Easy mathematics! 15 points!

1. Any number with trailing digits 001 can be expressed as 1000N + 1.
1000N + 1 = 7*142N + 6N + 1
By inspection, when N = 1, 1000N + 1 is divisible by 7. Other values for N is 8, 15, 22, …, 7k +1, …


2. Let’s label the 6 persons arbitrarily as A, B, C, D, E, and F.
Assume that we cannot find 3 persons who know each other and we cannot find 3 persons who do not know each other… (X)
Any person, for example A, either knows 3 or more of the other 5 persons or do not know 3 or more of the other 5 persons.
When A knows at least 3 persons B, C and D, then B, C and D do not know each other because of assumption (X). [A knows B and A knows C => B and C do not know each other, etc] This is contradiction to (X).
When at least A does not know 3 persons B, C and D, then B, C and D do know each other because of the same assumption.
Since there are contradictions in both cases, assumption (X) is wrong and we must be able to find 3 persons who know each other or 3 persons who do not know each other.


3. Let the number of hours of study per day be S1, S2, … S30 where Si>1
Let the cumulative study hours from day 1 be Ck = ∑1 to k Sk, and C30 = 45
All these 30 Ck’s are different as all Si<>0, so we have 30 distinct Ck’s.
Assume that we cannot find any successive days that add up to 14 hours, that is,
Cj – Ci <>14 for all i,j
Cj <> 14 + Ci for all i,j … (A)
Since all Cj’s are distinct, all 14 + Ci are distinct, and the max is 14 + C30 = 59.
30 distinct Cj’s and 30 distinct 14 + Ci’s having a maximum of 59 => there must be some Cj equals to 14 + Ci which contradicts (A)
Therefore the assumption (A) is wrong and we can find some successive days that add to up 14 study hours.