536 Puzzles and Curious Problems

260 Answers

is less than 10. To form the three-figure divisor, one factor at least must be 5,

and therefore the last figure must be 5 or O. The subtraction from the single 0

near the bottom shows that it is a 5 and at once gives us the 5,000. The factor

2 being excluded from the divisor (or it could not end in a 5), the final figure

in the quotient must be 8 (2 X 2 X 2), and the divisor 625, making x a

fourth 5. The rest is quite easy.

Here is an answer:

146. SIMPLE MULTIPLICATION

4539281706

2

9078563412

If you divide the first number into pairs-45, 39, etc.-these can be arranged

in any order so long as the 06 is not at the beginning or the 45 at the end.

147. AN ABSOLUTE SKELETON

It can soon be discovered that the divisor must be 312, that 9 cannot be in
the quotient because nine times the divisor contains a repeated figure. We
therefore know that the quotient contains aJl the figures 1 to 8 once, and the
rest is comparatively easy. We shall find that there are four cases to try, and
that the only one that avoids repeated figures is the following:

312) 10,114,626,600(32,418,675.

148. ODDS AND EVENS

249)764,752,206(3,071,294.

249)767,242,206(3,081,294.

245)999,916,785 (4,081,293.

245)997,466,785 (4,071,293.

248)764,160,912(3,081,294.

248) 761,680,912(3,071,294.