would add 3 * (x + y) * (x*y).

So, x + y = -1, and x*y = -2. Using that data and what I stated above: x^3 + y^3 = (-1)^3 - 3*(-1)*(-2) = -1 - 6 = -7.

[AuSmith]

[SumitG]

Though I very much agree with AuSmith's statement about taking advantage of 0's, and I like his methods for these problems specifically, I might have broken down these three digit multiplication problems a little differently, just because of the following idea's general efficiency. You can take advantage of the fact that the digits in three digit multiplication problems tend to be small. NOTE: This idea only works well if you are fairly fast at two-digit multiplication.

Similarly to what David did, you would do a FOIL. However, instead of dealing with three term FOILing, you can simply treat the hundreds and tens digits together as a single term. So, for your first problem:

213 x 411 = (21)(41) + (21x1 + 41x3) + (1x3) = 87543