Question

among 5 integers there are always 3 with sum divisible by 3 prove this statement​

Answer 1

Answer:

Step-by-step explanation:

Suppose the numbers are x1,x2,x3,x4, and x5. For each i, let ri be the unique number ri∈{0,1,2} such that ri≡xi(mod3). Then we need to show that every possible instance of the list r1,r2,r3,r4,r5 contains three numbers whose sum is congruent to 0 modulo 3

.

The first thing to notice is that all three numbers 0,1

, and 2 cannot occur in the list at the same time, since 0+1+2≡0(mod3)

So you must have a list of 5 numbers chosen from one of the sets {0,1},{0,2}

, or {1,2}. By the pigeon-hole principle one of those numbers must occur at least three times. Since 0+0+0≡1+1+1≡2+2+2≡0(mod3), then there will always exist a subset of three numbers whose sum is a multiple of three.

Answer 2

Answer:

five integers:

-1,-2,0,6,9.

sum = 6+9 = 15

sum = -2+(-1) = -3

now 15-3 = 12.

12 / 3 = 4.

hence proved it.

please mark brillint to it.

or atleast one thanks to it.