**Puzzle 51**

Find a four-digit number that is divisible by 15 and whose digits are in ascending order from left to right.

Show Solution

**Puzzle 52**

A man spent half the amount left on his debit card and noticed that the remaining balance had exactly as many cents as there had been dollars before his purchase and exactly half as many dollars as there had been cents before his purchase. How much had been the amount on his debit card before his purchase?

Show Solution*d*= the number of dollars on the debit card before the purchase and let

*c*= the number of cents on the debit card before the purchase. The information given in the puzzle can be represented algebraically as . Multiplying both sides of the equation by 2 and then transposing to isolate the two variables on opposite sides of the equation results in 98

*d*= 99

*c*. One obvious solution to this equation is

*d*= 99 and

*c*= 98, which leads to the puzzle solution of $99.98, half of which is $49.99, thereby satisfying the given conditions. All values of the form

*d*= 99

*k*and

*c*= 98

*k*are also solutions to the equation, but, except for

*k*= 1, do not yield solutions to the puzzle. Why not?

**Puzzle 53**

Assuming that the given word-arithmetic puzzle follows the general rule of such puzzles that different letters must represent different digits and all occurrences of the same letter must represent the same digit, prove that the given puzzle has no solution. (As usual, the fact that your first million attempts did not produce a correct solution doesn’t prove anything other than that you’re extremely obsessive and have great stamina.)

. S P E N D

. – L E S S

. M O N E Y

Show SolutionSubtraction in the tens place either requires borrowing from the hundreds place or does not require borrowing from the hundreds place.

If subtraction in the tens place requires borrowing from the hundreds place, then the E in SPEND will be changed into one less than the E in LESS, so the N in MONEY will be 9, but if the N in SPEND is 9, the S in LESS will be less than 9 and subtraction in the tens place will *not* require borrowing from the hundreds place.

If subtraction in the tens place does not require borrowing from the hundreds place, then the E in SPEND will stay the same as the E in LESS, so the N in MONEY will be 0, but if the N in SPEND is 0, the S in LESS will be greater than 0 and subtraction in the tens place *will* require borrowing from the hundreds place.

Either assumption about borrowing implies its opposite. QED

**Puzzle 54**

If four squirrels and three chipmunks form a queue to buy tickets to see *Dr. Doolittle*, what is the probability that both end positions will be occupied by chipmunks?

**Puzzle 55**

Arrange 24 toothpicks (or match sticks or chopsticks or worms with rigor mortis) to form the given diagram. Then remove eight of them, without repositioning the ones remaining, so as to leave a 16-toothpick diagram consisting of exactly the following number of squares with no extraneous toothpicks remaining. A) Two, B) Three, C) Six, D) Four, E) Five. ( A, B and C each has two distinct solutions, D has three distinct solutions and E has four distinct solutions. In case you forgot or never knew, distinct solutions cannot be transformed into each other by rotation and/or reflection.)

Show Solution*A, B, C, D.*The four points in the second row are

*E, F, G, H.*The points in the third row are

*I,*

*J, K, L.*The bottom row of points are

*M, N, O, P.*The eight toothpicks that should be removed to leave the given number of squares are:

A) *BF, CG, GH, KL, KO, JN, IJ, EF.* or *EF, FG, GH, BF, FJ, JN, KL, KO.*

B) *CG, GH, GK, FG, IJ, FJ, JK, JN, *or *EF, FG, GH, JK, BF, FJ, CG, GK*.

C) *AB, BC, CD, AE, BF, CG, DH, JK* or *AB, BC, CD, AE, BF, CG, DH,* *KL.*

D) *BC, FG, JK, NO, EI, FJ, GK, HL, *or *AB, BC, AE, BF, EF, EI, KL, KO,* or *AB, BC, AE, BF, EF, EI,** GK, JK*.

E) *AB, AE, CD, DH, IM, MN, LP, OP, *or *AB, BC, CD, AE, BF, CG, DH,* *FG,* or *BC, CD, FG, GH, CG, GK, DH, HL, *or *EI, IM, IJ, MN, CD, DH, GH, HL.*

I believe that any other solutions are equivalent to one I have listed, but if you have another distinct solution, Id be delighted to hear from you.