**Puzzle 4**

If *x* is 50% larger than *z* and *y* is 25% larger than *z*, then what percent larger than *y* is *x *?

And what percent smaller than *x* is *y* ?

And why are these two answers different ?

Show SolutionDon’t be fooled by the presence of

*x*,*y*and*z*into thinking that you have to use algebra to solve this puzzle. The easiest way to solve it is to choose easy numbers to replace*x*,*y*and*z*. One possibility is to let*z*= 100. That would mean*x*= 150 and*y*= 125.*x*is 25 more than*y*, so*x*is 25*/*125 or 20% larger than*y*. Also,*y*is 25 less than*x*, so*y*is 25*/*150 or 16 and 2*/*3 % smaller than*x*. The two answers are different, because to get the two answers, we are comparing the same number, 25, to two different numbers, 125 and 150. If you prefer to work with smaller numbers, let*z*= 4, so*x*= 6 and*y*= 5.

*Puzzle* 5

If a chicken and a half lay an egg and a half in a day and a half, how many eggs does a chicken lay in a day?

(No credit for pointing out that half a chicken can’t lay an egg. This is math, not biology.)

Show SolutionHow would you solve this if it were “If two chickens lay twelve eggs in three days, how many eggs does a chicken lay in a day?” Wouldn’t you divide twelve by two and the result by three (or divide twelve by three and the result by two) to get the rate for one chicken in one day? If we divide the number of eggs by the number of chickens and by the number of days, in either order, in the puzzle that was actually given, we get 1½ divided by 1½ = 1 and 1 divided by 1½ = 2

*/*3, so the answer is 2*/*3 of an egg. Both puzzle 4 and puzzle 5 can be solved algebraically, but why bother?

**Puzzle 6**

Write down all the integers from 1 through 60 to form the number 123456789101112131415….495051525354555657585960. Delete 100 digits from this number. What is the largest possible value of the number formed by the remaining digits *without* rearranging them ?

The first number that was formed has 111 digits (nine one-digit numbers and 60 – 9 = 51 two-digit numbers), so if we delete 100 digits from it, there are eleven digits left in the new number. To get the largest possible value for the new number, we want to use as many nines as we can, starting from the left side of the number, where the place-value is greatest. There are only six nines in the old number. We can’t use all of them as the first six digits of the new number, because there are only two digits in the old number to the right of the sixth nine, so we use the first

*five*nines of the old number as the first five digits of the new number and then form the largest six-digit number we can from what’s left of the old number to the right of the fifth nine, namely, from 5051525354555657585960. Since that number is 785960 (because seven is the greatest digit we can use as the sixth digit and still have at least five digits to choose from left in the old number), the solution to the puzzle is 99999785960.