How would you show algebraically that this multiplication "trick" does/doesn't work?

almostshelby

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How would you show algebraically that this multiplication "trick" does/doesn't work?

Thanks so much for trying to help. I have a really tricky question here, hoping someone can help me with it. My professor asked us to show algebraically if this 'trick' works with all numbers. It involves multiplying two numbers which share the same first digit (e.g. 57 and 53). Take the first number of the first term (5 from 57) and multiply it by the first number of the second term plus one (6 by adding 1 to the 5 in 53), in this example, =30. Then multiply the second number of the first term (7 from 57) by the second number of the second term (3 from 53), in this example, =21. Visually, these two products will accurately represent the correct answer to the original problem. For example, 30 followed by 21 equals the product of 57 x 53 = 3,021.

He stated this would only work for two numbers which begin with the same first digit. How would you go about proving algebraically that this does/doesn't work for all numbers?


Example:
57 x 53 = 3021

Trick:
5 x (5+1) = 30
7 x 3 = 21
 
Thanks so much for trying to help. I have a really tricky question here, hoping someone can help me with it. My professor asked us to show algebraically if this 'trick' works with all numbers. It involves multiplying two numbers which share the same first digit (e.g. 57 and 53). Take the first number of the first term (5 from 57) and multiply it by the first number of the second term plus one (6 by adding 1 to the 5 in 53), in this example, =30. Then multiply the second number of the first term (7 from 57) by the second number of the second term (3 from 53), in this example, =21. Visually, these two products will accurately represent the correct answer to the original problem. For example, 30 followed by 21 equals the product of 57 x 53 = 3,021.

He stated this would only work for two numbers which begin with the same first digit. How would you go about proving algebraically that this does/doesn't work for all numbers?


Example:
57 x 53 = 3021

Trick:
5 x (5+1) = 30
7 x 3 = 21

To examine the trick algebraically, you have to represent it algebraically.

You have two (two-digit?) numbers with the same first digit, so maybe we can call them "ab" = 10a + b and "ac" = 10a + c, where the first digit of each is "a", and the second digits are "b" and "c".

Where you say "first number", I assume you mean "first digit", and so on. So here is what you said, with my variables replacing the example numbers:

Take the first digit of the first term (a from "ab") and multiply it by the first digit of the second term plus one (a+1 by adding 1 to the a in "ac"), in this example, a(a+1). Then multiply the second digit of the first term (b from "ab") by the second digit of the second term (c from "ac"), in this example, bc. Visually, these two products will accurately represent the correct answer to the original problem. For example, a(a+1) [as a 2-digit number] followed by bc equals the product of (10a+b)(10a+c).

So, you need to show that the first two digits of (10a+b)(10a+c) are the number a(a+1), and the last two digits are the number bc:

100[a(a+1)] + bc

Is that true?

If you can't see algebraically that it is true, then you can solve the equation that claims it is true, to find conditions under which it is true, and so show why it does work for a=5, b=7, c=3, but not for some other cases. This will also provide you with counterexamples, to show directly that it is not always true.
 
Thanks so much for trying to help. I have a really tricky question here, hoping someone can help me with it. My professor asked us to show algebraically if this 'trick' works with all numbers. It involves multiplying two numbers which share the same first digit (e.g. 57 and 53). Take the first number of the first term (5 from 57) and multiply it by the first number of the second term plus one (6 by adding 1 to the 5 in 53), in this example, =30. Then multiply the second number of the first term (7 from 57) by the second number of the second term (3 from 53), in this example, =21. Visually, these two products will accurately represent the correct answer to the original problem. For example, 30 followed by 21 equals the product of 57 x 53 = 3,021.

He stated this would only work for two numbers which begin with the same first digit. How would you go about proving algebraically that this does/doesn't work for all numbers?


Example:
57 x 53 = 3021

Trick:
5 x (5+1) = 30
7 x 3 = 21
An alternative way to approach this kind of problem is to do some numerical experimentation, which is way easier today, what with calculators and computers

50 * 50 = 2500, which does not = 5 * 6 0*0 = 30 00 = 3000

50 * 51 = 2550, which does not = 5 * 6 0 * 1 = 30 00 = 3000.

If you kept doing this you would get to

50 * 59 = 2950, which does not = 5 * 6 0 * 9 = 30 00= 3000.

So it is obviously not a trick that works all the time or even often. Then you could start with

51 * 51 =2601, which does not = 5 * 6 1 * 1 = 30 01 = 3001.

And with a spread sheet, you would eventually find some that do work.

51 * 59 = 3009, which does = 5 * 6 1 * 9 = 30 09 = 3009.

52 * 58 = 3016, which does = 5 * 6 2 * 8 = 30 16 = 3016.

53 * 57 = 3021, which does = 5 * 6 3 * 7 = 30 21 = 3021.

54 * 56 = 3024, which does equal 5 * 6 4 * 6 = 30 24 = 3024.

55 * 55 = 3025, which does equal 5 * 6 5 * 5 = 30 25 = 3025.

Now you may see a pattern.

3025 - 0 = 3025
3025 - 1 = 3024
3025 - 4 = 3021
3025 - 9 = 3016
3025 - 16 = 3009

What does 0, 1, 2, 9, 16 remind you of?

Once you have some ideas, it is way easier to attack such a problem algebraically.
 
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