Replace the letters $A$, $B$, $C$, $D$, $E$ with the numerals $1$, $2$, $3$, $4$, $5$ such that the following equation is true:
$$AB*C=DE$$
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5 Answers
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$13*4=52$ is the only solution. We note that we must have $D=5$, since $E = 5$ forces $B$ or $C = 5$, and vice versa, while $A \leq D$ prevents $A = 5$.
Now factorising the possibilities for $DE$:
$51 = 17 * 3$
$52 = 13 * 4 = 26 * 2$
$53$ is prime
$54 = 27 * 2 = 9 * 6 = 3 * 18$
Only $52 = 13 * 4$ is of the correct form.
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The only two numbers that multiply together $B*C$ to another (E) is $3*4 = 12$
That leaves 1 and 5 remaining, so obviously $DE = 52$.
52 doesn't divide into 3, so $AB = 13, C = 4$ and
$13 * 4 = 52$
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You really don't have many cases to analyze, because of two pretty strict conditions that come from the puzzle's hypothesis.
Since the result must be smaller than $54$ because it's the greatest two digit number you can write with $1,2,3,4,5$. The only $AB$ combinations that multiplied for another number $C \in \{1,2,3,4,5\}\setminus\{A,B\}$ give a result $DE$ that is lower than $54$ are $12,13,14,15,21$.
We also can exclude couples in which $C=1$ because in that case
$AB*C=AB*1=DE \implies AB=DE$
The possible combinations for the LHS reduce to
$\{12,3\},\{12,4\}$
$\{13,2\},\{13,4\}$
$\{14,2\},\{14,3\}$
$\{15,2\},\{15,3\}$
A bit of try and error will lead to the solution:
$13*4=52$
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Answer
$13 * 4 = 52$
Proof
C cannot be 1 because then $AB = DE$ which cannot be. Also, C cannot be 5 since it would mean $D >= 5$
If A is 2, and because C cannot be 1 it means C has to be at least 3. This is not possible because then D would be 6 or more. Same goes for A>2.
So $A = 1$.
For $B = 2$.
$12 * 3 = 36$ - not good
$12 * 4 = 48$ - not good
Stop here because C cannot be 5
For $B = 3$.
$13 * 2 = 26$ - not good
$13 * 4 = 52$ - we might be on to something here
Stop here because C cannot be 5
For $B = 4$
$14 * 2 = 28$ - not good
$14 * 3 = 42$ - Since 42 is the answer for everything I'm tempted to count this one also as a good solution. But I won't for now
For $B = 5$
$15 *2 = 30$ - wrong
$15 * 3 = 45$ - close but we are missing a 2
$15 * 4 = 60$ - wrong.
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3*3=9?
AB*C=DE
12*3=45
(1+2)*3=(4+5)
3*3=9
I'm guessing this is the inferred solution as it has come from a school - not a pure mathematics handbook - although it would appear that the question is poorly written (originally, not by the OP)
AB\times C=DEwhich becomes $AB\times C=DE$. $\endgroup$