Help with Algebra Problem Please: My special number has a 9 in the units column...

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LateDeveloper

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Good morning all. Can anyone help with this problem please. I'm sure there is an algebraic solution but I just can't see it.

My special number has a 9 in the units column. If I remove the 9 from the units column and place it at the left hand end of the number, but leave all the other digits unchanged, I get a new number. This new number is four times my special number. What is my special number?
 
LateDeveloper said:
Good morning all. Can anyone help with this problem please. I'm sure there is an algebraic solution but I just can't see it.

My special number has a 9 in the units column. If I remove the 9 from the units column and place it at the left hand end of the number, but leave all the other digits unchanged, I get a new number. This new number is four times my special number. What is my special number?
Click to expand...
Is this homework from a computer-programming course or a math class?

If so, please read the forum guidelines, follow the instructions, and provide the requested information.

I'm thinking of an algorithmic hint for you that involves representing the two numbers symbolically (using place value) and eliminating possibilities by taking cases.

I wait to see your thoughts or efforts, first. :cool:
 
Thank you mmm4444bot. I apologise for not familiarising myself with the protocol. Not only is this my first time in this forum, it's my first time in any forum.

So, firstly, this is a question from an exam paper for 11 year olds. it's the last question, and, for sure, way the hardest question. I think the examiners expect only a very clever few to get the answer.

Evidently, I am not one of the very clever few, and neither is my 18 year old son who has recently taken his A-level (taken right at the end of school, before going to university).

As for my own efforts, I tried something like this:

(n digits)9 x 4 = 9(n-1 digits)6

But even if that's correct, I can go no further. Your help would be appreciated.
 
LateDeveloper said:
My special number has a 9 in the units column. If I remove the 9 from the units column and place it at the left hand end of the number, but leave all the other digits unchanged, I get a new number. This new number is four times my special number. What is my special number?
Click to expand...
Are you told how many digits are in the "special number"? I think this information is needed, if we're to know how far the 9 moved. Thank you! ;)
 
No, we're not told that. The complete question is as I posted, copy-and-pasted from the exam.
 
LateDeveloper said:
No, we're not told [how many digits the special number is].
Click to expand...
Gosh, considering that Denis (one of our resident programmers) already determined that it's six digits, that would take some time to reason out. Not really fair, in my opinion, for 11-year-olds taking an exam (unless it's a take-home exam, and they're motivated to spend a couple hours or more).

I only checked the 3-digit case:

ab9 (where a and b are unknown digits)

Digit a is in the hundreds place and digit b is in the tens place.

We can symbolically represent the value of number ab9 like this:

100a + 10b + 9

We are told that four times this value equals the new number. Hence:

4(100a + 10b + 9) = 900 + 10a + b

This equation can be rearranged to:

390a + 39b = 864

Now check cases, to see if you can find multiples of 390 and 39 that sum to 864. It doesn't take long to discover there are none.

So, now you start over with abc9, and follow the same strategy.

With abcde9, it can be done manually, but not quickly!

By the way, when I saw "Developer" in your username, I had thought your were a software developer. :)
 
Many thanks everybody for your help with this, I really appreciate it.

Now I understand, mmm4444bot, why you thought I was a software developer. I chose the name because I'm a late developer in maths (with an "s" as we shorten "mathematics" in the UK).

I am just trying to give a helping hand with maths, and this is the first question that has really stumped me. It's not a take-home exam, it's the last question in an invigilated exam, and the time allowed for the whole exam of 35 questions is one hour! But it is for sure one of the most academically demanding schools, and as I said before, I think the school would not expect a big majority of the kids to get the answer.

Now I'm going to sit down later when I have time - unfortunately not before tomorrow morning - and try to understand all your explanations.

Thanks again to all!
 
This is time consuming, but not hard, if attacked in a different way. I want to find the smallest integer with the given attribute. Call that i.

\(\displaystyle 0 < 10^a < i < 10^{(a+1)} \text { and } i \in \mathbb N^+ \text { and } i = 10J + 9 \text { and }\)

\(\displaystyle 9 * 10^a + J = 4i = 4(10J + 9) = 40J + 36 = 40J + 30 + 6. \)

Obviously a cannot be 0.

\(\displaystyle a = 1 \implies i = 10J + 9 \implies 40J + 36 = 90 + J \implies 39J = 54 \implies\)

\(\displaystyle J \not \in \mathbb Z \implies a > 1.\)

\(\displaystyle a \ge 2 \implies i = 100K + 10j + 9 \implies\)

\(\displaystyle 4i = 400K + 40j + 30 + 6 = 9 * 10^a + 10K + j \implies j = 6.\)

\(\displaystyle a = 2 \implies i = 100K + 69 \implies\)

\(\displaystyle 4i = 400K + 276 = 900 + 10K + 6 \implies 390K = 630 \implies\)

\(\displaystyle K \not \in \mathbb Z \implies a > 2.\)

\(\displaystyle a \ge 3 \implies i = 1000M + 100k + 69 \implies\)

\(\displaystyle 4i = 4000M + 400k + 200 + 70 + 6 = 9 * 10^a + 100M + 10k + 6 \implies k = 7\).

