MHB Problem on Simultaneous equations

[mathlearn]

There is a number between 100 and 1000. Its middle digit is 0.sum of the rest , first and last digits is 11.The number which gets by exchanging the first digit and the last digit is greater than by 495 from the previous number

I. Build up an equation for the sum of first digit and the last digit.

II.Build up an equation for the number which gets by exchanging the first digit and the last digit

III. Find the number

I'm stuck on this problem so far i think and I'm not sure whether its correct my equation would be

f(first digit)+l(last digit)-r(rest digits)=11

an I'm completely unable to think a solution for the second and third problem.

Can you help me on this sum and would you be kind enough to explain this in a little detailed manner

Many thanks

 

[MarkFL]

I would let $F$ be the first digit and $L$ be the last digit, and so the number $N$ is:

$$N=100F+L$$

Now, we are told:

$$F+L=11\tag{1}$$

and:

$$(100L+F)-(100F+L)=495$$

If we simplify this last equation by combining like terms, we find:

$$99L-99F=495$$

Dividing through by 99, we obtain:

$$L-F=5\tag{2}$$

What do we get if we add (1) and (2)?

 

[mathlearn]

Then we get 2L=16 and $$\therefore L=8$$

- - - Updated - - -

So Is the number 8?

Many Thanks

 

[MarkFL]

Then we get 2L=16 and $$\therefore L=8$$

Correct! (Yes)

So, what must $F$ be, and then what must $N$ be?

 

[mathlearn]

Can you tell me how did you arrive at

"I would let FF be the first digit and LL be the last digit, and so the number NN is:

N=100F+LN=100F+L"

Many Thanks

 

[mathlearn]

L-F=5
8-F=5
-F=5-8
F=3

and

N=100*3+8
N=308

So is the number 308 or just 8

 

[MarkFL]

I chose $F$ to be the First digit and $L$ to be the Last digit so it would be obvious which is which. How you decide to name your variables when solving a problem is up to you, as long as it makes sense to you.

Now, a base 10 (decimal) number is represented by the numerals (0-9) multiplied by powers of 10...for example the number 512 is:

$$5\cdot10^2+1\cdot10^1+2\cdot10^0$$

So, if we have a 3 digit number, where the first digit is $F$, the s econd digit is 0 and the last is $L$, then the number $N$ is:

$$N=F\cdot10^2+0\cdot10^1+L\cdot10^0=100F+L$$

Does this make sense?

 

[MarkFL]

L-F=5
8-F=5
-F=5-8
F=3

and

N=100*3+8
N=308

So is the number 308 or just 8

$$N=308$$, so the number you are asked to find is 308, and indeed we find:

$$803-308=495$$ as required. :D

 

[mathlearn]

so to sum up then,

F=3
L=8
N=308

Many Thanks