This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.
Let the Passenger Train Through!
The goods train reverses into the siding, leaving three trucks, then reverses out of the way.
The passenger train attaches the three trucks to its front then reverses out of the way.
The goods train backs into the siding.
The passenger train drives past the siding.
The goods train drives forwards out of the way.
The passenger train reverses, then drives forward into the siding, leaving the three trucks. It then drives forward out of the way.
The goods train reverses into the siding, picking up the trucks. The trains have now passed each other.
Extension
1. If the goods train can drive forwards into the siding, how could it let the passenger train through?
2. If the goods train had 8 trucks, how could it let the passenger train through?
Cooked Turkey
Let the first digit be \(a\) and the final digit be \(b\). The cost of the 72 turkeys (in pence) is \(10000a+6790+b\). This must be divisible by 72.
$$10000a+6790+b=(138\times 72+64)a+(94\times 72+22)+b$$
$$=(138a+94)\times 72+64a+22+b$$
\(10000a+6790+b\) is divisible by 72, so \(64a+22+b\) is divisible by 72. This means that \(b\) must be even (as 64,22,72 are all even). Let \(b=2c\).
Dividing by two, we find that \(32a+11+c\) is divisible by 36. \(c\) must be odd, so that \(32a+11+c\) is even. Let \(c=2d+1\) and so \(b=4d+2\).
Dividing by two again, we find that \(16a+6+d\) is divisible by 18. \(d\) must be even, so that \(16a+6+d\) is even. Let \(d=2e\) and so \(b=8e+2\). But \(b\) is a single digit number, so \(e=0\), \(b=2\) and \(16a+6\) is divisible by 18.
Dividing by two yet again, we find that \(8a+3\) is divisible by 9. \(a\) must be divisible by 3. Let \(a=3f\).
Dividing by three, we find that \(8f+1\) is divisible by 3. \(8f+1=6f+2f+1\) so \(2f+1\) is divisible by 3. This will be true when \(f=1,4,7,10,...\). \(a\) must be a single digit, so \(f=1\) and \(a=3\). And so the price of the 72 turkeys is £367.92, and one turkey will cost £5.11.
Extension
Which numbers could 72 be replaced with in the original problem so that the problem still has a unique solution?
Mrs. Coldcream
Let \(G\) be the number of girls, and £\(x\) be the amount each girl was originally to receive. This means that there are \(19-G\) boys who were to receive £\(x+30\). The total scholarship is £1000, so:
$$xG+(x+30)(19-G)=1000$$
This simplifies to:
$$19x-30G=430$$
\(G\) is less than 19, so the only integer solution of this equation is \(x=40\) and \(G=11\).
After the funds are redistributed, each girl therefore receives £48. Let \(B\) be the amount each boy will now receive. The total is still £1000, so:
$$48\times 11+8B=1000$$
Which can be solved to find \(B=59\).
Each girl receives £48 and each boy receives £59.
Extension
Mrs. Coldcream still feels that this is unfair. How much more should be added to each girl's scholarship (resizing the boys' accordingly) so that the girls get at least as much as the boys?
Ten Digit Number
6210001000 has 6 zeros, 2 ones, 1 two, 0 threes, 0 fours, 0 fives, 1 six, 0 sevens, 0 eights and 0 nines.
Extension
Are there any more numbers like this? Prove that you have them all.
Here's another simple solution to the cooked turkey problem:
ReplyDeleteConsider the number \(X679Y\), a multiple of 72. Since 4 divides 72, we see that \(X679Y\) must be either 12 or \(16 \mod 20\), and hence \(Y\) is either 2 or 6. From this we deduce that \(X679Y \mod 40\) is either 12, 16, 32, or 36. Since 8 also divides our number, we may exclude all but 32. Therefore, \(Y = 2\).
Since our number is also a multiple of 9, we may use a well-known property of multiples of 9 to say \((X + 6 + 7 + 9 + Y) \mod 9 = 0\). Substituting 2 for \(Y\), we get \((X + 24) \mod 9 = 0\), implying \(X = 3\). Therefore, the number we seek is 36792, or $367.92, meaning each turkey is $5.11.
Sorry, had some trouble with LaTeX there.