Yeah... so I was too busy/tired to post the answers to last week's Maths Madness. Never fear! I will do it today. I will also put pictures in for your enjoyment :)
1. Take your house number and double it. Add 5. Multiply by 50. Then add
your age, the number of days in the year, and subtract 615. The last
two digits are your age, and the first two are your house number. Can
you explain why?
If you turn this question into algebra, it is so much easier to understand. Say that your house number is x and your age is y. Then the question tells you to do this:
(2x + 5)*50 + y + 365 - 615. But this simplifies down to:
100x + 250 + y + 365 - 615 = 100x + y.
Hopefully it is easier to see now why you are left with a number that has your age as the first two digits and your house number as the last two. When you multiply a whole number by 100, the last two digits will be zero, so adding your age (provided you are under 100...) will not change the digits of your house number. Of course this only works if your house number is a 2 digit whole number. Sorry Sherlock, won't work for you.
2. If all the stars stand for the same number, can you complete: (*/*) - (*/6) = (*/12)?
* = 4.
Again, simple algebra will help out here. You know that */* = 1 no matter what * is, so you are left with:
1 - (*/6) = (*/12)
1 = (*/12) + (2*/12) = (3*/12) = (*/4)
So * = 1x4 = 4
3. If you start with the number 2 and stick an extra digit 1 at the
beginning you get 12, which is exactly six times the number you started
Can you find a number so that when an extra digit 1 is written at the beginning you get 3 times the number you started with?
Can you find one so that the number you get is 5 times the number you started with?
Can you find one so that the number you get is 9 times the number you started with?
Again, there is more than one solution to some or all of these. Comment away!
4. In the multiplication 6 x 2 = 3 all the digits are correct, but they
are in the wrong places. The equation should be 2 x 3 = 6 (or 3 x 2 =
6). In each of the next three multiplications all the digits are
correct, but some of them are in the wrong places. Can you put them
28 x 1 = 44
43 x 2 = 14
76 x 8 = 41
There are (apparently, I haven't really had time to check) a few different answers to some of these. Here are the ones that the back of the book says. Feel free to comment any others that you find!
(a) 21 x 4 = 84
(b) 14 x 3 = 42
(c) 17 x 4 = 68