All the Digits

Well done to everybody who worked out the correct solution to this problem. We received a lot of very good solutions to this activity and unfortunately we don't have space to include all of them here.

Samson and Ollie from ACPS found the solution 5694 × 3 = 17082, and they described how to multiply those numbers together to check the solution in this video:

Thank you for sharing this with us - it was very interesting to watch how you checked the calculation step-by-step.

Lots of children started by working out that the 4-digit number must include the digits 4, 5, and 6. Hugh from Sha Tin Junior School 5.3 in Hong Kong sent in this solution:

Explain the solution

???? x 3=?????

The first thing I noticed was that 3 was already used and 0, 1, 2 cannot be on the first 4 digits as you multiply them by 3 it will still be too small to reach 5 digits, so they must be the consecutive numbers at the bottom. 

Then I realised that the third digit is a sum of 2 consecutive numbers so it ruled out the numbers 7 8 9 because if you add them together they will equal to a two- digit number.

So the only numbers left are 4 5 6 but 5+6 and 4+6 equals a 2 digit number so only 4 and 5 will work because 4+5 equals 9 so the number will be 5694.

Then all you have to do is multiply it and you should, to check it you can look if the first third and fifth digit are consecutive numbers which are not in order.

    5694

   x__ 3

   17082

Then you will get the number 17082 because 1,0,2 are consecutive numbers!

Imeth from Pristine Private School, Dubai in the UAE had some similar ideas:

For the four digits, the only possible 3 consecutive numbers which don't have three and can make a sum that is neither 3 nor one of the others, is 4,5,6. The Third digit of the 4-digit is 9. For the five digit, 5 does not work, as 5*3=15 which has a 5 in the last digit. But 4 and 6 do work, and give 2 and 8 respectively. But both carry a 1, and 1 + 3 * 9 = 28, with an 8. So the last two digits of the two numbers, for 4-digit and 5-digit respectively, are 9 4, and 8 2. For the First 2 digits of 4-digit, they can be 6 5 or 5 6. Using trial and error, only 5 6 works, and the numbers are: 5694 and 17082.

These are all good ideas, Hugh and Imeth!

Quite a few children used trial and error to narrow down the final order of the digits. Katherine and Charmaine from Citipointe Christian College in Australia sent in this explanation of their method:

We know that the first digit of the four digit number cannot be 1,2 or 3, because if 3 was multiplied by these numbers, it would not give a two digit number. Also, three is already used.

Next we know that the three consecutive numbers in the four digit number must be 4,5 and 6, because if we add the two smallest numbers, it will give

nine, allowing us to fufil the rule of having two of the consecutive numbers added up to become the third digit of the four digit number. The next possibility of consecutive numbers would be 5,6 and 7, however, the two smallest numbers would add up to 11, which is a two digit number, therefore it would not work as the third digit of the four digit number. So the three consecutive numbers are 4,5,6 in the four digit number.

This concludes that 9 is the third digit of the four digit number.

After that, we can eliminate 5 from the last digit of the four digit number as 5 multiplied by 3 is 15, and we will use the number 5 twice. Thus, 5 is placed in either the first or second digit of the four digit number.

By knowing that 4, 5 and 6 and 9 are for the four digit number, the remaining numbers are 0,1,2,7,8. To figure out the product of this equation, we will use the facts stated on the website. That is that the first, third, and fifth are three consecutive numbers, and the second and fourth digits are also consecutive numbers. This allows us to find out that the first, third and fifth digit is 0,1,2 (in any order) and the second and fourth digits are seven and eight (in any order).

Now we do trial and error with the information found out by us above.

Trial and Error:

  5496

x       3

_______

16488             WRONG!

  4596

x       3

_______

13788             WRONG!

  5694

x       3

_______

17082            CORRECT!

Jake from Illogan School, Cornwall in the UK sent in this description of how they worked out the answer:

1) I wrote a number line and crossed out 3

2) The three consecutive numbers in the 4-digit number cannot be 0, 1 and 2 if the third digit of this number is the sum of two of them, and each digit is only used once (0 + 1 = 1, 1 + 2 = 3)

3) For the third digit to be a single number under 10 and the sum of two consecutive numbers above three, it must be nine and the sum of 4 plus 5.

