Stumped on an alphanumeric puzzle

Question

My 5th grade daughter gave me this problem, and I am stumped! Each letter, $A,B,C,D,E$ or $F$ represents one of the digits, $1,2,3,4,5$ or $6$. You are given the following problem. $D \cdot BCEA=F8B0$. Solve for what each letter represents. Any thoughts?

Answer 1

A brute force approach yields $ A = 5, B = 3, C=4, D=2, E=1, F=6$. The point is that we need to check that $$ D(1000B+100C+10E+A) = 1000F+800+10B. $$

Answer 2

Laying the multiplication by $D$ we get $\begin{cases} DA=0&\text{with carry }r_1\\ DE+r_1=B&\text{with carry }r_2\\ DC+r_2=8&\text{with carry }r_3\\ DB+r_3=F \end{cases}$

The two constraints we have here is all numbers $\le 6$ and also all different.

The first line is particularly interesting, only way to get a $0$ on right side, i.e. a multiple of $10$:

• $2\times 5$ with carry $r_1=1$
• $4\times 5$ with carry $r_1=2$.

So $A,D$ are taken among $2,4,5$.

But then the last line is also a strong constraint, we must have $DB\le 6$

since $DB\le DB+r_3\le F\le 6$.

• Let examine $D=4$

This would force $A=5,r_1=2$ and $B=1$ to get $DB\le 6$

The second line would become $DE+2=1$ with some carry

Possible values for $E$ are $2,3,6$ since $1,4,5$ already taken by $B,D,A$

leading to $DE+2=4E+2=10,14,26\ $ but none of them ends up with digit $1$.

This hypothesis is wrong.

• Let examine $D=5$

This would also force $B=1$ to get $DB\le 6$

but since $F$ and $D$ cannot be the same we need $F=6$ and $r_3=1$

This means $5C+r_2=18$ which is only possible for $C=3$ and $r_2=3$

Continuing propagating this means $5E+r_1=31$

Since $\begin{cases}A=2\implies 2\times 5=10\implies r_1=1\\A=4\implies 4\times 5=20\implies r_1=2\end{cases}$

We get to solve $5E=30$ or $5E=29$, only $5E=30$ is possible with $E=6$, but this digit is already taken by $F$.

This hypothesis is also wrong.

• The only remaining possibility is $D=2$

This leads to $A=5$ and $r_1=1$

The remaining digits for $E$ are $1,3,4,6$

Therefore $DE+r_1=2E+1=3,7,9,13$

Notice that $7,9$ are too large to be admissible digits so $B=3$ is forced for $E=1,6$.

The value of $B$ in turn forces $r_3=0$ and $F=DB=6$

Since $E\neq F$ then $E=1$ is forced.

$C$ takes the remaining available value $4$.

Summary $A=5, B=3, C=4, D=2, E=1, F=6$

Verification: $D\times BCEA=2\times3415=6830=F8B0$

Answer 3

We can cut down on trials by using "multiplication facts".

To get a product ending with 0 with the given range for A through F, either D=5 or A=5. Suppose the former. Then B=1 (to avoid a five- digit product) and F=6 (the product is greater than 5000 and we have to make F different from B). So the product is 6810, but then 6810/5=1362 which gives a wrong digit for E (E=F=6).

So A=5 and D is even. Both D=4 and D=6 would force B=1, which leads to contradictions: D=4, B=1 gives a product that's not a multiple of 4; and D=6, B=1 gives a product less than 6000 (we can't have F=D=6).

So D=2, A=5. To get a product less than 7000 we must accept B=1 or B=3. B=1 forces F=3 and the product is 3810, but this gives wrong digits when divided by 2 for BCEA (1905). The alternative, B=3, gives F=6 and a product of 6830, and this time we get proper digits for BCEA:

2×3415=6830

Answer 4

We can start with discussion:

1. $B \neq 6$, because multiplying of a four-digit number (starting with the number $6$) by the number $2 \le D \le 5$ yields to five-digit number(s), and in the case $D=1$ we will have $B=D$ what is not allowed in this puzzle.
   
2. Similarly we can claim $B \neq 5$.
   
3. For a similar reason we know $B\neq 4$. Ok, there is one different situation with previous two cases, when $D = 2$, but in this situation $F$ should be equal to $8$ or greater than $8$, not allowed as well.
   
4. $B \neq 2$, because multiplying a four-digit number (starting with the number $2$) by the number $4 \le D \le 6$ yields either to five-digit number(s) or to four-digit number which starting with numbers $8$ or $9$. Next, we should exclude $D=3$, since in that case there is no $A$, such that $D \cdot A$ will result with a number which ends with $0$. Finally, once again, $D=1$ is not allowed.
   
5. Trivially $B \neq 1$.
   
6. It is mandatory to have: $B=3$.
   
7. Now, only possible value for $D$ is $2$, what implies $F=6$, $A=5$, $E=1$ and $C = 4$.

Answer 5

$D \cdot BCEA=F8B0\Leftrightarrow$ $D(1000B+100C+10E+A) = 1000F+800+10B$

Now, $1000B\times D+100C\times D+10E\times D+A\times D = 1000F+800+10B$(clearly RHS is not more than $1000F + 860$)

$B,D,F \in\{1,2,3,4,5,6\}$ and $B,D,F$ are diffrent numbers. This the puzzle to find $B\times D=F$ there only possible in the range B,D are $3$ or $2$ but F is no doubt $6$.

Now, $100C\times D+10E\times D+A\times D = 800+10B$(clearly RHS is not more than $850$)

So, $C\times D=8$ i.e C,D are $2$ or $4$ and previously we get B,D are $3$ or $2$.

So, $D=2,C=4,B=3$ only left E,A and $5,1$

Now, $20E+2A = 30$ hence $E=1,A=5$