With thanks to Nancy Wilkes for contributing lots of the solutions.
400 centimetres of blue glass.
If you want to put your trust in algebra: let the area of the circumscribing
circel be C. The area of a small circle is (R + 2Y) so C=4(R + 2Y).
The area is also 4(R + Y + B), so R + 2Y=R + Y + B, whence Y=B.
We need at least 43 fours to write the number 1000,000 as a
sum of numbers made up of the digit 4.
444444 + 444444 + 44444 + 44444 + 4444 + 4444 + 4444 + 4444 + 4444 + 4
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One way to get the solution, is to fould open a dice like shown in the picture. The top dice in the question showed 3 faces. Put those three faces into the foulded out version of the dice. Then it shouldn't be a problem to finish the complete dice by simple deduction. The answer is 1. |
What middle volume? There are 101588 books - an even number.
There will be only a middle volume if the number of books is odd.
To see how many books there are, imagine re-numbering each book by
subtracting 1 from Tomus's number and deviding by 7. The new numbers
will go:
1, 2, 3, 4, 5, 6 ....101588.
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the answer is: |
N/4, 4N, N-4, N+4. Adding these up we get 6N + N/4, which has to
equal 400. That is, |
| 459 |
|
| 495 + | |
| 954 |
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We can add an equation Z = A+B+C+D+E+F, then rewrite the others as (e.g.) A(Z-A) = 184, etc. This makes C(Z-C)=301, which has factors 7 and 43. Since each of the transformed equations is a quadratic in the named variable, we can solve it in terms of Z (ex.: C = (Z +/- sqrt( Z^2 - 1204 ) )/2 ) ... this is integer only when Z = 50, and we can solve the remaining quadratics A = 4 or 46 B = 5 or 45 C = 7 or 43 D = 9 or 41 E = 10 or 40 F = 15 or 35 In general, Z must be larger than twice the square root of the greatest product. And looking for a product with a simple prime factorization can make the problem almost trivial. |
For a number to be divisible by both 6 and 8 does not require that it be
divisible by 48, since 6 and 8 have 2 as a common factor. The least common
multiple of 6 and 8 is 3 * 2 * 2 * 2 = 6 * 4 or 3 * 8 = 24.
Likewise, the least common multiple (LCM) of 6 and 9 is not 54,
but 2 * 3 * 3 = 6 * 3 or 2 * 9 = 18.
Similarly, the LCM of 6, 8, and 10 is not 6 * 8 * 10. These have 2 as a
common factor, and their LCM is 2 * 2 * 2 * 3 * 5 = 90, not 480.
The LCM of 2, 3, 4, 5, 6, 7, 8, 9, and 10 is
2 (ensures divisibility by 2)
* 3 (ensures divisibility by 3)
* 2 (ensures divisibility by 4)
* 5 (ensures divisibility by 5)
[don't need 6 here, because 2 and 3 are already factors]
* 7 (ensures divisibility by 7)
* 2 (ensures divisibility by 8 -- we already know it's divisible by 4)
* 3 (ensures divisibility by 9 -- we already know it's divisible by 3)
[don't need 10 here, because 2 and 5 are already factors]
ANSWER
2 * 3 * 2 * 5 * 7 * 2 * 3
= 6 * 10 * 14 * 3
= 60 * 42
= 2520 horses
Alphametics or Cryptarithems
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