Question

August has four integers. When he adds them three at a time, the sums are 46,50,54 and 57. Compute the largest of the original four integers.

Answer 1

So what I tried so far was setting up

A+B+C=46
A+B+D=50
A+C+D=54
B+C+D=57

not sure how to solve this though

Answer 2

@Jhannybean @Nnesha @Loser66

Answer 3

@agent0smith or even @zepdrix ? A little help here :)

Answer 4

I'm in a math research class so every question we get is supposed to be challenging lol

Answer 5

well you can solve them as simultaneous equations using matrices

or just the hard slog of the elimination or substitution methods

Answer 6

i have never done matrices before

Answer 7

would substitution be a lot longer?

Answer 8

this should be linear algebra u have 4 equations to solve 4 unknowns, have u tried ?

Answer 9

what i'm not feeling okay about is this condition lol
"Compute the largest of the original four integers"

Answer 10

i would rather just do this with substitution even if it took a while but im still a little stuck.

Answer 11

well good substitution is also away to solve i'll give you some hints

Answer 12

Using symmetry, you can solve it as follows.

Answer 13

subtract (4) from (3)
a-b=-3
subtract (3) from (2)
b-c=-4
subtract (2) from (1)
c-d=46-50=-4

observe the three equations, a<b<c<d, so d is the largest of the 4.

Now we know that adding all four equations together gives by symmetry
3(a+b+c+d)=207
a+b+c+d=69
substitute c=d-4, b=d-4-4=d-8, a=d-4-4-3=d-11
we get
4d-23=69
4d=92
d=23 voila

Answer 14

got to go, talkk later!

Answer 15

finally solved it with subsitution and elimination, thanks anyways @mathmate

Answer 16

this is what i wrote on paper btw

4 integers=A,B,C,D
"When he adds them three at a time"
A+B+C=46
A+B+D=50
A+C+D=54
B+C+D=57
You can add them 3 at a time in 4 different ways, which is correctly shown in the system above.

Step Two: Now, solving for just 1 of the 4 variables would easily lead the answers of the other three. For example, we will eliminate the variable A.
1. We can take the first two equations, and subtract them, to eliminate A.
(We would get C-D=-4)
2. Next, we want to get another equation, without the variable A in it. We can use elimination for the 2nd and 3rd equation. We could get (B-C=-4)
3. Finally, we can use the last equation (also without the variable A). (B+C+D)=57
4. Now, we have the 3 equations:
B+C+D=57. C-D=-4, B-C=-4

Beginning the elimination process, we can add the first 2 equations together and get B+2C=53. Now, we need either variable B or C in terms of the other, such as B-C=-4. We can say that B=-4+C, which we will then plug back into -4+2C+2C=53. -4+4C=53. C=19. We have now solved one of the 4 variables.

Step Three: Now that we know C=19, here comes the easy part, to solve for the other 3 variables, we simply plug into equation that contains the variable C. How about that equation C-D=-4 we received from the first elimination? (19)-D=-4, D=23.

Now, we know that C=19 and D=23. We now have a even easier time solving for the last 2 variables.

B+C+D=57, B+19+23=57, B=15

A+B+C=46, A+15+19=46, A=12

Step Four: Now, we have solved for all 4 variables. A=12, B=15, C=19, D=23. The question asks for the largest integer which is D=23, which is our final answer.

Answer 17

actually u can add them in 4! different way but it won't be a problem as there is only one set solution