Based on the information below, what numbers should replace A-D? Everything you need to know is there.

A, B, C, D are all integers. A and B are consecutive integers. B and C, C and D are not consecutive integers.

A < B < C < D

AB + CD = _ _

B x C x D ≠ A x B x C x D = _ 1 _


2 Answers 2


The answer is...

The first 4 prime numbers – 2, 3, 5, 7!

How'd I deduce that?

I didn't just blindly guess the first 4 prime numbers 🙂 Here's how:
It immediately seemed strange that the multiplication you showed $(A\times B\times C\times D)$ uses the $\times$ symbol (or rather an x).
There must be reason why – after thinking, it could be because $AB$ and $CD$ are used to denote $\overline{AB}$, or $A$ and $B$ conjoined to form a 2-digit number. Because of this, $A,B,C,D$ must all be single-digit numbers!

From here, I worked out the answer:

As we know $A,B$ are consecutive and $B,C$ aren't, $C$ must be at least $A+3$. Why's that important? 🤔
Well, it has to do with how $\overline{AB}+\overline{CD}$ is a 2-digit number. If $A+C$ is $9$ or more, then $B+D$ is even larger than 9 – that'd mean $\overline{AB}+\overline{CD}$ has 3 digits, and we can safely say $A+C<9$.
Now, if $A=3$, the minimum possible value of $C$ would be $6$, and that doesn't meet our criteria. Since $A\neq1$, that leaves us with only one possible value for $A$: $2$. This also means that $B=3$!
Also, note that because $A+C<9$, $C$'s maximum value here is $6$.

We can now focus on the last equation:

$A\times B\times C\times D$ fits the form $\_1\_$. Since $A=2$ and $B=3$, this is basically $6\times C\times D$. All we have to do is test the $C$ :)
$C=6$ makes it $36\times D$, but that yields no results for $D$ as a single-digit number. $36\times8=288$, and $36\times9=324$.
$C=5$ must be the case! So, $30\times D$ must have $1$ as its second digit, and we can quickly find that $D=7$.

Also, this is my first answer on this site. I hid the answers because everyone else does, hopefully that's good :)

  • 2
    $\begingroup$ Great deduction! Nice first answer, well done :) $\endgroup$ May 15 at 11:48
  • 1
    $\begingroup$ @Prim3numbah thanks, nice puzzle too! Very well-crafted 🙂 $\endgroup$ May 15 at 13:14

The numbers are:

A=2, B=3, C=5, D=9

  • 1
    $\begingroup$ Welcome to Puzzling SE! Please add explanation for your answer and how you got to it, to have a more complete answer. Additionally, if you are interested in learning more about this site (and earning another badge), check out the tour ! $\endgroup$
    – gabbo1092
    May 15 at 12:23

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