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Order the digits 1, 2, 3, 4 to make a number which is divisible by 4.
When the final digit is removed it becomes a 3 digit number which is divisible by 3.
And when the final digit is removed again it becomes a 2 digit number which is divisible by 2.
Then finally a 1 digit number which is divisible by 1.
Help!
oops
4312
No, hang on I’ve removed the wrong numbers
1234
Nope
assuming solutions in whole numbers something's off here
.
Trick question - any combo will work, it doesn't say the answer can't be a decimal.
As above.
I can't find an answer where the division yields a whole number.
4444 works, but isn't in the rules as I read them.
A number is divisible by 4 if the last two digits are a number that is divisible by 4
Is what you've written EXACTLY how the puzzle is written?
Could it be asking for a number where the number formed by combining (sum of) all 4 must be divisible by 4, the number formed by the sum of the first 3 is divisible by 3, and so on?
EDIT No because 10/4 isn't whole. I've seen this before but with more numbers and that's how it worked.
As I see it, the following are the only sequences which could be considered potential matches, and none of them meet the criteria.
1 2 3 4
3 2 1 4
1 4 3 2
3 4 1 2
The question is as worded, but starts...
Can you....
No!
As allthepies said, can't be done to produce whole numbers
(IMO)
if the 4 digit number has to be divisible by 4 then the last digit has to be 2 or 4
since the 2 digit number has to be divisible by 2 then the second digit has to be 4 or 2
once you have that there are very few possibilities and none of them works
Yes, straight from the website linked to, but the homework sheet is missing the last paragraph.
Keep em comin....
Just say it can't and explain why (nicking allthepies succinct explanation to do so). Top marks will surely follow.
is this really about the divisibility by... rules:
anything is divisible by 1
even numbers by 2
if the sum of the digits divides by three then the number will too (1+2+3 = 6 ... 123/3= 41 , etc)
if the last two digits divide by 4 (per all the pies)
etc
so if the question is can you? No I can't and here's why...
Unfortunatly the solution page https://nrich.maths.org/7218/solution must require a subscribtion or log on or something as it just shows the same as the problem page for me.
EDIT: Try this https://nrich.maths.org/7218&part=note
FFS what age group is Stage 2?It is not possible to find a four-digit or a five-digit number that follow these rules but you can expect the children to be able to convince you this is the case.
Teacher copied from> https://nrich.maths.org/7218 ?
But didn't check to see if there is actually a valid solution ?
Woffle words on "possible approach" https://nrich.maths.org/7218/note
Ah, and the woffle leads trhrough to proof that it can't be done (as we'd already shown, go us!)
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