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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?

  • 0
    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.2012-11-25
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    @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.2012-11-25

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