# Math Help - Problem

1. ## Problem

The last digit of a six-digit number is 2. If the 2 is moved to the start of the number, the new six-digit number is only a third of the original number. Find the original number.

2. Originally Posted by not happy jan
The last digit of a six-digit number is 2. If the 2 is moved to the start of the number, the new six-digit number is only a third of the original number. Find the original number.
Digitwise, the number is abcde2

the new number is 2abcde

so 3 times the new number equals the old number

$\begin{array}{lllllll}
&a&b&c&d&e&2\\
=&6&3a&3b&3c&3d&3e\\
\end{array}$

3e must have a ones digit of 2. So it could be equal to 2, 12, 22, 32, etc

But we know it is a digit, so it is less than 10, and since 3*9 = 27, it must equal 2, 12, or 22. And since 12 is the only one of those which is a multiple of 3, it must be 12. then 12/3=4. so e =4

$\begin{array}{lllllll}
&a&b&c&d&4&2\\
=&6&3a&3b&3c&3d&12\\
\end{array}$

we can't have a ones digit of 12, so we carry the one.

$\begin{array}{lllllll}
&a&b&c&d&4&2\\
=&6&3a&3b&3c&3d+1&2\\
\end{array}$

now 3d+1 returns a ones digit of 4. so options are 4, 14, and 24 (because d cannot be greater than 9, so 3d+1 cannot be greater than 28) subtracting one from each of these we get 3d = 3, 13, or 23. and clearly 3 is the only option returning an integer. so 3d=3. thus d = 1

$\begin{array}{lllllll}
&a&b&c&1&4&2\\
=&6&3a&3b&3c&4&2\\
\end{array}$

now we know that 3c has a ones digit of 1. so 3c equals 1, 11, or 21 (again, must be less than 27, because c < 9). 21 is the only multiple of 3. so 3c = 21. therefore c = 7

$\begin{array}{lllllll}
&a&b&7&1&4&2\\
=&6&3a&3b&21&4&2\\
\end{array}$

21 is not a digit, so we carry the 2.

$\begin{array}{lllllll}
&a&b&7&1&4&2\\
=&6&3a&3b+2&1&4&2\\
\end{array}$

now we have that 3b+2 must have a ones digit of 7. so 7, 17, or 27. Subtracting 2 we get that 3b must equal 5, 15, or 25. 15 is the only multiple of 3. so 3b=15. therefore b = 5

$\begin{array}{lllllll}
&a&5&7&1&4&2\\
=&6&3a&17&1&4&2\\
\end{array}$

17 is not a digit so carry the one.

$\begin{array}{lllllll}
&a&5&7&1&4&2\\
=&6&3a+1&7&1&4&2\\
\end{array}$

Now we get that 3a+1 has a ones digit of 5. so 5, 15, or 25. Subtracting one we get that 3a must be 4, 14, or 24. 24 is the only multiple of 3. so 3a=24. therefore a = 8

$\begin{array}{lllllll}
&8&5&7&1&4&2\\
=&6&25&7&1&4&2\\
\end{array}$

25 is not a digit, so carry the 2.

$\begin{array}{lllllll}
&8&5&7&1&4&2\\
=&8&5&7&1&4&2\\
\end{array}$

Now looking back at the original, equation we see that this becomes

$\begin{array}{lllllll}
&a&b&c&d&e&2\\
=&8&5&7&1&4&2\\
\end{array}$

Therefore the number is 857142, and the new number is 285714.

3. Hello, not happy jan!

Lucky for both of us, I'm familiar with this type of problem . . .

The last digit of a six-digit number is 2. If the 2 is moved to the front,
the new six-digit number is only a third of the original number.
Find the original number.
We have: . $ABCDE2$

Then: . $2ABCDE \:=\:\frac{1}{3}\times ABCDE2 \quad\Rightarrow\quad 2ABCDE \times 3 \;=\;ABCDE2$

And we have a multiplication problem . . .

. . $\begin{array}{cccccc}
^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\
2 & A & B & C & D & E \\
\times & & & & & 3 \\ \hline
A & B & C & D & E & 2 \end{array}$

In column-6, $3E \text{ ends in 2}\quad\Rightarrow\quad\boxed{ E = 4}\quad\hdots$ ("carry 1" to column-5)

. . $\begin{array}{cccccc}
^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\
2 & A & B & C & D & 4 \\
\times & & & & & 3 \\ \hline
A & B & C & D & 4 & 2 \end{array}$

In column-5, $3D + 1\text{ ends in 4}\quad\Rightarrow\quad\boxed{D = 1}$

. . $\begin{array}{cccccc}
^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\
2 & A & B & C & 1 & 4 \\
\times & & & & & 3 \\ \hline
A & B & C & 1 & 4 & 2 \end{array}$

In column-4, $3C\text{ ends in 1}\quad\Rightarrow\quad\boxed{C = 7}\quad\hdots$ ("carry 2" to column-3)

. . $\begin{array}{cccccc}
^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\
2 & A & B & 7 & 1 & 4 \\
\times & & & & & 3 \\ \hline
A & B & 7 & 1 & 4 & 2 \end{array}$

In column-3, $3B + 2\text{ ends is 7}\quad\Rightarrow\quad\boxed{B = 5}\quad\hdots$ ("carry 1" to column -2)

. . $\begin{array}{cccccc}
^1 & ^2 & ^3 & ^4 & ^5 & ^6 \\
2 & A & 5 & 7 & 1 & 4 \\
\times & & & & & 3 \\ \hline
A & 5 & 7 & 1 & 4 & 2 \end{array}$

In column-2, $3A + 1\text{ ends in 5}\quad\Rightarrow\quad\boxed{A = 8}$

. . $\begin{array}{cccccc}
2 & 8 & 5 & 7 & 1 & 4 \\
\times & & & & & 3 \\ \hline
8 & 5 & 7 & 1 & 4 & 2 \end{array}\qquad\hdots$
ta-DAA!

Finally took a look at your solution, angel.white . . .
Excellent job . . . nice explanations!

4. Originally Posted by Soroban
Finally took a look at your solution, angel.white . . .
Excellent job . . . nice explanations!
Thanks ^_^ your way was a bit clearer b/c it didn't rely on unconventional mathematical notations like mine. (I sort of figured it out / made it up as I went).