15 The double-square number problem |
1978
2307 take 4 variables a,b,c,d then your year is 1000a + 100b + 10c +d and 10a + b + 10c +d = 10b + c => 10a +d = 9(b-c) now putting 1978 on either sides , you will see tht such an year doesnt exist till 2000. put a = 20 and d such an number tht (10a + d )%9 = 0 d = 7 b=3 , c=0 2307 context |
3 Coins
I agree with ravi , the answer is 9/10. context |
Going to the Track
3 context |
Nice to meet you
10 context |
Going to the Track
divide into 5 groups equally race them => then make 5 groups of the containing three winning horses from each group. the all the first rankers out , race thm , the winner is the fastest. and the last and second last are slower then the 2nd and 3rd so they and their counterpart group is also out. then the horse which came 2nd , in its group the third one will be out because the best is 2nd , the 2nd is 3rd overall and so the 3rd one doesnt stand a chance. the group with 3rd ranker , its group is eradicated except him so you are left with five horses. race them , the winner is 2nd fastest and the 2nd one is 3rd fastest context |