4) The three consecutive numbers in the 4-digit number must be 4, 5 and 6

5) The 5-digit number must contain the 5 numbers left over - 0, 1, 2, 7 and 8. Only 7 or 8 can go in position 2 and 4 of the 5 digit number

6) I used trial an error to find the ones digit of the 4-digit number. Only 4 5 or 6 could go there. If 5 went there, the ones digit of the answer would be 5 (5 x 3 = 15) - not possible as I can only use each digit once and only 0 1 or 2 can go in this spot. If 6 was the ones digit of the 4-digit number, the ones digit of the answer would be 8 (6 x 3 = 18), but I know 8 can't go there. So the ones digit of the 4 digit number must be 4, making the ones digit of the answer 2 (4 x 3 = 12). I know that the tens digit of the 4-digit number is 9. 9 x 3 = 27, plus the one carried over from the 12 = 28. So 8 is the fourth digit of the answer.

7) This makes 7 the second digit of the answer as it can only be 8 or 7

8) I carried on using trial and error to put the 5 and 6 in place in the 4 digit number so that the answer was correct and made up of only the numbers that I had left

9) I ended up with 5694 x 3 = 17082

Well done to all of you for narrowing down the options and for being really descriptive about the steps you took to get there. I wonder if there is a way to work out that the first number is 5694 without using trial and error?

Liora from Yavneh Primary in the UK sent in this list of the steps they took to solve the problem:

Steps

1. Units of 5 digit number can't be 0 or 5 because they don't appear in the 3 times table if you use each number once and can't go up to 10 x 3

2. The first digit of the 5 digit number can only be 1 or 2 because otherwise it would be too large to be the answer to 3 x a 4 digit number 

3. Therefore the consecutive numbers used at the bottom are 0, 1, 2 (the three is already used)

4. The three consecutive numbers used in the four digit number are 4, 5, 6. So that two of them sum to a single digit number (4 + 5 = 9). The 10s digit of the 4 digit number is therefore 9.

5. We know where the 9 is. 3 x 9 = 27. The 10s. digit of the 5 digit would be a 7 or 8 (if a 1 is carried over). However we now know that a 2 is carried over to the 100s column. Three options are available (we already know the number at the bottom has to be a 0, 1, 2).

4 x 3 = 12 (add the carried over 2) 14 - not a possible solution

5 x 3 = 15 (add the carried over 2) 17 - not a possible solution

6 x 3 = 18 (add the carried over 2) 20 - correct solution

Therefore the 100s digit of the 4 digit number is a 6

6. Units digit of the 4 digit number is either a 4 or 5 (we've used the 6). The unit at the bottom is either a 0, 1, 2. Therefore it has to be 4.

7. Place the remaining 5 and 7 so the multiplication is correct

  5 6 9 4

x       3

1 7 0 8 2

You've thought really hard about this, Liora. I like the way you carefully considered what would be 'carried over' at each point and how that would affect the digits that could go in the different places in the 5-digit number.

Thank you as well to the following children who also sent in explanations of how they solved this problem: Dhruv from The Glasgow Academy in the UK; AE from ACPS; Jack from Cramlington Village Primary School; Emily, Yoonha, Bak Ying, Ryan, Rithisak from the Canadian International School of Phnom Penh in Cambodia; Fares from Pristine Private School in Dubai, UAE; Cecilia, Grace, Cynthia, Aarna, Freddy, Gloria, Elisha, Jerry, Ethan, Mira, Romeo and David from Citipointe Christian College in Australia; Zac in the UK; Eloise and Ben from Holy Trinity Pewley Down School in Guildford, UK; Group 1 at the Hamilton Academy; Chloe C, Ruby, Youyou, Sean, Jayden, Kathy and Vincent from Sha Tin Junior School 5.3 in Hong Kong; Shaunak from Ganit Manthan, Vicharvatika, India; and Hannah from Australia.