### Brain Teasers

# Math Reunion

Dear Jim:

Our twenty-fifth reunion went wonderfully last week. The Math Department put on a great spread. I'm sorry you couldn't make it. I ran into Ralph Jones there - remember him? It turns out he married Susie Jacobsen a couple of years after graduation - imagine that! They have three kids, all boys, and their ages make a nice puzzle. No two of them are the same age. Right now, Albert's age is the same as the sum of the digits of the ages of his two brothers. A year ago, Bill was in the same situation. Finally, six years from now it will be Charles' turn to have an age equal to the sum of the digits of his brothers' ages. I trust that's enough information to let you figure out their ages.

Give my regards to Sharon, and I hope I'll see you at our next reunion.

Jerry

Our twenty-fifth reunion went wonderfully last week. The Math Department put on a great spread. I'm sorry you couldn't make it. I ran into Ralph Jones there - remember him? It turns out he married Susie Jacobsen a couple of years after graduation - imagine that! They have three kids, all boys, and their ages make a nice puzzle. No two of them are the same age. Right now, Albert's age is the same as the sum of the digits of the ages of his two brothers. A year ago, Bill was in the same situation. Finally, six years from now it will be Charles' turn to have an age equal to the sum of the digits of his brothers' ages. I trust that's enough information to let you figure out their ages.

Give my regards to Sharon, and I hope I'll see you at our next reunion.

Jerry

### Hint

Any non-negative integer is congruent to the sum of its digits, mod 9. (They have the same remainder when divided by 9.)### Answer

Albert is 11, Bill is 6, and Charles is 5, 11 = 6+5. A year ago, they were 10, 5, and 4, 1+0+4 = 5. In six years, they will be 17, 12, and 11, 1+7+1+2=11.You can solve this one by trial and error, but you can also solve simultaneous congruences mod 9. Let a, b, and c represent the present ages of the three boys. Since any non-negative integer is congruent to the sum of its digits, mod 9, we have three congruences:

a = b + c (mod 9)

b-1 = (a-1) + (c-1) (mod 9)

c+6 = (a+6) + (b+6) (mod 9)

Substituting the value of a from the first congruence into the other two and simplifying, we get:

2c = 1 = 10 (mod 9), so c = 5 (mod 9), and

2b = -6 = 12 (mod 9), so b = 6 (mod 9),

so a = b + c = 2 (mod 9).

Age 2 is too young for Albert to be equal to the sum of digits of his brother's ages, and 20 is too old (it would require that both brothers be 19, contrary to the required congruences), so Albert must be 11, and the rest follows quickly.

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## Comments

I had fun cooking this one up, coming up with

a set of ages and year intervals that could make it work.

I hope you have as much fun solving it as I did

creating it!

a set of ages and year intervals that could make it work.

I hope you have as much fun solving it as I did

creating it!

Great one dewtell... good tough teaser, solvable with a method... or madness (I prefer the latter of course ;P).

How much do you want to bet that sometime madade will come around here and be like, there's no question.... well i don't think a teaser needs a question. good one...

I don't follow the math when you plug in the value for A into the third equation. It's the negative six number that throws me.

When you plug in the value of a into the third equation, you get: (c+6) = ((b+c)+6) + (b+6) (mod 9).

Subtracting c+6 from both sides and simplifying, you get: 0 = 2b+6 (mod 9).

Subtracting 6 from both sides, you get: 2b = -6 (mod 9).

Now since this is modular arithmetic, you can add or subtract any multiple of the modulus (9, in this case) from just one side of the equation. (Technically, this is a congruence that is supposed to be indicated by using a congruence sign with 3 parallel lines instead of equality, but the limitations of ASCII make it easier to just use an = sign instead when typing on a message board like this. When we say that x=y (mod m), what we mean is that the difference x-y is evenly divisibly by m. That's why you can add or subtract multiples of m on just one side of the congruence and keep it a true statement.) Also, in modular arithmetic, we can divide both sides of the equation/congruence by the same integer provided that integer is relatively prime to the modulus (for congruences mod 9, we can divide both sides by 2, but not 3, and still guarantee that the result is a valid congruence mod 9).

So we would like the right hand side to be a multiple of 2 so we can divide both sides by 2 to get b. We achieve that by adding 18 (a multiple of 9) to the right hand side,

yielding: 2b = 12 (mod 9). At this point, we can divide both sides by 2 to get b = 6 (mod 9).

What this means, in normal terms, is that b is 6 more than some multiple of 9. It could be 6, 15, 24, etc. But we've ruled out 8/9ths of the possible values for b (and likewise a and c), and we figure out the exact value by reasoning about a at the end. All the numbers from 0 to 25 except 19 have a digit sum between 0 and 9, so once we figure out that Albert is 11 now, then last year Bill's true age must have been 1 + 0 + a single digit sum of Charles' age, so the only age of the above list that works is 6.

Does this explanation help?

Subtracting c+6 from both sides and simplifying, you get: 0 = 2b+6 (mod 9).

Subtracting 6 from both sides, you get: 2b = -6 (mod 9).

Now since this is modular arithmetic, you can add or subtract any multiple of the modulus (9, in this case) from just one side of the equation. (Technically, this is a congruence that is supposed to be indicated by using a congruence sign with 3 parallel lines instead of equality, but the limitations of ASCII make it easier to just use an = sign instead when typing on a message board like this. When we say that x=y (mod m), what we mean is that the difference x-y is evenly divisibly by m. That's why you can add or subtract multiples of m on just one side of the congruence and keep it a true statement.) Also, in modular arithmetic, we can divide both sides of the equation/congruence by the same integer provided that integer is relatively prime to the modulus (for congruences mod 9, we can divide both sides by 2, but not 3, and still guarantee that the result is a valid congruence mod 9).

So we would like the right hand side to be a multiple of 2 so we can divide both sides by 2 to get b. We achieve that by adding 18 (a multiple of 9) to the right hand side,

yielding: 2b = 12 (mod 9). At this point, we can divide both sides by 2 to get b = 6 (mod 9).

What this means, in normal terms, is that b is 6 more than some multiple of 9. It could be 6, 15, 24, etc. But we've ruled out 8/9ths of the possible values for b (and likewise a and c), and we figure out the exact value by reasoning about a at the end. All the numbers from 0 to 25 except 19 have a digit sum between 0 and 9, so once we figure out that Albert is 11 now, then last year Bill's true age must have been 1 + 0 + a single digit sum of Charles' age, so the only age of the above list that works is 6.

Does this explanation help?

thank you. great teaser. i havent look at the answers. hope my 11, 6, 14 goes.

Excellent teaser. I mad an assumption that A was double digits and tried 5, 6, 11 and it worked. Lucky trial and error.

The scip model I used to solve this teaser can be found at https://pastebin.com/GC3Yw2nt

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