1.) The sum of the digits of a three - digit number is 17. The hundreds digit is thrice the tens digit. This number is 297 less than the number formed by reversing the order of the digits. What are the two numbers?

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- Sep 3rd 2006, 02:57 AM^_^Engineer_Adam^_^Help me on digit problem T_T
1.) The sum of the digits of a three - digit number is 17. The hundreds digit is thrice the tens digit. This number is 297 less than the number formed by reversing the order of the digits. What are the two numbers?

- Sep 3rd 2006, 03:12 AMRandom333
If the hundreds digit is twice the tens digit then the hundreds digit must be even.

The hundreds digit then can't be less than 6 because half of 4 is 2 and 2+4=6 which then wouldn't enable a third number to add up to 17.

Using 6 as the hundreds digit:

Therefore the tens digit must be three and the last 8.

The reverse then is 836, 836-638=198, so this is not the number.

Using 8:

The tens digit is 4

And the last digit must be 5

The reverse then is 548,

845-548=297

Numbers are 845 and 548. - Sep 3rd 2006, 06:00 AMSoroban
Hello, ^_^Engineer_Adam^_^!

[Random333, you missed the word "thrice".]

Quote:

1.) The sum of the digits of a three - digit number is 17.

The hundreds digit is thrice the tens digit.

This number is 297 less than the number formed by reversing the order of the digits.

What are the two numbers?

Since this is obviously an Algebra problem, I used an algebraic approach.

Let $\displaystyle H$ = hundreds digit, $\displaystyle T$ = tens digit, $\displaystyle U$ = units digit.

Then the number is: .$\displaystyle N \:=\:100H + 10T + U$

The sum of the digits is 17: .$\displaystyle H + T + U\:=\:17$**(1)**

The hundreds digit is three times the tens digit: .$\displaystyle H\:=\:3T\quad\Rightarrow\quad H - 3T\:=\:0$**(2)**

$\displaystyle N$ is 297 less than its reversal . . . Its reversal is: .$\displaystyle 100U + 10T + H$

Hence: .$\displaystyle 100H + 10T + U \:=\:(100U + 10T + H) -$$\displaystyle 297\quad\Rightarrow\quad H - U \:=\:-3$**(3)**

We have a system of equations: .$\displaystyle \begin{array}{ccc}(1)\\(2)\\(4)\end{array} \begin{array}{ccc} H + T + U \:= \:17 \\ H - 3T\qquad \:= \:0 \\ H\qquad - U \:= \:\text{-}3\end{array}$

. .Add (1) and (3): . . $\displaystyle 2H + T \:=\:14$

Subtract twice (2): .$\displaystyle -2H + 6T\:=\:0$

And we get: .$\displaystyle 7T = 14\quad\Rightarrow\quad T = 2$

Substitute into (2): .$\displaystyle H - 3(2)\:=\:0\quad\Rightarrow\quad H = 6$

Substitute into (1): .$\displaystyle 6 + 2 + U \:=\:17\quad\Rightarrow\quad U = 9$

Therefore: .$\displaystyle \boxed{N\:=\:629}$, and its reversal is $\displaystyle 926.$

- Sep 3rd 2006, 06:05 AM^_^Engineer_Adam^_^
Thanks you sir!!! :D