I solved this problem with first declaring sum and mul as integers. Then returning their values using modulus.
the question confused me, when did i have to multiply at any point???
@Ryan, the product of a and b means multiply a by b. Would this be clearer:
Multiply a by b and then get the remainder from dividing that product by the sum of a and b ?
Yeah, thank you. The product made me think addition but i guess that would be the sum. Maybe i should have taken more math classes lol. So far this has been the best teaching program i have used
online. Some of the ?'s are somewhat difficult but thats what learning is about. Thank you for the email and reply.
Excellent tutorial thus far. It is probably hard to understand the precise syntax that is needed to pose the problems.
i was so confused by the wording of product. to me it processed as the same thing as sum :/
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I read the challenge about 5 times and operated exactly what they wrote: return (a*b)%(a+b);
I hope this helps. Wording can be tricky but it's all going back to basic math terminology.
The remainder of 5 % 3 could have been -1 too.
This one tripped me too. It can be done this way as well:
int product = a * b;
int sum = a + b;
return product % sum;
PEMDAS is also known as BEDMAS (brackets instead of parentheses)
I am a math teacher so this wasnt to hard for me :-)
I have a question... I understand that you can just return the value of "(a * b) % (a + b);" ... But I also tried doing this one by assigning to value to a new variable, int C. I added an int
C in public static int, and set the equation equal to C, returned C, but the code didn't work. How would I store its value to a new variable?
i got confused, i got it when i read the question again.
Good way of learning..
that's really awsome man
My solutions was like ArtTric
I don't know where to input int product = a*b;
Int sum = a+b;
I got it correct after 3 tries. I may be wrong but I surmise that '%'is to be put between code that identifies what variables you are working with such as 'sum' and really means: "divide and also give me the remainder".
something is wrong here because how is 5*1 divided by (5+1) have a remainder of 5?
basic math says its .8 with a remainder of 2. so 5*1 is 5. well 5 divided by 6 equals 0.8 r2
I get the code, but it has a flaw because 5*1 divided by 6 does not have a remainder of 5. its 2
because basic math says that 5/6 is 0.8 with a remainder of 2.
guys i did this
int mul = a * b;
int sum = a+b;
int result = mul%sum;
What is going on in this solution:
Applying the rules for Math and Operator Order of Precedence
1. addition is performed first because of ( )
2. multiplication is the next operation
return ( a * b % ( a + b ));