Question

Suppose a,b, and c are positive integers such that a+b+c+ab+bc+ca+abc=1000. Find a+b+c.

Answer 1

You have posted the same question again. same hint applies - try to factor the expression on the left

Answer 2

i dont know how to do that.

Answer 3

and make an attempt to factor it - you will soon see a pattern evolve

Answer 4

*at least make an attempt

Answer 5

I can start you off...

Answer 6

a + b + c + ab + bc + ca + abc = 1000
a(1+b) + b + c + bc + ca + abc = 1000
a(1+b) + c(1+b) + b + ca + abc = 1000
a(1+b) + c(1+b) + ca(1+b) + b = 1000
a(1+b) + c(1+b) + ca(1+b) + b + 1 - 1 = 1000
a(1+b) + c(1+b) + ca(1+b) + (1+b) = 1001

can you continue from here?

Answer 7

let me try

Answer 8

no idea

Answer 9

can you there is a common factor of (1+b) in all the terms on the left of the equals sign?

Answer 10

yes

Answer 11

so first pull that common factor out - what do you end up with?

Answer 12

a(1+b) + c(1+b) + ca(1+b) + (1+b) = (1+b)( ... ) = 1001

Answer 13

(1+b)( a+c+ac+1) = 1001

Answer 14

then?

Answer 15

good - now notice the two terms that have a common factor of "c" - factor those next

Answer 16

ok

Answer 17

what do you end up with?

Answer 18

(1+a)(1+b)(1+c)=1001

Answer 19

perfect!

Answer 20

now notice the right hand side - split it into a product of prime numbers

Answer 21

how?

Answer 22

find the factors of the number 1001

Answer 23

hint - try dividing it by 11

Answer 24

91*11

Answer 25

now try factoring 91

Answer 26

7*13

Answer 27

perfect!

Answer 28

so you ended up with:

(1+a)(1+b)(1+c)=1001=7*11*13

Answer 29

since the product on the right involves primes numbers only, you can assert that each term on the left is equal to one of these primes

Answer 30

oh i see
thank you sir thank you very much

Answer 31

yw :)