Offering 30 if a plus b plus c equals 68 and ab plus bc plus ca equals 1121, where a, b, and c are all prime numbers, find the value of abc. the answer is 1978 but i need an explanation on how to get that.

This looks a lot harder than it is, because with some clever observations we can quickly eliminate most solutions.

The first thing that springs to mind is a,b,c cannot all be odd, because then the LHS is even and the RHS is odd. Thus at least one of them must be even and therefore equal to 2. So let’s break this into two cases:

Case 1: c=2.

Hence we get

a(a+b)=122=2×61

Now a must be a positive prime factor of this number, and a+b is clearly more than a, so the only solution in this case is (a,b,c)=(2,59,2). Note here that we must check 59 is prime; fortunately it is.

Case 2: c≠2.

Now the RHS is odd, so a certainly can’t be even. But then b must be even, so b=2. Then we have

a(a+2)=120+c

Adding one to each side and factorising shows

(a+1)2=121+c

from which

c=(a+12)(a−10)

(if you are confused about this step think about difference of two squares factorisations).

But we know c is a prime number, so it can’t have any non-trivial factors. Thus one of the brackets must be 1, so a=11, which yields the solution (a,b,c)=(11,2,23). Once more we must check this solution is valid — again, it is, since 11 and 23 are prime. And those are the only two solutions since our cases cover every possibility.

Step-by-step explanation:

1978

Step-by-step explanation:

This looks a lot harder than it is, because with some clever observations we can quickly eliminate most solutions.

The first thing that springs to mind is a,b,c cannot all be odd, because then the LHS is even and the RHS is odd. Thus at least one of them must be even and therefore equal to 2. So let’s break this into two cases:

Case 1: c=2.

Hence we get

a(a+b)=122=2×61

Now a must be a positive prime factor of this number, and a+b is clearly more than a, so the only solution in this case is (a,b,c)=(2,59,2). Note here that we must check 59 is prime; fortunately it is.

Case 2: c≠2.

Now the RHS is odd, so a certainly can’t be even. But then b must be even, so b=2. Then we have

a(a+2)=120+c

Adding one to each side and factorising shows

(a+1)2=121+c

from which

c=(a+12)(a−10)

(if you are confused about this step think about difference of two squares factorisations).

But we know c is a prime number, so it can’t have any non-trivial factors. Thus one of the brackets must be 1, so a=11, which yields the solution (a,b,c)=(11,2,23). Once more we must check this solution is valid — again, it is, since 11 and 23 are prime. And those are the only two solutions since our cases cover every possibility.

all it means is function 3