Last week in the evening class I teach we were discussing systems of linear equations. This is a highly enjoyable topic that, despite its initial simplicity, contains a wealth of truly fascinating mathematics (though we don’t cover the fascinating stuff in my class). I was delighted, therefore, when a friend sent me the following riddle this week, and I thought I’d post it up for the health and enjoyment of all. The first person to post the solution shall be rewarded with . . . a well-deserved sense of satisfaction. (The person who sent me the riddle — and he knows who he is — is excluded from the fun.)

Here we go:

It’s their birthday, you see,

The same for all three,

It’s strange but it’s perfectly true.

There’s Bertie and Ben

Who differ by ten;

Eight years older than one there is Sue.

Double one brother,

Plus treble the other,

Plus Sue’s age makes seventy-two.

From what’s on this page

You can find the girl’s age.

It’s really quite easy to do.

*[J.A.H. Hunter, Fun with Figures. Dover Pub. 1965. p. 67]*

Only integer solutions shall be accepted.

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This entry was posted on October 23, 2007 at 7:12 am and is filed under News, Poetry.

October 23, 2007 at 10:46 am

Solution?

Bertie is 7 years and 4 months old

Ben is 17 years and 4 months old

Sue is 15 years and 4 months old

I can’t make it work with even years/numbers.

3y + 2(y+10) + (8 + y) = 72

3y + 2y + 20 + 8 + y = 72

3y + 2y + y = 72 – 28

6y = 44

y = 7.333

.333 of a year is about 4 months.

October 23, 2007 at 12:09 pm

That’s a nice attempt, but it’s not the solution I am looking for. I want a solution in which all three ages are an integer number of years, with no months dangling off the end.

Bonus points to the person who can determine how many such integer-only solutions exist. I assure you there is at least one.

October 23, 2007 at 3:47 pm

I got:

Bertie = 14

Ben = 4

Sue = 22

Can I switch around Bertie’s and Ben’s ages for another solution? 😉 I haven’t found any other integer solutions, bah!

October 23, 2007 at 4:32 pm

You got it! And it’s true: you can switch the ages of the brothers for another solution. I suppose those mathematics-related advanced degrees gave you an advantage.

October 23, 2007 at 8:40 pm

I guess I didn’t drink enough coffee this morning!! Is there an algabraic way of expressing the problem to get the right answers?

(This is the type of advanced math thinking you get from a mediaevalist!!)

October 23, 2007 at 9:24 pm

Let s be Sue’s age.

Let b1 be Bertie’s age.

Let b2 be Ben’s age.

We’re told that b1 and b2 differ by 10, but we’re not told which is larger, so there are two possibilities:

(i) b1 – b2 = 10

(ii) b2 – b1 = 10

Likewise we’re told that Sue is 8 years older than one of them, but we’re not told which one. There are two possibilities again:

(iii) s = b1 + 8

(iv) s = b2 + 8

Finally, we’re told that twice one brother’s age added to thrice the other’s plus Sue’s age yields 72, but once again we’re not told which brother is which, so once again there are two possibilities:

(v) 2 b1 + 3 b2 + s = 72

(vi) 2 b2 + 3 b1 + s = 72

One integer-only solution is obtained with (i), (iii), and (vi), and another with (ii), (iv), and (v). All other combinations produce fractional solutions.

Thanks for playing!

October 24, 2007 at 10:13 am

Oh was I right in switching the brothers around? And here I thought I was cheating. No wonder I couldn’t find another integer solution 😛

October 24, 2007 at 11:25 am

The correct answer has already been found but I’ll post another solution that uses one variable. I want to note also that the reason only integer solutions are accepted is because it is their birthday. Although… celebrating birthdays more than once a year is a great excuse for enjoying more chocolate cake with espresso.

We aren’t told which brother is older so let x be the age of the younger brother. Then x + 10 is the age of the older brother and Sue is either ( x + 8 ) or ( x + 10 ) + 8. Now we want to find an integer solution to one or more of the following:

Twice the younger, thrice the older, Sue is a middle child (if she’s a sibling):

2x + 3(x+10) + x+8 = 72

==> 6x + 38 = 72

==> 6x = 34

Twice the younger, thrice the older, Sue is the oldest:

2x + 3(x+10) + x+18 = 72

==> 6x + 48 = 72

==> 6x = 24

Thrice the younger, twice the older, Sue is in the middle:

3x + 2(x+10) + x+8 = 72

==> 6x + 28 = 72

==> 6x = 44

Thrice the younger, twice the older, Sue is the oldest:

3x + 2(x+10) + x+18 = 72

==> 6x + 38 = 72

==> 6x = 34

We see that the only integer solution is x = 4 so the younger brother is 4, the older is 14 and Sue is 22. Now we are left only with guessing which brother is younger. I’ll hazard that the riddle gives clues. Since Bertie is a nickname for Bert that is more fitting a 4 year old than a 14 year old and the OED defines the word ben as inner or interior my vote is that Ben is in the middle at 14 and Bertie is 4.

October 24, 2007 at 11:44 am

Very nice method, Reg. You’ve reduced the problem to four cases rather than eight, which is a considerable savings in effort.

Your mapping of names to ages is also very persuasive. You prove yourself a sensitive and discerning hermeneut. Bravo!