The Coolest Math Puzzles Ever, Stimulating Your Mind

Arithmetical and Algebraical Problems
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Ok it your turn.

Can you solve these 91 year old, Math Puzzles from Amusements in Mathematics.

I give you a couple of easy ones, to get you started

Get work, let me know how you went.

A man recently bought two aeroplanes, but afterwards found that they would not answer the purpose for which he wanted them. So he sold them for $600 each, making a loss of 20 per cent, on one machine and a profit of 20 per cent, on the other. Did he make a profit on the whole transaction, or a loss? And how much?

A man had in his office three cupboards, each containing nine lockers, as shown in the diagram. He told his clerk to place a different one-figure number on each locker of cupboard A, and to do the same in the case of B, and of C. As we are here allowed to call nought a digit, and he was not prohibited from using nought as a number, he clearly had the option of omitting any one of ten digits from each cupboard.

Now, the employer did not say the lockers were to be numbered in any numerical order, and he was surprised to find, when the work was done, that the figures had apparently been mixed up indiscriminately. Calling upon his clerk for an explanation, the eccentric lad stated that the notion had occurred to him so to arrange the figures that in each case they formed a simple addition sum, the two upper rows of figures producing the sum in the lowest row. But the most surprising point was this: that he had so arranged them that the addition in A gave the smallest possible sum, that the addition in C gave the largest possible sum, and that all the nine digits in the three totals were different. The puzzle is to show how this could be done. No decimals are allowed and the nought may not appear in the hundreds place.

Seven men, whose names were Adams, Baker, Carter, Dobson, Edwards, Francis, and Gudgeon, were recently engaged in play. The name of the particular game is of no consequence. They had agreed that whenever a player won a game he should double the money of each of the other players-that is, he was to give the players just as much money as they had already in their pockets. They played seven games, and, strange to say, each won a game in turn, in the order in which their names are given. But a more curious coincidence is this-that when they had finished play each of the seven men had exactly the same amount-two shillings and eightpence-in his pocket. The puzzle is to find out how much money each man had with him before he sat down to play.

I wonder how many of my readers are acquainted with the puzzle of the "Dutchmen's Wives"-in which you have to determine the names of three men's wives, or, rather, which wife belongs to each husband. Some thirty years ago it was "going the rounds," as something quite new, but I recently discovered it in the Ladies' Diary for 1739-40, so it was clearly familiar to the fair sex over one hundred and seventy years ago. How many of our mothers, wives, sisters, daughters, and aunts could solve the puzzle to-day? A far greater proportion than then, let us hope.

Three Dutchmen, named Hendrick, Elas, and Cornelius, and their wives, Gurtrün, Katrün, and Anna, purchase hogs. Each buys as many as he (or she) gives shillings for one. Each husband pays altogether three guineas more than his wife. Hendrick buys twenty-three more hogs than Katrün, and Elas eleven more than Gurtrün. Now, what was the name of each man's wife?

I bet, you've just woke up some dormant brain cells.
The rule of the the brain is use it or lose it.

Did you find this fun or a challege?

The man must have paid $500 and $750 for the two machines, making together $1,250; but as he sold them for only $1,200, he lost $50 by the transaction.

The smallest possible total is 356 = 107 + 249, and the largest sum possible is 981 = 235 + 746, or 657 + 324. The middle sum may be either 720 =134+586, or 702 = 134 + 568, or 407 = 138 + 269. The total in this case must be made up of three of the figures 0, 2, 4, 7, but no sum other than the three given can possibly be obtained. We have therefore no choice in the case of the first locker, an alternative in the case of the third, and any one of three arrangements in the case of the middle locker. Here is one solution:-


107 134 235
249 586 746
356 720 981

Of course, in each case figures in the first two lines may be exchanged vertically without altering the total, and as a result there are just 3,072 different ways in which the figures might be actually placed on the locker doors. I must content myself with showing one little principle involved in this puzzle. The sum of the digits in the total is always governed by the digit omitted. 9/9 - 7/10 - 5/11 - 3/12 - 1/13 - 8/14 - 6/15 - 4/16 - 2/17 - 0/18. Whichever digit shown here in the upper line we omit, the sum of the digits in the total will be found beneath it. Thus in the case of locker A we omitted 8, and the figures in the total sum up to 14. If, therefore, we wanted to get 356, we may know at once to a certainty that it can only be obtained (if at all) by dropping the 8.

Puzzles of this class are generally solved in the old books by the tedious process of "working backwards." But a simple general solution is as follows: If there are n players, the amount held by every player at the end will be m(2n), the last winner must have held m(n+1) at the start, the next m(2n+1), the next m(4n+1), the next m(8n+1), and so on to the first player, who must have held m(2n-1n+1).

Thus, in this case, n = 7, and the amount held by every player at the end was 27 farthings. Therefore m = 1, and G started with 8 farthings, F with 15, E with 29, D with 57, C with 113, B with 225, and A with 449 farthings.

The money paid in every case was a square number of shillings, because they bought 1 at 1s., 2 at 2s., 3 at 3s., and so on. But every husband pays altogether 63s. more than his wife, so we have to find in how many ways 63 may be the difference between two square numbers.

These are the three only possible ways: the square of 8 less the square of 1, the square of 12 less the square of 9, and the square of 32 less the square of 31. Here 1, 9, and 31 represent the number of pigs bought and the number of shillings per pig paid by each woman, and 8, 12, and 32 the same in the case of their respective husbands.

From the further information given as to their purchases, we can now pair them off as follows: Cornelius and Gurtrün bought 8 and 1; Elas and Katrün bought 12 and 9; Hendrick and Anna bought 32 and 31. And these pairs represent correctly the three married couples.

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