I am taking from The Official Hunter College High School Test: problem 76 on page 20. We read the following.

In the expression below, each letter represents a one digit number. Where the same letter appears, it represents the same number in each case. Each distinct letter represents a

different number. In order to make the equation true, what number must replace C?

AAA

AAB

+ ABC

2012

A great start is to decode each AAA, AAB, and ABC. It helps to look at this problem wholly; particularly we look at the leading sum on the left wall (of the same types). We glean that either: (1) A + A + A = 20, (2) A + A + A + 1 = 20 or (3) A + A + A + 2 = 20: its very important to remember that given three numbers each less than ten, the sum of them which is great, is at most 2 in the tens place. This means that each row can only donate a 1 or 2 to the next. We can conclude that our line is line (3). That, A + A + A + 2 = 20 is the correct choice of the three since A may only equal an integer and…that 20/3 and 19/3 are not integers.

We have shown that A = 6.

If A = 6, then A+A+B, as a sum is almost solved for. The variable B is either a 3, 2, 1, or 0 depending on the sum on the right wall. See that A+B+C is bounded (between) 10 and 27. Recall our reminder. Thus we have two equations, where either A + A + B + 1 = 21 or A + A + B + 2 = 21. Substituting A = 6 we get B = 7 or B = 8. We now simply fill in our data into two scenarios and see which is correct. We simultaneously solve for C using the ones place in the bottom right of the equation.

Scenario 1:

666

668

+688

2012

Scenario 2:

666

667

+679

2012

Only scenario 2 is correct: we conclude that A=6, B=7, and C=9. The answer is 9.

In conclusion, we address the following question: did we encode or did we decode? You can actually say that we did either. We encoded into a numerical alphabet {0,….,10}. We decoded out of a three letter alphabet {A,B,C}.

Thanks for your attention, you guys.

In the expression below, each letter represents a one digit number. Where the same letter appears, it represents the same number in each case. Each distinct letter represents a

different number. In order to make the equation true, what number must replace C?

AAA

AAB

+ ABC

2012

A great start is to decode each AAA, AAB, and ABC. It helps to look at this problem wholly; particularly we look at the leading sum on the left wall (of the same types). We glean that either: (1) A + A + A = 20, (2) A + A + A + 1 = 20 or (3) A + A + A + 2 = 20: its very important to remember that given three numbers each less than ten, the sum of them which is great, is at most 2 in the tens place. This means that each row can only donate a 1 or 2 to the next. We can conclude that our line is line (3). That, A + A + A + 2 = 20 is the correct choice of the three since A may only equal an integer and…that 20/3 and 19/3 are not integers.

We have shown that A = 6.

If A = 6, then A+A+B, as a sum is almost solved for. The variable B is either a 3, 2, 1, or 0 depending on the sum on the right wall. See that A+B+C is bounded (between) 10 and 27. Recall our reminder. Thus we have two equations, where either A + A + B + 1 = 21 or A + A + B + 2 = 21. Substituting A = 6 we get B = 7 or B = 8. We now simply fill in our data into two scenarios and see which is correct. We simultaneously solve for C using the ones place in the bottom right of the equation.

Scenario 1:

666

668

+688

2012

Scenario 2:

666

667

+679

2012

Only scenario 2 is correct: we conclude that A=6, B=7, and C=9. The answer is 9.

In conclusion, we address the following question: did we encode or did we decode? You can actually say that we did either. We encoded into a numerical alphabet {0,….,10}. We decoded out of a three letter alphabet {A,B,C}.

Thanks for your attention, you guys.