A number problem

sharpe669

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Sep 9, 2010
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A number has 4 digits. No digits in the number are repeated. The digit in the tens place is three times the digit in the thousands place. The number is odd. The sum of the digits in the number is 27. What is the number?
 
sharpe669 said:
A number has 4 digits. No digits in the number are repeated. The digit in the tens place is three times the digit in the thousands place. The number is odd. The sum of the digits in the number is 27. What is the number?

Assume that the number is

wxyz

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.
 
I think the number in the ten's place is a 9 and the number in the thousands place is a 3 but i cant get a number to equal 27 and be a odd number. This is my sons math homework and Im not real sure what the number is
 
sharpe669 said:
I think the number in the ten's place is a 9 and the number in the thousands place is a 3 but i cant get a number to equal 27 and be a odd number. This is my sons math homework and Im not real sure what the number is

okay - so you got:

3x9z

you are correct - but why do you think it could not be 2x6z or 1x3z

Let's go back to your finding

3x9z

then

x + z = 15

then x and z can be

9 and 6 - no - we got 9 already
8 and 7

and that's it

Now you tell me how you can make odd number out of it?
 
This seems extremely hard for 4th grade math. Im not sure how to explain this to my son or how to show him how to figure this out.Could someone please point me in the right direction and help me solve this problem. All i could come up with is 3w9x = 12 27-12=15 I couldn't figure out two odd numbers that equal 15 without using 3 or 9 so im questioning wether 3 and 9 is correct or not
 
sharpe669 said:
This seems extremely hard for 4th grade math. Im not sure how to explain this to my son or how to show him how to figure this out.Could someone please point me in the right direction and help me solve this problem. All i could come up with is 3w9x = 12 27-12=15 I couldn't figure out two odd numbers that equal 15 without using 3 or 9 so im questioning wether 3 and 9 is correct or not

Did you read my response above?
 
yes I read your response but I dont understand how the number is odd and the digits add up to 27
 
sharpe669 said:
yes I read your response but I dont understand how the number is odd and the digits add up to 27

For a number to be odd - the number in the unit position must be odd.

I have almost worked it out for you....
 
How can four digits be a odd number and equal 27 . Im racking my brain and I cant come up with it
 
sharpe669 said:
I just cant figure out what the four digits are that equal 27

3x9z

then

x + z = 15

then x and z can be

9 and 6 - no - we got 9 already
8 and 7

and that's it
 
\(\displaystyle My \ way, \ using \ rref.\)

\(\displaystyle \ \ \ \ a \ \ \ \ b \ \ \ \ \ c \ \ \ \ d\)

\(\displaystyle 1 \ \ \ \ 1 \ \ \ \ 1 \ \ \ \ 1 \ \ 27\)

\(\displaystyle 3 \ \ \ \ 0 \ \ -1 \ \ 0 \ \ 0\)

\(\displaystyle This \ gives \ a \ = \ \frac{c}{3}, \ \implies \ c \ = \ 3a \ and \ also \ b \ = \ 27-d-\frac{4c}{3} \ \implies \ b \ = \ 27-d-4a.\)

\(\displaystyle Now, \ when \ a \ =1, \ c \ =3, \ no \ good \ and \ when \ a \ = \ 2, \ c \ =6, \ no \ good, \ hence \ a \ = \ 3, \ c \ = \ 9,\)

\(\displaystyle good.\)

\(\displaystyle Hence, \ we \ have \ 3b9d \ and \ b \ = \ 15-d \ \implies \ d \ \ge \ 7, \ in \ fact \ d \ = \ 7, \hence \ b \ = \ 8.\)

\(\displaystyle Therefore, \ we \ have \ 3897, \ QED.\)
 
sharpe669 said:
I think the number in the ten's place is a 9 and the number in the thousands place is a 3 but i cant get a number to equal 27 and be a odd number. This is my sons math homework and Im not real sure what the number is
(nice, BigGlen, but will probably only add to Mr Sharpe's confusion!)

Mr Sharpe, 3897 adds to 27 (3+8+9+7 = 27) and is an odd number (ends with 7):
so why are you (apparently) thinking that 3897 is not correct?
 
Hello, sharpe669!

A number has 4 digits. .No digits in the number are repeated.
The digit in the tens place is three times the digit in the thousands place.
The number is odd. .The sum of the digits in the number is 27.
What is the number?

Let the number be: .\(\displaystyle ABCD\) .(four different digits)

Since the number is odd, then \(\displaystyle D\) is odd.


We are told that: .\(\displaystyle C\) is 3 times \(\displaystyle A \quad \hdots\quad C = 3A\)

Since \(\displaystyle C\) is a digit (less than 10),
. . we see that \(\displaystyle A\) is 1, 2, or 3 and \(\displaystyle C\) is 3, 6, or 9.


The sum of the digits is 27: .\(\displaystyle A+B+C+D\:=\:27\)

Since \(\displaystyle C = 3A\), we have: .\(\displaystyle A + B + 3A + D \:=\:27 \quad\Rightarrow\quad 4A + B + D \:=\:27\) .[1]


Recall that \(\displaystyle A \:=\: 1, 2, 3\)

If \(\displaystyle A = 1\), [1] becomes: .\(\displaystyle 4(1) + B + D \:=\:27 \quad\Rightarrow\quad B + D \:=\:23\) . . . impossible!

If \(\displaystyle A = 2\), [1] becomes: .\(\displaystyle 4(2) + B + D \:=\:27 \quad\Rightarrow\quad B + D \:=\:19\) . . . impossible!


Hence: .\(\displaystyle A = 3,\;C = 9\) and [1] becomes: .\(\displaystyle 4(3) + B + D \:=\:27 \quad\Rightarrow\quad B + D \:=\:15\)

Since \(\displaystyle D\) is odd, we have these two possibilites: .\(\displaystyle (B,D) \;=\;(6,9)\text{ or }(8,7)\)
. . But \(\displaystyle 9\) is already assigned to \(\displaystyle B.\)
Hence, the only pair is: .\(\displaystyle B = 8,\:D = 7\)


Therefore, the number is: .\(\displaystyle 3897\)

 
\(\displaystyle Denis, \ Mr. \ Sharpe \ is \ already \ confused, \ hence \ I \ fail \ to \ see \ where \ my \ input\)

\(\displaystyle would \ alleviate \ his \ confusion.\)
 
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