$\Large \begin{array} {c c c } & \color{#3D99F6}A & \color{#D61F06}B & \color{#EC7300}C \\ \times& & & 8 \\ \hline \color{#3D99F6}A & \color{#EC7300}C & \color{#EC7300}C & \color{#EC7300}C \\ \end{array}$
In the cryptogram above, $A$ , $B$ , and $C$ are distinct digits. Find the value of $A+B+C+8.$
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Once you know that $C = 0$ , a much easier approach is to run through $\frac{ \overline{A000}}{8}$ to conclude that $A = 2$ .
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125 x 8 = 1000. So 1+ 2+ 5 + 8 = 16 . This is also valid !!
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@Calvin Lin – Lin: 125 x 8 = 1000. So 1+ 2+ 5 + 8 = 16 . This is also valid !! C = 0 !
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@Luis Rivera – Unfortunately 125x8 will not work because given A=1, B=2 and C=5 then ACCC=1555; but you have correctly identified that A=1 and C=0 and since other Number Bases are in General Use, (Binary, Hexadecimal and Sexagesimal etc.) then A=1, B=6 and C=0 with Base 12 ≡ 1000 = ACCC.
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@Sarah Louise Jennings – You said : """ Unfortunately 125x8 will not work because given A=1, B=2 and C=5 then ACCC=1555; """/// Huh !!. Remember its being multiplied by 8, so they end in 0, not 5 !! You guys might as well accept it. Either I demonstrated that the problem does not have a unique solution, or I demonstrated that the problem is flawed !!
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@Luis Rivera – @Luis Rivera Sorry for my typo, I hit 8 instead of 5. You have missed the point that Sarah and I are trying to make. The question asks for
$ABC \times 8 = ACCC$
If you substitute in $A = 1, B=2, C = 5$ , then the equation becomes
$125 \times 8 = 1555$
which is not a true statement.
If you substitute in $A = 1, B = 2, C = 0$ , then the equation becomes
$120 \times 8 = 1000$
which is not a true statement.
I agree that
$125 \times 8 = 1000$
but that doesn't satisfy the conditions.
IE You cannot let C sometimes be 0 and sometimes be 5.
As such, you have not demonstrated that the problem is flawed.
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@Calvin Lin – Can't you see that you are proposing 250. I am proposing 125. Which is exactly half. Problems are solved by simplifying to lowest terms Face it, your problem was ill thought out. I have seen you create very good problems. This was not one of them !!
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@Luis Rivera – Luis, you are still missing the point.
What happens to $ACCC$ when you replace $A = 1, B = 2, C = 5$ ? Do you obtain $1000$ or $1555$ ?
Note that I did not create this problem. I think it's a good problem (and that it's correct as phrased).
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@Calvin Lin – what an epic 4 years old debate LOL
@Luis Rivera – This is invalid... In my opinion.
About B, it also could not be zero, since it explicitly says they are distinct digits and C=0.
I agree with you, this is the Answer that Satisfies the Conditions of the Problem.
250 followers... all that was needed was some trial and error to see if my intuition was correct.
How do you know that the Number Base is 10, it could equally be Binary!
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In order for A, B, and C to be unique digits, it would have to be at least base 3; the number 8 couldn't exist in the problem unless it's at least base 9.
I think it's reasonable to assume base 10 unless told otherwise, but I like your impulse.
Why could C not equal 1? A=3; B=6; C=1 also satisfies the cryptogram. As does A=7; B=5; C=0.
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$361 \times 8 \neq 3111, 750 \times 8 \neq 7000$ .
Your solutions do not satisfy the equation.
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:
$\huge \begin{array} {c c c } & \color{#3D99F6}2 & \color{#D61F06}5 & \color{#EC7300}0 \\ & & \times & \color{#20A900}8 \\ \hline \color{#3D99F6}2 & \color{#EC7300}0 & \color{#EC7300}0 & \color{#EC7300}0 \\ \end{array}$
which is absolutely perfect!
How can 8C=C be correct if C=8? 8C=C is only true when C=0.
Challenge Master, what? You proved that C =/= 8 if 8C = C because 64 =/= 8. 8C = C iff C = 0.
Can't it also be 125 x 8 = 1000
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In that case, what is the value of $C$ ?
For $ABC$ , you used $C = 5$ . But for $ACCC$ , you used $C = 0$ .
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
$(1 \times 8)$ % $10 = (6 \times 8)$ % $10 = 8$
$(2 \times 8)$ % $10 = (7 \times 8)$ % $10 = 6$
$(3 \times 8)$ % $10 = (8 \times 8)$ % $10 = 4$
$(4 \times 8)$ % $10 = (9 \times 8)$ % $10 = 2$
$(5 \times 8)$ % $10 = (0 \times 8)$ % $10 = 0$
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
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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 $A+B+C+D=2+5+0+8=\boxed{15}$ .