WORD WEDNESDAY - Three of a kind
Today’s Puzzle
What do these three words have in common?
- JOB
- POLISH
- HERB
Monday’s Puzzle - Swipe left if you hate Math Monday
Find a six-digit number with all unique digits such that, if you take the last digit and move it to the front (i.e. take the rightmost digit and move it all the way to the left), you get a new number that is exactly five times the original.
[SPOILER] Answer to Monday’s Puzzle
142,857.
Well there’s no way around it - the solution to this puzzle relies on some good old fashioned algebra. It is Math Monday, after all. If you’re averse to algebra, we encourage you to read on anyway! We tried to make our explanation as clear as possible.
Before we get into it, let’s talk about digits. As you probably remember from school, every digit in a number has a place - the rightmost digit is in the ones place, the next digit is in the tens place, the next is the hundreds place, and so on. This means that we can break down any number (say, 536) by taking each component digit, multiplying it by the right place, and adding them all together (so, 536 = 1005+ 103 + 16).
Okay, now back to the puzzle. We’ve been given a six digit number. We don’t know what the number is, but if we give every digit in the number a name, say abcdef, we can break it down the same way we did with 536: 100,000a + 10,000b + 1,000c + 100d + 10e + 1f = abcdef. Breaking down abcde isn’t really interesting, since that part doesn’t move around, so we can express things even more succinctly: abcde10 + f1 = abcdef.
We’re now told that, by moving the last digit (f) to the front, we get a new number that’s exactly 5 times the original. We can also break down this new number (fabcde) using our places trick: 100,000f + abcde1 = fabcde.
So the puzzle is telling us that 5abcdef = fabcde. Or, using our broken down representations: 5(10abcde + 1f) = 100,00f + 1abcde.
One more trick before we can truly enter algebra land. Since the chunk of digits abcde isn’t moving around at all, let’s take this five digit number and give it a name. We’ll call it g. Now we’re ready to go. We start with the equation above, substituting g in for number abcde:
5(10g + 1f) = 100,000f + 1g.
Now we multiply out the left side:
50g + 5f = 100,000f + 1g.
Subtract 1g from both sides:
49g + 5f = 100,000f.
Subtract 5f from both sides:
49g = 99,995f.
Divide both sides by 7:
7g = 14,285f
Now, we know that g is a five digit whole number (abcde), and we know that f is a one digit whole number. In order to figure out which values of g and f give us the answer, we just need to test all possible values of f (1-9) and figure out which one gives us a value for g that’s a whole number. It turns out the correct answer is f=7:
7g = 14,2857
Thus, g = 14,285.
Now that we have our answers for g and f, we can combine them to get our original number: 142,857. We can check that this is correct by multiplying by 5 and we should see the 7 move to the front:
142,8575 = 714,285