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- Fun with Maths: Answers -

When you reverse the digits in a number, you can consider this to be the same as multiplying by 99 (100 - 1) for the difference between the end two digits. Algebraically, this can be expressed as:

• (100A + 10B + C) - (100C + 10B + A)

where A, B and C are the digits in your number.
Expanding and simplifying this gives us:

• 100A + 10B + C - 100C - 10B - A = 99A - 99C = 99(A - C)

It can be easily seen that the 10B term cancels itself out, hence the middle digit of the number you first thought of is irrelevant!
Your answer will be a three digit number, the first of which can legitimately be zero. Since A must be greater than C, and both must be between 0 and 9 inclusive, the ONLY possible results to the subtraction are:

• 099, 198, 297, 396, 495, 594, 693, 792 and 891
  (990 is not possible because A - C can never be 10).

Note that the middle digit is always 9, and that the first and last digits add up to 9. These are the first 9 terms of the 99 times table!
Therefore let's call our answer 100X + 90 + (9 - X).
The algebraic expression for the reverse of that number is given by

• 100(9 - X) + 90 + X

Adding these two values together gives us:

• (100X + 90 + (9 - X)) + (100(9 - X) + 90 + X)

which simplifies to (100X + 90 + 9 - X) + (900 - 100X + 90 + X).
The 100X and X terms cancel each other out, leaving us with:

• (90 + 9) + (900 + 90) = 1089

This one works by effectively taking the number of times you'd like to have chocolate a week, multiplying it by 100, adding in the current year, then subtracting the year of your birth. But it's done in a fairly obfuscated manner so you're not aware of what's going on until you work the maths out.

1. Let's call the number of times you'd like to have chocolate "x".
   
2. Multiplying by 2 gives you 2x.
   
3. Adding 5 gives you 2x + 5.
   
4. Multiplying by 50 gives us 50(2x + 5) which simplifies to 100x + 250.
   Now you can see where the 100 times the number you first thought of gets there!
   
5. Add 1750. Adding 1750 to 250 give you 2000. When this first came out it would only work in the year 2000. This is the reason why that was so.
   The 100x term remains unaltered by this step.
   
6. You need to subtract one if you have not yet had your brithday, because otherwise you may come out as being a year older than you are if you perform this calculation before your birthday in any given year.
   
7. Add 1 for every year since 2000 - necessary to make it work beyond 2000.
   After this step you now have the current year (less 1 if you have not yet had your birthday) + 100x.
   
8. Subtracting the year of your birth from the current year gives you your age. Adding this to 100x gives you the first digit as the number you first thought of, followed by your age.
   

Obvious when you know how, isn't it?

1. TWO-VARIABLE PROBLEM: SOLUTION
   Let the distance between his home and his mother's place be called A.
   Let the distance between his home and place of work be B.
   We know that A + B is 120: the distance between his mother's house and his place of work.
   The distance to work and back for a week is given by 2 * 5 * B.
   The distance to his mother's and back is given by 2 * A.
   We are told that these two distances are the same, so 2 * 5 * B = 2 * A.
   This simplifies to 5B = A.
   We now have 2 options to get the same answer, namely the Elimination method or the Substitution method.
   
   Elimination method:
   Re-arrange 5B = A to be 5B - A = 0.
   Add (5B - A = 0) to (A + B = 120) and we get (5B - A) + (A + B) = 120 - 0.
   Notice that A is eliminated in the subtraction so we get 6B = 120.
   Hence B (the distance to work each day) is 20 miles.
   
   Substitution Method:
   We have A on its own in the equation 5B = A.
   Substitute into A + B = 120, and we get (5B) + B = 120, or 6B = 120.
   Hence B (the distance to work each day) is 20 miles.
   
   
2. SIX-VARIABLE PROBLEM: SOLUTION
   Ok, 6 simultaneous equations. This should be fun (not!)
   Let's try to keep it simple, shall we? (Suggested answer: Yes)
   Let's use letters to avoid repeating everyone's names:
   
   • F = Father's age
     
   • M = Mother's age
     
   • A = Andrew's age
     
   • B = Bernard's age
     
   • C = Charles's age
     
   • D = Debbie's age
     
   
   Now let's convert the prose into algebra...
   
   • "The combined ages of the eldest son Andrew and their only daughter Debbie is the same as the age of their mother."
     
     • Mathematically, that is A + D = M......(a)
       
   • "The combined ages of the middle two sons (Bernard and Charles) is the same as the age of their father."
     
     • Mathematically, that is B + C = F......(b)
       
   • "The father is 2 years older than the mother."
     
     • Mathematically, that is F = M + 2......(c)
       
   • "The age gap between Andrew and Debbie is 10 years."
     
     • Mathematically, that is A - D = 10......(d)
       
   • "Bernard is 4 years older than Charles."
     
     • Mathematically, that is B = C + 4......(e)
       
   • "The sum of everyone's ages is 236."
     
     • Mathematically, that is A + B + C + D + F + M = 236......(f)
       
   
   Let's start by eliminating the age of the mother by equating (a) and (c):
   
   • M = A + D......(a)
     
   • M = F - 2......(c)
     
   • Hence A + D = F - 2......(g)
     
   
   Now let's get rid of the age of the father by equating (b) and (g):
   
   • B + C = F......(b)
     
   • A + D + 2 = F......(g, re-arranged)
     
   • Hence B + C = A + D + 2......(h)
     
   
   Now modify (f) to get rid of F (= B + C) and M (= A + D) in one go:
   
   • A + B + C + D + (B + C) + (A + D) = 236
     
   • which simplifies to A + B + C + D = 118......(i)
     
   
   We want to ignore any equations for now that reference M or F. Thus the only equations we want to use now are (d), (e), (h) and (i). Using these equations let's get rid of A:
   
   • A = 10 + D......(d, re-arranged)
     
   • A = B + C - D - 2......(h, re-arranged)
     
   • Hence 10 + D = B + C - D - 2
     which simplifies to 12 + 2D = B + C......(j)
     
   
   While we're at it let's get rid of any reference to B, ensuring that no reference to A creeps back in:
   
   • B = C + 4......(e)
     
   • B = 12 + 2D - C......(j, re-arranged)
     
   • Hence C + 4 = 12 + 2D - C
     which can be tidied up to give C = 4 + D......(k)
     
   
   Now change (i) to get rid of the references to A and B:
   
   • (10 + D) + (C + 4) + C + D = 118.
     Thus we can deduce that 2C + 2D = 104, or even better:
     C + D = 52......(l)
     
   
   Now (k) and (l) are two simple simultaneous equations involving only two variables. It gets easy now, honest.
   
   • C = 52 - D......(l, re-arranged)
     
   • C = 4 + D......(k)
     
   • So therefore 4 + D = 52 - D
     i.e. 2D = 48, or D = 24.
     
   It's now a simple matter of plugging values back into the original or derived equations in order to solve the problem. It all unfolds itself very quickly now.
   
   • D = 24, thus Debbie is 24 years old.
     
   • Using (k) tells us that Charlie is 4 + 24 = 28 years old.
     
   • From (d), Andrew is 10 years older than Debbie, so he is 34.
     
   • From (e), Bernard is 4 years older than Charles, so he is 32.
     
   • The mother is as old as Andrew and Debbie put together (a), so she is 58.
     
   • The father is as old as Bernard and Charles put together (b), thus he is 60.