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# How do I solve this mathematical problem? watch

1. 1. Think of a positive two digit number.
2. Calculate the digit sum.
3. Subtract the digit sum from the number that you thought of.
4. What did you get?

Example:
I think of the number 68.
The digit sum is therefore 6+8=14
68-14=54

Now:

I. Think of a new two/three-digit number and do the same thing again.

II. Re-do the calculation with new numbers from the one before until you find that they have something in common. What does the answer have in common?

III. Show that your foundings is valid for all two-digit numbers, the value of the two digit number can be written as 10*a+b

IV. Investigate whether your foundings also applies to positive three-digit numbers

84-12=72
72-9=63
63-9=54
54-9=45
45-9=36
36-9=27
27-9=18
18-9=9
9-9=0

I see that all answers are multiples of 9. How do I explain in words?
3. I. Do it again...
II. Do it again multiple times... you should get a series of results you can see a pattern to.
III. So you have 10a + b and you have to subtract a+b as per the question... (This will answer part II)
IV. You've been given that 10a + b = a 2 digit number so perhaps come up with the equivalent for a 3 digit number?
4. (Original post by manny103)

84-12=72
72-9=63
63-9=54
54-9=45
45-9=36
36-9=27
27-9=18
18-9=9
9-9=0

I see that all answers are multiples of 9. How do I explain in words?

You should do more general examples.

E.g. 77 - 14 = 63 = 7 x 9.

At that point you only need to say that the result is a multiple of 9.

Can you now work with the suggestion of using 10a+b?
5. (Original post by BuryMathsTutor)

You should do more general examples.

E.g. 77 - 14 = 63 = 7 x 9.

At that point you only need to say that the result is a multiple of 9.

Can you now work with the suggestion of using 10a+b?
Spoiler:
Show

When you do it repeatedly the answers will of course give you multiples of 9 since the digit sum of multiples of 9 is always 9. So OP wasn't choosing multiples of 9 on purpose.
6. (Original post by uponthyhorse)
When you do it repeatedly the answers will of course give you multiples of 9 since the digit sum of multiples of 9 is always 9. So OP wasn't choosing multiples of 9 on purpose.
The result is a multiple of 9 regardless of the initial value. The OP's initial values were all multiples of 9. If they only consider these initial values then they cannot say much about what happens in general.
7. (Original post by BuryMathsTutor)

You should do more general examples.

E.g. 77 - 14 = 63 = 7 x 9.

At that point you only need to say that the result is a multiple of 9.

Can you now work with the suggestion of using 10a+b?
Is there any other words that I can explain my findings with? Besides being a multiple of 9? Is there a certain formula we are following here that I can give a validity to by doing this experiment?

10a+b, so a = 3 and b = 4
30+4 = 34

34-7 = 27
27-9 = 18
18-9 = 9

It seems that any number I put in, into the formula where we have two digit numbers, gives me a multiple of 9?
8. (Original post by BuryMathsTutor)
The result is a multiple of 9 regardless of the initial value. The OP's initial values were all multiples of 9. If they only consider these initial values then they cannot say much about what happens in general.
yep but the "initial" values they were using were the results of the previous calculation, they weren't chosen by the OP as far as I can tell

(am I getting confused here?)
9. (Original post by BuryMathsTutor)
The result is a multiple of 9 regardless of the initial value. The OP's initial values were all multiples of 9. If they only consider these initial values then they cannot say much about what happens in general.
By initial values do you mean the first result I get when giving a and b values?
a=5 b=3
10a+b
50+3=53
Digit sum is 8? Which initial values did you refer to?
10. (Original post by uponthyhorse)
yep but the "initial" values they were using were the results of the previous calculation, they weren't chosen by the OP as far as I can tell

(am I getting confused here?)
Their instructions said, "Re-do the calculation with new numbers."

I didn't even notice that they were taking the last output as the new input.
11. (Original post by manny103)
Is there any other words that I can explain my findings with? Besides being a multiple of 9?
Yes, there is a little more that you can say. Which multiple of 9 do you get? How is it related to the number you start with?
12. (Original post by BuryMathsTutor)
Yes, there is a little more that you can say. Which multiple of 9 do you get? How is it related to the number you start with?
18, 27 etc.. the multiple of the first digit sum.
(Original post by BuryMathsTutor)
Yes, there is a little more that you can say. Which multiple of 9 do you get?
The digit sum minus the original number gives you a multiple of 9, correct? So how can I explain this with a formula? If we have 36 and I have the digit sum 9 from it I will get 27, which is a multiple of 9 and one multiple below 36. But does it generally apply to every original number?

(Original post by BuryMathsTutor)
The digit sum
13. (Original post by manny103)
18, 27 etc.. the multiple of the first digit sum.

The digit sum minus the original number gives you a multiple of 9, correct? So how can I explain this with a formula? If we have 36 and I have the digit sum 9 from it I will get 27, which is a multiple of 9 and one multiple below 36. But does it generally apply to every original number?

The digit sum
You really need to look at the general case.

This will show that the result is a multiple of 9 and also which multiple of 9 it is.
14. (Original post by BuryMathsTutor)
You really need to look at the general case.

This will show that the result is a multiple of 9 and also which multiple of 9 it is.
30+3 = 33
33 - (6) = 27

Right?
15. (Original post by manny103)
30+3 = 33
33 - (6) = 27

Right?
Yes, that is correct, but can you finish this ?
16. (Original post by BuryMathsTutor)
Yes, that is correct, but can you finish this ?
A=3 b= 3

30+3-(a+b) = 33-a-b = 27

Same thing?
17. (Original post by manny103)
A=3 b= 3

30+3-(a+b) = 33-a-b = 27

Same thing?
Try it again but don't replace a and b with specific values.
18. (Original post by BuryMathsTutor)
Try it again but don't replace a and b with specific values.
10a+b-a-b=....
10a-a=....
10a=....+a
19. (Original post by manny103)
10a+b-a-b=....
10a-a=....
10a=....+a
You're OK as far as 10a-a.

Just simplify that a little more.

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