Messing with numbers

Question

 

 The mischievous leprechaun’s promised anyone entering their forest a pot of gold if they can solve this riddle:

Divide the number 45 into 4 parts such that when 2 is added to the first part, 2 is subtracted from the second part, 2 is multiplied by the third pot, and the fourth part is divided by 2, the same number results.

Find their number to win your pot of gold.

Answer 1

Let the 4 parts be a, b, c, d. We can write: a+2=b-2=2c=d/2 and a+b+c+d=45, so b+c+d=45-a.

We can now write b, c, d in terms of a: b=a+4; 2c=a+2; d=2a+4; double the equations for b and d: 2b=2a+8, 2d=4a+8.

Now add the equations together:

2b=2a+8+

2c=a+2+

2d=4a+8

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2(b+c+d)=7a+18; 2(45-a)=7a+18; 90-18=9a, so a=72/9=8.

b=a+4=12; c=(1/2)(a+2)=5; d=2a+4=20. Therefore, the subdivisions of 45 are: 8, 12, 5 and 20, and the common number is 10 (8+2, 12-2, 2*5, 20/2).

(Now, where's that pot of gold?!)