The first is 8; to which add 2, the sum is 10 The second is 12; subtract 2, the remainder is 10 The third is 5; multiplied by 2, the product is 10 The fourth is 20; divided by 2, the quotient is 10 -- 45 Required to subtract 45 from 45, and leave 45 as a remainder. SOLUTION.--9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45. 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45. -------------------------------------- 8 + 6 + 4 + 1 + 9 + 7 + 5 + 3 + 2 = 45. The Astonished Farmer. A and B took each 30 pigs to market. A sold his at 3 for a dollar, B at 2 for a dollar, and together they received $25. A afterwards took 60 alone, which he sold as before, at 5 for $2, and received but $24: what became of the other dollar? This is rather a catch question, the insinuation that the first lot were sold at the rate of 5 for $2, being only true in part. They commence selling at that rate, but after making ten sales, A’s pigs are exhausted, and they have received $20; B still has 10, which he sells at “two for a dollar,” and of course receives $5; whereas had he sold them at the rate of 5 for $2, he would have received but $4. Hence the difficulty is easily settled. The Expunged Figure. In the first place we desire a person to write down secretly, in a line, any number of figures he may choose, and add them together as units; having done this, tell him to subtract that sum from the line of figures originally set down; then desire him to strike out any figure he pleases, and add the remaining figures in the line together as units (as in the first instance), and inform you of the result, when you will tell him the figure he has struck out. 76542 -24 24 ----- 76518 Suppose, for example, the figures put down are 76542; these added together, as units, make a total of 24; deduct twenty-four from the first line, and 76518 remain; if 5, the center figure, be struck out, the total will be 22. If 8, the first figure, be struck out, 19 will be the total. In order to ascertain which figure has been struck out, you make a mental sum one multiple of 9 higher than the total given. If 22 be given as the total, then 3 times 9 are 27, and 22 from 27 show that 5 was struck. If 19 be given, that sum deducted from 27 shows 8. Should the total be equal multiples of 9, as 18, 27, 36, then 9 has been expunged. With very little practice, any person may perform this with rapidity: it is therefore needless to give any further examples. The only way in which a person can fail in solving this riddle is when either a number 9 or a 0 is struck out, as it then becomes impossible to tell which of the two it is, the sum of the figures in the line being an even number of nines in both cases. Mysterious Addition.