Question

Find all pears (a,b,c,d) such that
2a + 3b + 5c + 7d = 100
for a,b,c,d is natural number

Answer 1

@mukushla

Answer 2

note that these numbers are restricted...

for example d can not be greater than 12

Answer 3

if the equation given 2 or 3 variables, ithink ican to find the solution but this 4 variable
hmmm, not eassy for me ...

Answer 4

Is there a way to do it other than by going through all the options? What area of maths is this? Diophantine equations or something? Or partition theory?

Answer 5

maybe related to both

Answer 6

what should the first i do?
let take d less than 12?

Answer 7

going through all of the options will be an ugly answer lets think of something else

Answer 8

You could write a computer program to go through all options. Less mentally satisfying though.

Answer 9

d=12 has no solutions. d=11 has at least one solution I've found. 2,3,5,7 are all the primes less than 11. I wonder if that has anything to do with it.

Answer 10

yea, i got one of solution is (1,2,2,11)
but i use trial and error method, but
i need algebraic methode...

Answer 11

sorry, (1,2,3,11)

Answer 12

You could subtract (2+3+5+7) from 100 = 83 so that then a, b, c, d can be any non-negative integer. Then you can find quite a lot of solutions. For example (41,2,1,1).

Answer 13

can u explaind to me how u got (41,2,1,1) ?

Answer 14

Well, a,b,c,d are natural numbers. So you at least have 2+3+5+7 in your sum adding to 100 (because a>=1, b>=1, etc). So you may as well subtract that from 100. 100 - (2+3+5+7) = 83. Now you have no restrictions on "how much" of 2, 3, 5, or 7 goes into the sum to make up the remaining 83, except that you need at least non-negative portions of each. So, I just said let there be 40 more 2's, to get to 80, and then 1 more 3 to get to 83. Then, you have (1+40, 1+1, 1, 1) as (a,b,c,d). (the 1's are from the initial allocation of 1 each of 2,3,5,7)

Answer 15

And then you just need to find all the ways of making 83 from non-negative integer amounts of 2,3,5,7, and there are quite a few ways. 39*2 + 0*3 + 1*5 + 0*7, etc... You can pretty much just adapt them from the last from this point, I think.

Answer 16

Not sure that's the best way though...

Answer 17

yeah i think better to do it in that way... !!

Answer 18

Ok thank u very much @scarydoor
i'll try get all solution according ur explaind