The one guessing a person’s age has me stumped. The algorithm they give works.
Age 32
3 (1st digit) x 5 = 15 + 3 = 18 x 2 = 36 + 2 (2nd number in age) = 38
The person says 38 and you tell them they are 32. By subtracting 6. applause
Age 57
5 (1st digit) x 5 = 25 + 3 = 28 x 2 = 56 + 7 (2nd digit) = 63
Minus 6 they are 57. applause
It works with various ages I tried, but how?
This one is even more confusing. I understand you are working backwards from the answer. Page 108 and 9th entry. So you need a arithmetic operation that yields 1089
The random number at the beginning has me puzzled. How can this algorithm work when it must start with a random number picked by the person?
I’m tempted to write a quick program to see if it really works with all 3 digit number (with no duplicate digits) combinations. But I suspect it does. Somehow it yields 1089 every time.
741 reverse it 147
Subtract 741-147= 594 Reverse it 495
594 + 495= 1089. It works
As for the second example, suppose we have the three digit number ABC with smaller reverse CBA. Since this is smaller, it must be that C < A. Subtracting the smaller from the larger gives us (A - C)0(C - A). Except C is less than A, so there must be a borrow for the last digit’s subtraction: we actually get (A - C - 1)9(10 + C - A). Finally, adding this to its reverse gives 9(9 + 9)9 = 1089.
Indeed. Read literally, the math in your OP contains a number of statements, one for each equal sign:
3 x 5 = 15 + 3
15 + 3 = 18 x 2
18 x 2 = 36 + 2
36 + 2 = 38
Of course, all of these but the last are incorrect.
What do you use instead of equals sign for “execute” in such a simple case as above? (Although I would have said “gives,” were I orally stepping through the symbols).
I say “simple case” because no doubt simple opertions/demonstrations in other formalist endeavors are governed by unique symbologies.
Here’s an easier version of the trick that’s too obvious and lots of people would see right through it:
Take the first digit of your age, multiply it by ten, then add one, then add the second digit of your age and tell me the result. The result is 25? Aha! I know that your age is really 24.
To make it a little more sneaky, they added six instead of one, and they broke up the multiplying by ten into two steps, first asking you to multiply by five and then later multiply by two. And the icing on the cake is that instead of asking you to add six at the end, they had you add three just before you multiply by two, which has the same effect.
If AB is your age, i.e. A is the first digit of your age and B is the second digit…
“Multiply A by ten then add B” gives you the exact same answer as
“Multiply A by ten then add B then add six then subtract six” which is the same as
“Multiply A by ten then add six then add B then subtract six” which is the same as
“Multiply A by five then multiply again by two then add six then add B then subtract six” which is the same as
“Multiply A by five then add three then multiply by two then add B then subtract six”.
Understood; and in this context, I’m inclined to forgive you. In an educational context, however, you should bend over backward to avoid the “running equals sign,” since thinking of = as “gives” or “execute” can be a serious stumbling block to kids’ understanding of math. See “Students’ Understanding Of The Equal Sign Not Equal” (or more discussion here, here, or here.
One symbol commonly used is “=>” which is pronounced “yields” or “gives.”
Actually it proves how dirt-simple these puzzles are. Now that you know the tricks you can make up your own private version and baffle your friends.
As has been said, it all starts from the fact that all 2 digit numbers are actually in the form of 10a+b. So as long as you hide a multiply by 10 and a single addition in amongst a bunch of do-nothing steps you’ve got a working puzzle. e.g. take a digit, add 3 then double the result, then subtract 6, then divide by 2. Magically the same digit pops out. etc.
