A logic problem by Christopher Boo

Relevant wiki: Cryptogram

ABC + BCA+CBA = 100(A+B+C) +10(A+B+C) +1(A+B+C) = 111(A+b+C) so answer is multiple of 111 so option is 666

All three columns are effectively the same, with a sum $A+B+C$ .

There is no carry-over digit from the first column, since there is no column in front of it, so there cannot be any carry-over in any of the other columns either.

Therefore all the digits must be the same.

The only answer satisfying that condition is $\boxed{666}$ .

This solution can be actually realized, for example by setting $A=1, B=2, C=3$ .

Since A, B and C appear in each column, then their total is 111*(A+B+C) and 111 divides 666

Each column is the same so the sum must also be the same (assuming no sum larger than 9 as evidenced in the proposed solutions). Process of elimination reveals 666 is only possible answer.

ABC=100A+10B+1C ,BCA=100B+10C+1A ,CAB=100C+10A+1B . SO,NOW,

ABC+BCA+CAB=

100(A+B+C)+10(A+B+C)+1(A+B+C)=

(100+10+1)(A+B+C)=111(A+B+C)

So the answer must be a multiple of 111 Hence,the anwer is 666.

Relevant wiki: Cryptogram

You can rearrange this cryptogram to $\begin{array} { rrrr } & & A & A & A \\ & & B & B & B \\ &+ & C & C & C \\ \hline & & &\phantom0 & \end{array}$ Than each column has A+B+C, so the answer is 666

If you rearrange the digits of a number, when take its remainder when it is divided by 3, the remainder stays the same. Therefore, ABC, BCA, and CAB must have the same remainder when divided by 3. No matter if this remainder is 0, 1, or 2, when we add the same remainders three times, we will get a multiple of 3, because 0+0+0=0, 1+1+1=3, and 2+2+2=6. Therefore, the answer has to be a multiple of 3, and the only answer choice which satisfies this is 666.

Relevant wiki: Divisibility Rules (2,3,5,7,11,13,17,19,...)

The digital root of each of the three numbers is, by definition, A+B+C (modulo 9)

And so the sum of the three numbers has digital root 3(A+B+C) (modulo 9).

So the digital root of the total is a multiple of 3.

This is a well known test for divisibility by 3, and the only choice satisfying this criterion is $\boxed{666}$

100A+10B+C+100B+10C+A+100C+10A+B=111A+111B+111C=111(A+B+C) 666 is the only possible answer.

In addition, associative property holds. So, no matter what the order is the answers will be same in each row vertically as A,B,C are present in each vertical row.

If it is helpful then pls upvote.

Since every column sums the same 3 digits the only way that the answer is not 3 identical digits is if A+B+C > 9 but if that were true the answer would be 4 digits. The only answer with 3 digits all the same is 666.

If you separate each character with a line, you get a MS Excel table. This leads you to a different point of view: know you can see rows and columns of characters. Whatever, when you read the equations in every possible direction up down and down up, left to right and right to left, you always get some kind of palindromic equations of this shape: (A+B+C)=(B+C+A)=(C+A+B)=6 And then I'm becomes x=x=x=6 And x=6 So x=6/(3 equal valued single-digits )=6/3=2

Check, 2+2+2=6 in 3 columns of this visual palindromic equation I told before. The result is definitely 666 with this demonstration 👹

Let me know you comments! But please forgive my poor English🙏

I found the phrase "not necessarily all distinct" confusing. What is that saying?

I like all the solutions given below, but a combination of Scott Rodham's and Marta Reece's answers come closest to the solution I found. I just looked at the possible answers and they all started with 6. So, that left 1 + 2 + 3 as the only possible combination for the columns making 666 the only possible answer.

As a side note, this kind of highlights why multiple choice questions lack when it comes to assessing student's knowledge.

A, B, and C are limited within digits 1 - 9. One can eliminate digits - 4, 5, 6, 7, 8, and 9 for A, B, or C; since the 1st two columns add up to 6. So the inequality: 1 < = A or B or C < = 3 holds. I have zero background in this subject - I'm just an signatures/electric engineer. Cheers!

Since the first row has to add up to 6 and there's 3 letters 6 divided 3 is 2 So each letter is 2

If we assume A=1 B=2 & C=3, Most probable answer is 666

I'm not a math person, at all (lol - that is why I am here!). I got the correct answer, but my brain is . . . odd.

A/B/C are single-digit integers, and they do not have to be distinct from each other. So the negative whole numbers from -9 through to 0 up to +9 on the positive whole number side.

Seeing that the rows & columns will work out exactly the same, the 3 numbers at the bottom will have to be identical, as well as the sum of A+B+C cannot be greater than 9. The only fitting answer is 666.

So, A/B/C could be any variation of ... what? (0,0,6), (0,3,3), (0,2,4), (1,0,5), (1,1,4), (1, 2, 3), and (2,2,2).

Negatives can't be used the way it is set up (as an addition problem); otherwise, things like (-3, 0, 9) would also work.

The Given Answers were an aide to solving the Cryptogram. And using the Commutative Property, Order of Addition states the #s could be added in any order. The Problem Statement said 3 INTEGERS, that eliminated ZERO. So the 3 Digits had to be 123. 1+2+3 is going to equal 6 any order. The Sum is 666.

A + B + C =600 so to have all the digits differ, the best choice would be 1, 2, 3. Each letter is a different number (1,2,3).

Because of the commutative property of addition, each column should have the same sum value so 666 (number of the beast lol) is an obvious solution. There is, of course, the possibility of a carry, but none of the solutions fit that. If A + B + C = 15 then you'd get the answer 1,665.

We know that the solution will present itself under the form of

$(A+B+C)\cdot 100 + (B+C+A)\cdot 10 + (C+A+B) =(A+B+C)\cdot 111$

Thus the solution is a multiple of $111$ , thus it has to be $666$ .

Note that it could also have been $111,222,333,444,555,666,777,888,999$

I assigned values to A, B and C. Since I saw the sum has 6 in any of its digits so my values should add up to 6. I said let A=1, B=2 and C =3. Thus, the possible value for $XYZ=666$