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Given:

${AA}\times{BC}=BDDB$

Find $BDDB$:

  1. $1221$
  2. $3663$
  3. $4884$
  4. $2112$

The way I solved it:

First step - expansion & dividing by constant ($11$): $AA\times{BC}$=$11A\times{BC}$

  1. $1221$ => $1221\div11$ => $111$
  2. $3663$ => $3663\div11$ => $333$
  3. $4884$ => $4884\div11$ => $444$
  4. $2112$ => $2112\div11$ => $192$

Second step - each result is now equal to $A\times{BC}$. We're choosing multipliers $A$ and $BC$ manually and in accordance with initial condition. It takes a lot of time to pick up a number and check whether it can be a multiplier.

That way I get two pairs:

$22*96$=$2112$

$99*37$=$3663$

Of course $99*37$=$3663$ is the right one.

Is there more efficient way to do this? Am I missing something?

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There are simple divisibility tests that can help here. A number is divisible by $2$ if and only if its last digit is; the same is true for $5$. A number is divisible by $3$ if and only if the sum of its digits is divisible by $3$. There's also a rule for divisibility by $7$, but it's a bit more complicated. –  Michael Joyce Nov 25 '12 at 20:20
    
@MichaelJoyce I know these rules. The problem is that it takes a long time. This question is from some book and it is said that a solution for this question can be found within one minute. That's why I'm curious whether there is a more efficient of doing this. –  artyh Nov 25 '12 at 20:56

2 Answers 2

I'm not sure what you mean by "pick up a number and check whether it can be a multiplier".

You can factorize the numbers $111$, $333$, $444$, $192$:

$$ \begin{align} 111&=3\cdot37\;,\\ 333&=3^2\cdot37\;,\\ 444&=2^2\cdot3\cdot37\;,\\ 192&=2^6\cdot3\;. \end{align} $$

From these factorizations it's straightforward to find the factorizations into one single-digit and one double-digit number:

$$ 111=3\cdot37\;,\\ 333=9\cdot37\;,\\ 444=6\cdot74\;,\\ 192=2\cdot96\;,\\ 192=3\cdot64\;,\\ 192=4\cdot48\;,\\ 192=6\cdot32\;,\\ 192=8\cdot24\;. $$

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Don't consider my solution. Can you show me a simple and straightforward solution to the initial problem? I think that my solution is a bit overcomplicated/irrelevant. Forget it. –  artyh Nov 25 '12 at 20:50

I think your division by $11$ is a good first step. Now maybe you know $111=3 \cdot 37$ and that is the key to factoring three of your four choices. Note that now you need a single digit $A$ times a two digit $BC$, so $333=9\cdot 37$ and $444=6 \cdot 74$ are the only acceptable ones and should be quick to find from the first. For $333$ you have to add in a factor $3$ and $3 \cdot 37$ has three digits, so it has to pair with the $3$. For $444$, you need to add in a factor $4$ and neither number can take the whole thing, so they have to share. Your possibilities are $33\cdot 37=1221, 99 \cdot 37=3663, 66 \cdot 74=4884$. Seeing that the second works, we are done.

Alternately, you might just notice that $3663+37=3700$, too big a coincidence to ignore and say $3663=(100-1)37$

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