This is a timed quiz. You will be given 400 seconds to answer all questions. Are you ready?

#### Readers who remember the Ribbon Problem No. 83 in The Canterbury Puzzles, may be glad to have this slightly easier variation of it: What number is it that can be multiplied by I, 2, 3, 4, 5, or 6 and no new figures appear in the result?

The number is 142,857. This is, of course, the recurring decimal fraction of one-seventh.

#### They were discussing mental problems at the Crackhams' breakfast table when George suddenly asked his sister Dora to multiply as quickly as possible IX2X3X4X5X6X7X8X9X10 How long would it have taken the reader?

Dora was not to be caught by George's question. She, of course, immediately gave the correct answ

#### what is the largest square number that contains all the ten digits (I to 9 and 0) once, and once only?

#### Take nine counters numbered I to 9, and place them in a row as shown. It is required in as few exchanges of pairs as possible to convert this into a square number. As an example in six pairs we give the following: 78 (exchanging 7 and 8), 8 4, 4 6, 6 9, 9 3, 3 2, which gives us the number 139,854,276, which is the square of 11,826. But it can be done in much fewer moves.

In four moves 73,34,48,25, we can get 157,326,849, which is the square of 12,543. But the correct solution is I 5, 84, 46, which gives us the number 523,814,769, the square of 22,887, which is in three moves only.

#### It will be found a very good puzzle to try to discover a number which, together with its square, shall contain all the nine digits once, and once only, the zero disallowed. Thus, if the square of 378 happened to be 152,694, it would be a perfect solution. But unfortunately the actual square is 142,884, which gives us those two repeated 4's and 8's, and omits the 6, 5, and 9. There are only two possible cases, and these may be discovered in about a quarter of an hour if you proceed in the right way.

The only two solutions are 567, with its square, 321,489; and 854, with its square, 729,316. We need only examine cases where the digits in the root number sum to 9, 18, or 27; or 8, 17, or 26, and it can never be a lower sum than 317 to form the necessary six figures.

#### Colonel Crackham placed eight numbered cards on the breakfast table, as here shown, and asked his young friends to rearrange them, moving as few as possible, so that the two columns should add up alike. Can it be done?

You need only make the 8 and 9 change places, first turning the 9 round so as to change it to a 6. Then each column will add up 18.

#### Can you find two numbers composed only of ones which give the same result by addition and multiplication? Of course 1 and 11 are very near, but they will not quite do, because added they make 12, and multiplied they make onlyll.

The two numbers composed of I's that sum and multiply alike are II and 1.1. In both cases the result is 12.1.

#### It will be found that 32,547,891 multiplied by 6 (thus using all the nine digits once, and once only) gives the product 195,287,346 (also containing all the nine digits once, and once only). Can you find another number to be multiplied by 6 under the same conditions? Remember that the nine digits must appear once, and once only, in the numbers multiplied and in the product

The number 94,857,312, multiplied by 6, gives the product 569,143,872, the nine digits being used once, and once only, in each case. [Victor Meally supplies two other solutions: 89,745,321 X 6 = 538,471,926, and 98,745,231 X 6 = 592,471,386.