1) SO + MANY = SUMS What is the value of the product O * Y? (letters represent significant decimal digit; different letters correspond to different digits). ----------------------------------------------------------------------------------------------- 2) There are 2004 sets, each one of 40 elements such that, in any way you take 2 sets, they have exactly one common element. Show that there is an element that is common to all the 2004 sets. ------------------------------------------------------------------------------------------------ Good luck.
Right but... half right. There's another possible value (and it's not 8 Tank1337 ) And it would be nice if you could explain us the method you used... SUGGESTION FOR #2 : Take a set and show that one of its elements is contained in at least 50 other sets. Then show that this element must be contained also in all the other sets.
im to tired to post right now but if someone starts a conversation with me I will tell them besides someone else might want to try and solve it
Mmm, no because the result must be related to the formula: SO + MANY = SUMS... And btw chickadee already gave an exact response which is 35. But there is another possible and correct result that is still missing. HINT: If you look at the thousands and the hundreds you'll notice that it must be M+1= S, A=9, U=0.
Ok, seems like none has solved it hence I'm gonna give you solutions. #1 Looking at the thousands and hundreds, it must be M + 1 = S, A = 9 and U = 0. If O + Y = S, then S + N = 10 + M, but this (along with M + 1 = S) implies N = 9 that is impossible (because A = 9). So it must be O + Y = 10 + S and S + N + 1 = 10 + M, hence N = 8. As the digits not yet occupied are 1, 2, 3, 4, 5, 6, 7, necessarily Y · O is less than (or equal to) 6 + 7 = 13. Then S = 1 or S=2 or S=3. If S = 1, we have M = 0 that is impossible. If S = 2, we have O + Y = 12 that is O*Y = 35. If S = 3, then O + Y = 13 that is O*Y = 42. #2 Fix a set and call it S. There exists at least one element of S that is contained in at least 51 sets (counting S itself) otherwise the total number of the sets couldn't be greater than 50*40=2000, which is absurd. Let's call this set S1, S2, ..., S50. Let "x" be this element. We want to show that x is contained in all the other sets. Suppose there exists a set, say A, which does not contain x. If we intersect A with one of the 50 sets above, we obtain another 50 sets (we call them B1, B2, ..., B50) that are non empty. Furthermore if we intersect two of this sets we obtain nothing, because A does not contain x. But this is absurd because it would imply that A has at least 51 element. Hence such A can not exists and x is contained in every set.
I don't know. I suppose it is when they teach you how sums work and they teach you the notion of a set. So probably at the elementary school.
Its mathematical reasoning. which I probably studied in class 10-12. Remember to implement your realtime common sense while solving these! Also, you gonna need this even after ur Ph.D, trust me!
