We shall also close round 366 then. The answer - and congratulations to those who found it - is one, four, seven.
The crux to solving it is to note that:
(i) If Charlie's "rectangle" (which we shall see in a moment must turn out to be a square) has x rows and y columns, then the bottom layer of Oliver's triangle has (x+y) rows.
(ii) Charlie's two triangles then have (x+y)^2 balls. So xy must divide (x+y)^2.
(iii) If xy divides (x+y)^2, then x=y, so Charlie has a square, and Oliver has 4 times as many balls as Charlie.
(iv) Hence there are 4 crates of red balls - and this we know without even considering the total number of balls there are, nor have we yet considered Gordon. Also the number of balls in each crate is x^2 = y^2, hence a square.
(v) Then look at the highest square that divides 147 million to find how many balls in a crate. Our first try - 49 million - is too big, so look at the next highest square, by dividing by 4. Et voila.
Round 368 to follow....
The crux to solving it is to note that:
(i) If Charlie's "rectangle" (which we shall see in a moment must turn out to be a square) has x rows and y columns, then the bottom layer of Oliver's triangle has (x+y) rows.
(ii) Charlie's two triangles then have (x+y)^2 balls. So xy must divide (x+y)^2.
(iii) If xy divides (x+y)^2, then x=y, so Charlie has a square, and Oliver has 4 times as many balls as Charlie.
(iv) Hence there are 4 crates of red balls - and this we know without even considering the total number of balls there are, nor have we yet considered Gordon. Also the number of balls in each crate is x^2 = y^2, hence a square.
(v) Then look at the highest square that divides 147 million to find how many balls in a crate. Our first try - 49 million - is too big, so look at the next highest square, by dividing by 4. Et voila.
Round 368 to follow....
Originally Posted by davis_greatest
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