\(\displaystyle a = 3 \implies i = 1000M + 769 \implies\)

\(\displaystyle 4i = 4000M + 3076 = 9000 + 100M + 76 \implies 3900M = 6000 \implies\)

\(\displaystyle M \not \in \mathbb Z \implies a > 3.\)

\(\displaystyle a \ge 4 \implies i = 10000N + 1000m + 769 \implies\)

\(\displaystyle 4i = 40000N + 4000m + 3000 + 76 = 9 * 10^a + 400m + 76\)

\(\displaystyle m = 0 \text { or } m = 5.\)

\(\displaystyle a = 4 \text { and } m = 0 \implies i = 10000N + 769 \implies\)

\(\displaystyle 4i = 40000N + 3076 = 90000 + 1000N + 76 \implies 39000N = 87000 \implies\)

\(\displaystyle N \not \in \mathbb Z \implies a \ne 4.\)

\(\displaystyle a = 4 \text { and } m = 5 \implies i = 10000N + 5769 \implies\)

\(\displaystyle 4i = 40000N + 20000 + 3076 = 90000 + 1000N + 576 \implies 39000N = 67500 \implies\)

\(\displaystyle N \not \in \mathbb Z \implies a \ne 4.\)

\(\displaystyle a \ge 5 \text { and } m = 0 \implies i = 100000P + 10000n + 769 \implies\)

\(\displaystyle 4i = 400000P + 40000n + 3076 = 9 * 10^a + 1000n + 76 \implies n = 3.\)

\(\displaystyle a = 5 \implies i = 100000P + 30769 \implies\)

\(\displaystyle 4i = 400000P +\ \)\(\displaystyle 123076 = 900000 + 10000P + 3076 \implies 390000P = 780000 \implies\)

\(\displaystyle P = 2 \implies i = 230769.\)

Let's check.

\(\displaystyle 4 * 230769 = 923076.\)

ABSURD PROBLEM
 
I wonder, if a student wrote out Jeff's method on that exam, whether the instructor would understand it. :p
 
LateDeveloper said:
Good morning all. Can anyone help with this problem please. I'm sure there is an algebraic solution but I just can't see it.

My special number has a 9 in the units column. If I remove the 9 from the units column and place it at the left hand end of the number, but leave all the other digits unchanged, I get a new number. This new number is four times my special number. What is my special number?
Click to expand...

Workings:

4 x 9 = 36 so, 6 is the last number in the 'new number'.
Why? Because you are multiplying by 4 a number that ends in a 9: it must result in a number that ends with a 6.

You might think about the number 6 9 But it must be at least 3 digits long. They say move the 9 in front of other digits (i.e. plural).
Nevertheless, here is some proof of that:
6 9 x 4 = 276 so, no. the answer must have a 9 at the front anyway, and the original number must be more than two digits.

NOW: try doing the reverse action and divide by 4; so, here is the division:

answer: 2 3 0 7 6 9

1 3 2 3
4 9 2 3 0 7 6

  • Set up the division box.
  • Put the numbers in (that you know - the 'divisor 4' and the 9s and 6s you now know about)
  • 4 into 9 = 2 Not only do you have the first digit in the ‘special number’, you can enter it as the 2nd digit in the ‘larger number’ that you are dividing by 4.
  • Keep going… so 12 divided by 4 = 3. Work your way through…

So: 230769 x4 = 923076 Answer = 230769
 
Sara Richmond said:
Workings:

4 x 9 = 36 so, 6 is the last number in the 'new number'.
Why? Because you are multiplying by 4 a number that ends in a 9: it must result in a number that ends with a 6.

You might think about the number 6 9 But it must be at least 3 digits long. They say move the 9 in front of other digits (i.e. plural).
Nevertheless, here is some proof of that:
6 9 x 4 = 276 so, no. the answer must have a 9 at the front anyway, and the original number must be more than two digits.

NOW: try doing the reverse action and divide by 4; so, here is the division:

answer: 2 3 0 7 6 9

1 3 2 3
4 9 2 3 0 7 6

  • Set up the division box.
  • Put the numbers in (that you know - the 'divisor 4' and the 9s and 6s you now know about)
  • 4 into 9 = 2 Not only do you have the first digit in the ‘special number’, you can enter it as the 2nd digit in the ‘larger number’ that you are dividing by 4.
  • Keep going… so 12 divided by 4 = 3. Work your way through…

So: 230769 x4 = 923076 Answer = 230769
Click to expand...
Nice Sara, but a little late.
 
Actually, it's even easier than that!

Don't do a division; do a multiplication. You know the last digit of the multiplicand, so write it in and do the first multiplication:

Code: 
    3
... _ 9
     *4
-------
...   6

4*9 = 36, so write the 6 and "carry" the 3. You now know the next-to-last digit is 6, so write it in and multiply again:

Code: 
    2 3
... 6 9
     *4
-------
... 7 6

4*6 = 24, adding 3 makes 27, so write the 7 and "carry" the 2. Keep going!

I recall a problem like this that I did a long time ago where you did the same thing with a division; that probably involved moving a digit the other direction.

Furthermore, trying finding the decimal value of 3/13. You'll be surprised.
 
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