Can anyone solve this maths problem ?

[Her Ivory Tower]

You've only got 4 spare numbers per sum. Be really f***ing hard without the 2.

I think it is supposed to be digits used once only in the 8 spaces, that's why there is precisely 8 digits available to you between 2 and 9?

 

[muggboots]

I think it is supposed to be digits used once only in the 8 spaces, that's why there is precisely 8 digits available to you between 2 and 9?

Exactly.

 

[Lewberry pie]

6+7x5 =

 

[The Bridge Lad]

Nobody got this yet? If i had a calculator handy i would get in in no time

 

[Le Chuck]

4876 = 92 x 53
5394 = 93 x 58 or 62 x 87

Either way you use the 2 or the 3 twice. They are the only 2 digit combos.

 

[Her Ivory Tower]

Exactly.

Eh? I mean I think you should only use each of the 8 available digits once across BOTH sums. So all you have left for the bottom sum is 4 6 7 8.

 

[Wilfy]

4876 = 92 x 53
5394 = 93 x 58 or 62 x 87

Either way you use the 2 or the 3 twice. They are the only 2 digit combos.

This.

 

[Her Ivory Tower]

4876 = 92 x 53
5394 = 93 x 58 or 62 x 87

Either way you use the 2 or the 3 twice. They are the only 2 digit combos.

But you've used the 9 twice too.

 

[hypersonic]

is this one of those questions teachers set to get the kids thinking?

the answer is probably that it's impossible, but only accepted with the proof that has been posted above

 

[Her Ivory Tower]

@Keyser Soze what did your child do for the answer?

 

[Keyser Soze]

@Keyser Soze what did your child do for the answer?

She did the same as Muggboots and got 92 x 53 and 62 x 87 using the 2 twice. She hasn't been able to find any other way, so maybe the question is misleading and it is actually 8 numbers per question ?

 

[Her Ivory Tower]

She did the same as Muggboots and got 92 x 53 and 62 x 87 using the 2 twice. She hasn't been able to find any other way, so maybe the question is misleading and it is actually 8 numbers per question ?

Yes maybe.....I don't know, it feels like the 4 should be used somewhere so that all 8 available digits are used within the answer otherwise what was the purpose of restricting it to those specific 8 digits and saying use them once only..... hmmm maybe I'm looking too hard for something which isn't there! Do let us know the correct answer from her teacher!

 

[muggboots]

 

[Tadger]

Is there any practical application of this mathematical exercise?

 

[Keyser Soze]

Yes maybe.....I don't know, it feels like the 4 should be used somewhere so that all 8 available digits are used within the answer otherwise what was the purpose of restricting it to those specific 8 digits and saying use them once only..... hmmm maybe I'm looking too hard for something which isn't there! Do let us know the correct answer from her teacher!

Will do - I'm intrigued myself

 

[Astra]

Maybe the question is written ambiguously because if you use 87 x 62 for the second one then each sum uses the numbers 2-8 once only?

 

[peachy]

Complete the following so that you create four 2 digit numbers that use the digits 2 to 9 (once and once only)
to get the following four digit answers when multiplied together :

** x ** = 4876
** x ** = 5394

Any ideas ?

53 x 92 = 4876 (all digits used 2-9)
62 x 87 = 5394 (all digits used 2 -9)

 

[Her Ivory Tower]

53 x 92 = 4876 (all digits used 2-9)
62 x 87 = 5394 (all digits used 2 -9)

Yes I think this is how it is supposed to be but the question from the OP was worded wrongly.

So the answer is.........


OP's FAULT.

 

[Keyser Soze]

Yes I think this is how it is supposed to be but the question from the OP was worded wrongly.

So the answer is.........


OP's FAULT.

Thanks a lot !

Yes it probably was - as I say my daughter was doing it for her homework and I decided to also have a go whilst watching us not win, yet again ! The maths problem was probably more exciting that the match