Sec 3 Math Olympiad Question

Sec3math

THE SOLUTION

OUT of the 10 days, some numbers are repeated, only “18” and “19” are listed once.

If Cheryl’s birthday happened on either of these two days, then Bernard (who was given the day but not the month) would have known immediately when it was.

As Albert is sure that Bernard can’t possibly know the birthday yet, then her birthday cannot fall on the months where the “18” or “19” occur. So it can’t happen in May or June, and must fall in July or August.

With this new information, Bernard can now pin down Cheryl’s birthday for two reasons:

Two of the remaining five days are dated the 14th – as this number is repeated, it cannot be Bernard’s choice, and they can both be ruled out.

With three remaining dates – July 16, Aug 15 and Aug 17 left, Bernard knows the correct date.

By this time, Albert (who was given only the month) also knows the birthday.

As there are two dates in August, Albert would be sure of the date only if Cheryl’s birthday was in July.

Therefore, Cheryl’s birthday is on July 16.

Source: Singapore and Asian Schools Math Olympiad

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