Algebra Play

Think of a number. Double it. Add four. Divide by two. Subtract the number you first thought of. Whatever number was chosen, the answer always comes out as 2. Tracking this in algebra removes the mystery completely: starting number x, doubled is 2x, plus four is 2x+4, divided by two is x+2 (since both 2x and 4 divide by 2 separately), and subtracting the original x leaves (x+2)-x = 2. The x cancels out at the very last step, no matter what it was, which is exactly why the answer never changes. A trick like this is really just an equation-building machine, engineered so the unknown disappears at the end on purpose.

Ask a friend to think of a memorable date, multiply the month number by 5, add 6, multiply by 4, add 9, multiply by 5, then add the day number, and tell you only the final total. If the total is 291, the original date can be worked out exactly. Calling the month M and the day D, the steps build the expression 5M, then 5M+6, then 20M+24, then 20M+33, then 100M+165, then finally 100M+165+D. Subtracting 165 from 291 gives 126, and since a day number is at most 31 (needing only two digits), the last two digits, 26, must be the day, and whatever remains, 1, must be the month: 26th of January. The trick works for any date because the steps deliberately spread the month and day into two separate digit-groups of the final number, ready to be split apart again.

In a number pyramid, each number is the sum of the two numbers directly below it. Given a pyramid with top value 60, and bottom row 12, c, 8 (with c unknown in the middle), the missing values a and b directly above can be written as a = 12+c and b = c+8. Since a and b must add to the top value, 60, this gives (12+c)+(c+8) = 60, which simplifies to 20+2c = 60, so c = 20. Once c is known, a = 32 and b = 28, and the whole pyramid is filled in. This is genuinely two unknowns hiding behind one letter-number, c, because a and b were both written in terms of c before ever writing the final equation, the same trick used for two-unknown word problems.

In any 2 by 2 square picked from a calendar, calling the top-left date a, the square reads a, a+1 on top and a+7, a+8 below, since the next row down always adds 7. Adding all four numbers gives a + (a+1) + (a+7) + (a+8) = 4a+16. If a friend adds up a 2x2 square and reports the total as 36, the original four dates can be found by solving 4a+16 = 36: subtracting 16 from both sides gives 4a = 20, and dividing both sides by 4 gives a = 5, so the square must be 5, 6 on top and 12, 13 below. This is the very same 2x2 grid seen before with the DIAGONAL sums (both equal to 2a+8), but here a completely different question, the total of all four numbers, leads to a completely different, equally provable, expression, 4a+16.

Given three different digits, say 2, 3 and 5, and one multiplication slot (a 2-digit number times a 1-digit number), which arrangement gives the largest product? Trying every option shows 32 x 5 wins, and it turns out the pattern is general: put the LARGEST digit as the 1-digit multiplier, and arrange the other two in decreasing order as the 2-digit multiplicand. Calling the three digits p < q < r, comparing the two strongest candidates, qp x r and rp x q, by expanding each using place value, qp x r = (10 x q x r) + (p x r) and rp x q = (10 x r x q) + (p x q), shows the first terms are identical, so the comparison comes down to comparing p x r and p x q. Since r > q, qp x r must be the larger one, confirming the rule works for any three digits, not just 2, 3 and 5.

Pick any 2-digit number with different digits, reverse its digits, and subtract the smaller from the larger: the result always divides evenly by 9. Writing the number as ab (meaning 10a+b) and its reverse as ba (meaning 10b+a), and supposing b is bigger, the difference is (10b+a)-(10a+b) = 9b-9a = 9(b-a), which is obviously a multiple of 9 for any digits a and b. Adding the number and its reverse instead of subtracting gives a different guaranteed fact: (10a+b)+(10b+a) = 11a+11b = 11(a+b), always a multiple of 11. Both tricks work for exactly the same reason: writing numbers in expanded place-value form turns a mysterious pattern into a straightforward algebra fact.

In a story, a genie offers to double the coins in Karim's pocket every time he walks around a tree, but charges 8 coins after each round. After three rounds, Karim is left with exactly 8 coins, the same amount he owes. Working forwards from an unknown starting amount x is possible but clumsy; working BACKWARDS from the known ending is far cleaner. Before the last charge, Karim had 8+8=16 coins, which was double his amount before that round's doubling, so before round 3 he had 16/2=8 coins. That 8 was itself the result of the previous round's charge, so before charge 2 he had 8+8=16, meaning before round 2's doubling he had 16/2=8. Continuing this backward chain all the way to the very start recovers Karim's original number of coins, by patiently undoing each operation in reverse order, exactly the same skill used to solve any ordinary equation.