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#### If we multiply 64253 by 365 we get the product 23452345, where the first four figures are repeated. What is the largest number that we can multiply by 365 in order to produce a similar product of eight figures with the first four figures repeated in the same order? There is no objection to a repetition of figures-that is, the four that are repeated need not be all different, as in the case shown.

Multiply 273863 by 365 and the product is 99959995. Working the problem backwards, any number whatever consisting of eight figures with the first four repeated is divisible by 73 (and by 137) without remainder, and if it ends in 5 or 0 is divisible by 365 (or by 50005). Knowing this, the highest possible product can be written down at once.

#### If the nine digits are written at haphazard in any order, for example, 4 1 2 5 3 97 6 8, what are the chances that the number that happens to be produced will be divisible by 11 without remainder? The number I have written at random is not, I see, so divisible, but if I had happened to make the 1 and the 8 change places it would be.

To be divisible by 11, four of the alternate digits must sum to 17 and the remaining five to 28, or four to 28 and five to 17. Thus, in the example I gave (482539761),4,2,3, 7, 1 sum to 17, and 8, 5, 9, 6 to 28. Now four digits will sum to 17 in 9 different ways and five to 17 in 2 ways, making 11 together. In each of the 11 cases 4 may be permuted in 24 ways and 5 in 120 ways, or together in 2880 ways. So that 2880 X 11 = 31,680 ways. As the nine digits can be permuted in 362,880 ways, the chances are just 115 to 11 against a haphazard arrangement being divisible by 11

#### A number is formed of five successive digits (not necessarily in regular order) so that the number formed by the first two multiplied by the central digit will produce the number expressed by the last two. Thus, if it were 1 2 8 9 6, then 12 multiplied by 8 produces 96. But, unfortunately, I, 2, 6, 8, 9 are not successive numbers, so it will not do.

1 ,3, 4, 5, 2. The successive numbers are 1, 2, 3, 4, 5; 13 X 4 = 52. [Victor Meally supplies the following solution for a six-digit number: 947,658. 94 X 7 = 658.-

#### A grocer proposed to put up 20 Ibs. of China tea into 2-lb. packets, but his weights had been misplaced by somebody, and he could only find the 5-lb. and the 9-lb. weights. What is the quickest way for him to do the business? We will say at once that only nine weighings are really necessary

ANs With the 5 lb. and 9 lb. weights in different pans weigh 4 lb. (2) With the 4 lb. weigh second 4 lb. (3) Weigh third 4 lb. (4) Weigh fourth 4 lb., and the remainder will also be 4 lb. (5), (6), (7), (8), (9) Divide each portion of 4 lb. in turn equally on the two sides of the scales.

#### This is a curious subject for investigation-the search for square numbers the figures of which read backwards and forwards alike. Some of them are very easily found. For example, the squares of 1, 11, Ill, and 1111 are respectively 1, 121, 12321, and 1234321, all palindromes, and the rule applies for any number of l's provided the number does not contain more than nine. But there are other cases that we may call irregular, such as the square of 264 = 69696 and the square of 2285 = 5221225. Now, all the examples I have given contain an odd number of digits. Can the reader find a case where the square palindrome contains an even number of figures?

The square of 836 is 698896, which contains an even number of digits and reads backwards and forwards alike. There is no smaller square number containing an even number of figures that is a palindrome

#### Find two whole numbers with the smallest possible difference between them which, when multiplied together, will produce 1234567890.

The factors of 1234567890 are 2 X 3 X 3 X 5 X 3607 X 3803. If we multiply 3607 by 10 and 3803 by 9 we get two composite factors 36070 and 34227, which multiplied together produce 1234567890 and have the least possible difference between them.

#### What are the factors (the numbers that will divide it without any remainder) of this number-l 0 0 0 0 0 0 0 0 0 0 0 I? This is easily done if you happen to know something about numbers of this peculiar form. In fact, it is just as easy for me to give two factors if you insert, say, one hundred and one noughts, instead of eleven, between the two ones. There is a curious, easy, and beautiful rule for these cases. Can you find it

If the number of noughts enclosed by the two ones is 2 added to any multiple of 3, two factors can always be written down at once in this curious way. 1001 = 11 X 91; 1000001 = 101 X 9901; 1000000001 = 1001 X 999001; 1000000000001 = 10001 X 99990001. The last is our required answer, and 10001 = 73 X 137. The multiple of 3 in 11 is 3: therefore we insert 3 noughts in each factor and one more 9. If our number contained 10 1 noughts, as I suggested, then the multiple of 3 is 33 and the factors will contain 33 noughts and 34 nines in the form shown. If the number of noughts in the number be even you can get two factors in this way: 1001 = 11 X 91; 100001 = 11 X 9091; 10000001 = 11 X 909091, and so on.

#### An American correspondent asks me to find a number composed of any number of digits that may be correctly divided by 2 by simply transferring the last figure to the beginning. He has apparently come across our last puzzle with the conditions wrongly stated. If you are to transfer the first figure to the end it is solved by 3 I 5 7 8 9 4 7 3 6 84 2 I 0 5 2 6, and a solution may easily be found from this with any given figure at the beginning. But if the figure is to be moved from the end to the beginning, there is no possible solution for the divisor 2. But there is a solution for the divisor 3. Can you find it?

We may divide 8 5 7 I 4 2 by 3 by simply transferring the 2 from the end to the beginning. Or 4 2 8 5 7 1, by transferring the 1.

#### I want to know whether the number 49,129,308,213 is exactly divisible by 37, or, if not, what is the remainder when so divided. How may I do this quite easily without any process of actual division whatever? It can be done by inspection in a few seconds-if you know how.

#### What is the smallest square number that terminates with the greatest possible number of similar digits? Thus the greatest possible number might be five and the smallest square number with five similar digits at the end might be 24677777. But this is certainly not a square number. Of course, 0 is not to be regarded as a digit.

If a square number terminates in similar digits, those digits must be 4, as in the case of 144, the square of 12. But there cannot be more than three equal digits, and therefore the smallest answer is 1444, the square of 38.