So for the three dice with the top faces adding up to $8$ then the bottom faces must add up to $13$ as $21 - 8 = 13$. Key 1: The top and bottom of a die always add up to $7$, so if you had two dice the top and bottom faces would always add up to $14$, three dice would add up to $21$ and so on. Here is Lilith, Bradley and Emily's work: The methods they have used to tackle the problems in the second and fifth rooms are particularly interesting. Lilith, Bradley, Emily and Jake from Great Torrington Junior clearly communicated their thinking too. I really like the way Hannah checks her solutions. Here is the Word document which Hannah sent us. Hannah from Leicester High School for Girls gave lots of detail about how she tackled each room's challenge. The fourth key number is $8 + 6 + 7 + 3 - 10=14$ There cannot be three even numbers because an even + an even + an even = an even. The $8$ must go with the $6$ and $1$ because the one has to go with the two highest even numbers. So this means that there either has to be the number $8$ and two numbers that add up to $7$ or vice versa. The distance between the circles therefore is $4$cm The distance between the centres is $9$cm - twice the radius so $9 - 5 = 4$. Therefore the radius of the circle is $2.5$cm. The radius of the circle is half the height of the rectangle The number nine is the key to the second room To do this we need to use trial and improvement To work out what the number on the last card is we need to find out what the number on every card is. They all equal $13$ so it doesn't matter which set of numbers is on the dice Opposite sides of a die add up to $7$ below are the bottom numbers taking into consideration this fact The numbers on the top of the dice add up to $8$ below are possible numbers Sam from Shrewsbury House used some slightly different reasoning. $19$ will leave a remainder of three when divided by four and a remainder of one when divided by three, so it must by the number of the key. $18$ will leave a remainder of two when divided by four, so it cannot be the correct number. $17$ will leave a remainder of one when divided by four, so it cannot be the correct number. $16$ will leave a remainder of zero when divided by four, so it cannot be the correct number. We need the remainder to be three when divided by four and a remainder of one when divided by three. It also is divided by $3$ to leave a remainder of nothing. $15$ cannot be the number of tarts as $15$ is divided by $4$ to leave a remainder of $3$. These two sentences suggest that the key number is between $14$ and $28$ (not including $14$ and $28$). Because there are two circles, you must double the distance and subtract that from $9$ (the key to the second room). The centre point of both circles is $2.5$cm away from the edge of the diagram. Fourth card and fifth card add up to $16$. Third card and fourth card add up to $11$. Because the sum of the numbers on the top of the dice is eight, you must subtract $8$ from $21$ to reveal the sum of the numbers that you can't see.įirst card and second card add up to $13$ (key to the first room). Since all opposite sides add up to seven, you must multiply seven by the number of dice there were (three). For example, if someone puts a dice on the floor and asks what number is on the other side, you can immediately tell them that the number is six if the number on the top of the die is one. To solve this room, you must know that opposite sides of a dice always add up to seven. Many of you sent good explanations for each part.ĭaniel from Kings School, New Zealand sent a very clear solution: We were overwhelmed with solutions for this problem - thank you everyone.
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