Posted 15th May

hello, m son has a maths question which im sure there must be a method to working out that isnt trial and error.

A child has some cards 23578 and needs to multiply the cards together to total 15,741. But i dont know how to work it out ..

A child has some cards 23578 and needs to multiply the cards together to total 15,741. But i dont know how to work it out ..

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An obvious point is that anything multiplied by an even number will be even, anything multiplied by 5 will end in 0 or 5.

So its a given that anything multiplying those numbers 2,3,5,7,8 together will end in 0. To get a number ending in 1 you need to have a multiple a number ending with 3, 7 or 9 (which is obvious really since its a multiple of 3).

Hence I think you need to clarify the rules.

Edited by:"mas99" 15th May"he uses all of the cards to make a three -digit number and a two-digit number number, He multiples the two numbers togehter and the answer is 15,741" thats all it says

but that seems a lot of trial and error and guess work ( my son is 11) I would have thought there would be a method for this or a maths equation that works in all instances.

yes - exactly. i added a pic, he got frustrated and screwed it up

also that pic is missing half of the cryptic clue!

Edited by:"mutley1" 15th MayNow using this new base number of 5247, again an even number or 5 cannot be used and as we have used 3 only 7 remains to use and 5247 divided by 7 does not produce a whole number so 3 MUST be used again and 5247 divided by 3 equals 1749. Again 1749 can only be divided by 3 to result in a number without a remainder and this is 583 which cannot be divided any further by the digits provided without leaving a remainder.

We have now divided the base number 3 times in sequence every time by 3 i.e, (((15741/3)/3)/3). So 3 times 3 times 3 is 27 and 15741/27 = 583. From here it can be seen how the digits of 23578 have been used to arrive at 15741. The rest should be easy!

Edited by:"tardytortoise" 15th MayThere is no general method to find factors beyond trial and error as far as I'm aware. Internet security is based on the fact that you can't factorise a very large number to primes without checking all possibilities.

You can certainly reduce the possibilities though.

As we know that one factor is two digits long and if we assume that each number has to be used once and once only that means there are 20 possibilities to check as we only have to find the two digit number - the three digit one will come automatically once we divide the target by the correct two digit one.

And as mentioned above we also know that a target number ending in 1 means that both numbers have to end in three or seven as multiplying by 5 or an even number can't produce that result, which reduces the possible numbers to six:

So the candidates are:

23

53

83

27

57

87

There's probably a way to reduce the size of the pool further but at this point you'd just try it.

In fact once you try a candidate you'll realise that most of the list is far too large. For example 15741 / 53 is 297 which is among the smallest of three digit numbers you can make with those digits. So you know it's not going to be eighty-something that would be way under 235/257.

Edited by:"EndlessWaves" 15th MayTo start with you have 5 numbers and 5 blank boxes giving you 120 (5x4x3x2x1) possible permutations, far too many to solve using trial an error, you need to eliminate the majority of the possibilities.

The total is an odd number not divisible by 5 therefore the last digit on the top and bottom line must be 3 and 7 (although you don’t know which is which)

The lowest possible value on the bottom line is 23, this means the first line cannot have an 8 as it’s first digit as the total would always be above 16,000

If the value on the bottom line starts with an 8 the total would always be over 16000 regardless of what’s in the top line so the first value here cannot be an 8 either

By process of elimination this gives us our first confirmed number the second digit of the top row must be an 8 as it doesn’t fit anywhere else and the first digit of both the first and second line must be either 2 or 7

This narrows down the initial list of 120 options to just 8 which you can then individually check

Edited by:"spoo" 15th Maylol, now the OP knows the answers to all the questions

two right hand digitsmust by 3 and 7 (don't know yet which is top and which bottom).That leaves 5 2 and 8 available as the

two leftmost digitsfor the two numbers.The two selected from this multiplied together must give an answer less than 16 given that the final answer is less than 16000. That means the two numbers must be a 2 and 5 (again don't know which is top and which is on bottom).

So the middle digit of top number must be the remaining value of 8 so think there are now only 4 possible calculations :

583 x 27

587 x 23

283 x 57

287 x 53

thanks, we had the answer, but i wanted to see how you work it out.

Edited by:"t3rm3y" 15th May

maths have got a lot harder to when i was at school!

The best in rapid technique on this page (though needed a minor change for typo).

Edited by:"splender" 15th MayThose unfamiliar with it might enjoy this little anecdote about a rather clever 8 year old :

nrich.maths.org/2478

15741 has to be two numbers multiplied together to end in 1. The only two numbers that multiply to end in 1 are 3 and 7.

Using the 5 numbers you only options for the two digit number are ones that end in 3 and 7. It also elimates either the 3 or 7 as the first number as they both have to be used. So the numbers left are 23,53,83, 27, 57, 87.

Try dividing all 6 of those into 15741 and the only ones that works are 53 and 27. They give the answers 297 and 583. It can't be 297*53 as there is no 9, so it has to be 583*27

Edited by:"m1keyp1key" 15th MayI'm also struggling with how they'd get the answer to challenge 3 .

What we need is a primary school teacher to describe the process and explain how a 7 to 11 year old would be expected to work this out.

I suspect many parents who are home schooling will be able to see the answers, but unable to explain how they got there.

Multiple of 5 means it ends in 5 or 0 & multiple of 4 means it ends in an even number so he must be _4 this year and _5 next year which means 24 or 34 - but 34 isn't divisible by 4.

You're missing my point. It's the 7 to 11 year old process I'm interested in. How I'd do it (by thinking multiplication tables and checking) is irrelevant.