If we take the three-digit number 999 and write it in words as NINE HUNDRED AND NINETY NINE, we can see that it’s four elements have 4, 7, 3, and 10 letters. If we multiply 4 x 7 x 3 x 10, we get 840, which is some way from the 999 that we started with. I invite you to find a three-digit number such that if you multiply together the numbers of it’s four elements, you get the number that you started with. What is that three-digit number?

I thought about this, and looked at it logically, trying to spot a pattern that might be applied, to reach the solution. I could, of course, fall back on the old trial and error method, working from 101-999, but this would be time consuming, and not what the creator of the enigma had in mind. This is what I found:

Firstly, the elements which are constant, are HUNDRED and AND, which gives us 7, and 3. Well, 7 x 3 = 21, so 21 is a constant multiple.

There are 49 numbers between 101-999 which are divisible by 21.

ONE, TWO, and SIX are three letters, which gives us 3 x 21 = 63

FOUR, FIVE, and NINE are four letters, which gives us 4 x 21 = 84

THREE, SEVEN, and EIGHT are five letters, which gives us 5 x 21 = 105

105 x the final element would most likely give an answer too high

84 x the final element would also likely give an answer too high

Therefore, only ONE, TWO, and SIX are left. However, 100 and 200 are likely to be too low, which then leaves 600.

Numbers between 601-699, which are divisible by 21, are 630 and 693.

630 works out as 3 x 7 x 3 x 6 = 378 – This does not match.

693 works out as 3 x 7 x 3 x 11 = 693 – This matches.

Therefore 693 is the answer :0)

For those who are less mathematically-minded, I also worked it out in a lengthier, more explanatory way.

You’ll find this here:

Firstly, we know that we have constants of the elements HUNDRED and AND. This is because whatever answer we reach, it has to contain these elements. There are 7 letters in the word HUNDRED, and 3 letters in the word AND, which means that 3 and 7 are constant multiples. 3 x 7 =21, so 21 is an easier way of dealing with the constant multiple.

The only other possible multiples in the first element are 3, 4, and 5. This is because ONE, TWO, and SIX = 3 letters. FOUR, FIVE, and NINE = 4 letters. THREE, SEVEN, and EIGHT = 5 letters.

So,

3 x 21 = 63

4 x 21 = 84

5 x 21 = 105

This means that now, our only possible multiples are 63, 84, and 105. This makes it much easier to handle the rest of the problem.

Now for the final element. Most of the final elements are 6 letters or more (ELEVEN upwards), which means that the answer is likely (not definitely) 600 upwards. This is because

63 x 6 = 342 as an absolute minimum

84 x 6 = 504 as an absolute minimum

105 x 6 = 630 as an absolute minimum

This effectively (almost) rules out any numbers beginning with ONE, TWO, THREE, or FOUR (because the answers would be too high to match their descriptions). But we can quickly and easily explore these, as there are only 3 possible multiples, per first element (3, 4, and 5 again, ONE, TWO, SIX = 3, FOUR, FIVE, NINE = 4, THREE, SEVEN, EIGHT = 5).

So,

63 x 3 = 189 (ONE HUNDRED AND EIGHTY NINE) = 3 x 21 x 10 = 630

63 x 4 = 252 (TWO HUNDRED AND FIFTY TWO) = 3 x 21 x 8 = 504

63 x 5 = 315 (THREE HUNDRED AND FIFTEEN) = 3 x 21 x 7 = 735

84 x 3 = 252 (TWO HUNDRED AND FIFTY TWO) = 3 x 21 x 8 = 504

84 x 4 = 336 (THREE HUNDRED AND THIRTY SIX) = 3 x 21 x 9 = 945

84 x 5 = 504 (FIVE HUNDRED AND FOUR) = 3 x 21 x 4 = 336

105 x 3 = 315 (THREE HUNDRED AND FIFTEEN) = 5 x 21 x 7 = 735

105 x 4 = 420 (FOUR HUNDRED AND TWENTY) = 4 x 21 x 6 = 504

105 x 5 = 525 (FIVE HUNDRED AND TWENTY FIVE) = 4 x 21 x 10 = 840

As you can see, most of the answers on the far right are much higher than their decriptions. This also eradicates FORTY, FIFTY, and SIXTY as possible elements predecessing the final element (as this would make the answer too low to match it’s higher description).

This leaves FIVE, SIX, SEVEN, EIGHT, and NINE HUNDRED, as possibilities. But the final element gives us a lot of possible multiples – anything from 3 to 11 (3 being ONE, TWO, SIX, 11 being for example THIRTY SEVEN). So, if we try this,

(FIVE and NINE both equal 4)

4 x 21 x 3 = 252 – No need to figure out as it begins with TWO

4 x 21 x 4 = 336 – No need to figure out as it begins with THREE

4 x 21 x 5 = 420 – No need to figure out as it begins with FOUR

4 x 21 x 6 = 504 (FIVE HUNDRED AND FOUR = 4 x 21 x 4 = 336)

4 x 21 x 7 = 588 (FIVE HUNDRED AND EIGHTY EIGHT = 4 x 21 x 11 = 924)

4 x 21 x 8 = 672 – No need to figure out as it begins with SIX

4 x 21 x 9 = 766 – No need to figure out as it begins with SEVEN

4 x 21 x 10 = 840 – No need to figure out as it begins with EIGHT

4 x 21 x 11 = 924 – (NINE HUNDRED AND TWENTY FOUR =4x21x10=840)

SIX equals 3

3 x 21 x 3 = 189 – No need to figure out as it begins with ONE

3 x 21 x 4 = 252 – No need to figure out as it begins with TWO

3 x 21 x 5 = 315 – No need to figure out as it begins with THREE

3 x 21 x 6 = 378 – No need to figure out as it begins with THREE

3 x 21 x 7 = 441 – No need to figure out as it begins with FOUR

3 x 21 x 8 = 504 – No need to figure out as it begins with FIVE

3 x 21 x 9 = 567 – No need to figure out as it begins with FIVE

3 x 21 x 10 = 630 (SIX HUNDRED AND THIRTY = 3 x 21 x 6 = 378)

3 x 21 x 11 = 693 (SIX HUNDRED AND NINETY THREE = 3 x 21 x 11 = 693)

SEVEN and EIGHT both equal 5

5 x 21 x 3 = 315 – No need to figure out as it begins with THREE

5 x 21 x 4 = 420 – No need to figure out as it begins with FOUR

5 x 21 x 5 = 525 – No need to figure out as it begins with FIVE

5 x 21 x 6 = 630 – No need to figure out as it begins with SIX

5 x 21 x 7 = 735 (SEVEN HUNDRED AND THIRTY FIVE = 5x 21 x 10 = 840)

5 x 21 x 8 = 840 (EIGHT HUNDRED AND FORTY = 5 x 21 x 5 = 525)

5 x 21 x 9 = 945 – No need to figure out as it begins with NINE

5 x 21 x 10 = 1050 – Not possible as it has to be between 101-999

5 x 21 x 11 = 1155 – Not possible as it has to be between 101-999

You will see from my workings that only one of the possibles actually matches up – 693, and when you write it out, it works:

SIX HUNDRED AND NINETY THREE = 3 x 7 x 3 x (6 + 5) = 693.

693 is the answer :0)