250 followers problem!

Lets crack this using simple congruence:

Lets examine the unit's place to get:

$C \times D \equiv C \pmod{10} \\ \Rightarrow 8C\equiv C \pmod{10} \\ \Rightarrow 7C\equiv 0 \pmod{10}$

This is only possible when $\boxed{C=0}$ .

Since there is no carry over on $B$ , lets examine the ten's place to get:

$B \times D \equiv C \pmod{10} \\ \Rightarrow 8B\equiv 0 \pmod{10}$

This is possible when $B=(0,5)$ .

But when $B=0$ then $\overline{A00} \times D = \overline{A000}$ which means $D=10$ which is a contradiction.

Hence , $\boxed{B=5}$ .So $B \times D= 5 \times 8= 40$ which means there is a carry over of $4$ on $A$ .

Now lets examine the hundred's place where we can form an equation:

$A\times D + 4=\overline{AC} \\ \Rightarrow A\times 8 + 4=10A+C \\ \Rightarrow 8A + 4 =10A \\ \Rightarrow 2A=4 \\ \Rightarrow \boxed{A=2}$

So we have $A+B+C+D=2+5+0+8=\boxed{15}$ .

According to the question, the last digit of $8C$ is $C$ . If this is true, $C$ must be equal to $0$ ,

• So we have concluded that $\boxed{C=0}$ .

• Now, the problem becomes: $\huge \begin{array} {c c c } & \color{#3D99F6}A & \color{#D61F06}B & \color{#EC7300}0 \\ & & \times & \color{#20A900}8 \\ \hline \color{#3D99F6}A & \color{#EC7300}0 & \color{#EC7300}0 & \color{#EC7300}0 \\ \end{array}$

We can write this as: $(100A+10B)×8=1000A \\ 800A+80B=1000A \\ 80B=200A \\ \boxed{2B=5A}$

• We find that the possible it comes are, $B=5, A=2$ or, $B=10, A=4$ and so on. . .

• But as $A$ and $B$ are one digit numbers the only option is $\boxed{B=5}$ and $\boxed{A=2}$

Let's now verify whether the values are correct:

• Substituting the values of $\boxed{A=2}$ , $\boxed{B=5}$ and $\boxed{C=0}$ in the question we get,

$\huge \begin{array} {c c c } & \color{#3D99F6}2 & \color{#D61F06}5 & \color{#EC7300}0 \\ & & \times & \color{#20A900}8 \\ \hline \color{#3D99F6}2 & \color{#EC7300}0 & \color{#EC7300}0 & \color{#EC7300}0 \\ \end{array}$

which is absolutely perfect!

From the ones column,

$8C = 10k+C$ where $k$ is a non-negative integer.

$7C = 10k$

Since $7$ and $10$ are coprime, $C$ must be divisible by $10$ but for $C$ to be single-digit, the only possibility is $C = 0$ .

Now we have $AB0 \times 8 = A000$ or 10(AB) \times 8 = 1000A)

$8AB = 100A$

$8(10A+B) = 100A$

$20A = 8B$

$5A = 2B$ .

For $A$ and $B$ to both be single-digit positive integers, $A =2$ and $B = 5$ . $A+B+C+D = 2+5+0+8 = \boxed{15}$ .

Simple inspection of the 8's multiplication table gives all digits.

The only one-digit number for which 8 * C ends in C is C = 0.

The problem is now AB0 * 8 = A000. This requires that 8 * B ends in 0, which means B = 0 or 5. Since 0 is already taken, this forces B = 5.

The problem is now A50 * 8 = A000. Since 8 * 5 carries a 4, 8 * A + 4 must end in 0, or 8 * A must end in 6. This requires A = 2 or A = 7. 750 * 8 = 6000, and 6 =/= 7, so A must be 2. This gives 250 * 8 = 2000, which fits the cryptogram. 2+5+0+8 = 15.

The only number C can be is 0, because C has be something, when multiplied by 8, gives a number that's last digit is also C. 0 is the only 1-digit number that works. So, C = 0.

Next, B has to be a number, that when multiplied by 8, a number that's last digit is 0. Since 0 was already used, we have to use 5, the only other number that fits. So B = 5.

A has to be a number, that when multiplied by 8, gives a number that ends in 6 (because of the carry-over) and it's first digit is A. 2 is the only number that works. So A = 2

2 + 5 + 0 + 8 = 15

1. So, for all integer values where $0 \leq C \leq 9$ , $f(C) = (8C) \% 10$ AND $g(C) = C$ ,

$(1 \times 8)$ % $10 = (6 \times 8)$ % $10 = 8$

$(2 \times 8)$ % $10 = (7 \times 8)$ % $10 = 6$

$(3 \times 8)$ % $10 = (8 \times 8)$ % $10 = 4$

$(4 \times 8)$ % $10 = (9 \times 8)$ % $10 = 2$

$(5 \times 8)$ % $10 = (0 \times 8)$ % $10 = 0$

The only number where $f(C) = g(C)$ is where $C = 0$ .

This means that $B$ is either $0$ or $5$ .

But if it was $0$ , then $A$ would also have to be $0$ or $5$ , but it cannot be $5$ because $5 \times 8 = 40$ , which does not satisfy the equation; and it cannot be $0$ because they are all distinct digits. So $B$ must be $5$ .

So now we must find where

$10A = 8A + 4$

$2A = 4$

$A = 2$

Thus,

$A + B + C + 8 = 2 + 5 + 0 + 8 = 15$

using multiplication facts we know there is no single digit number multiplied by 8 that is the same digit. Knowing this, means that the digit (or c) must be 0. B multiplied by 8 has to equal 0. Which means B equals 0 or 5. Since 0 is already C, B must equal 5. You then carry on the 4 to A. Using multiplication facts A must be 2. 8 times 2equals 16 + 4 = 20

8×c=c possible only if c=0, 8×B=c/0 possible only if B=5or0. 8×A=A0 possible only when we get a remaider 4 ,therefore B=5 &A=2