SOLUTION: Please help me solve this problems : 1. On Dave's birthday, his brother Harry is 17 yrs younger than 3 times his age. The boy's father , Tom is 12 yrs. older than twice Harry's a

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Edwin's complete solution:
Please help me solve this problems :
1. On Dave's birthday, his brother Harry is 17 yrs younger than 3 times his age. The boy's father , Tom is 12 yrs. older than twice Harry's age. If Dave is 7 years younger than his brother , how many candles are on Dave's cake?

Please help me solve this problems :

>>...his brother Harry is 17 yrs younger than 3 times his age...<<

 
H = 3D - 17

>>...The boy's father , Tom is 12 yrs. older than twice Harry's age...<<

 
T = 2H + 12

>>...Dave is 7 years younger than his brother...<<

 
D = H - 7
 

 
-3D +  H     = -17
    - 2H + T =  12
  D -  H     =  -7
 
Solve that and get D=12, H=19, T=50

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2. When Alvin divides each of 5 consecutive integers by his age, the sum of the 5 remainders he gets is 32. When Susan , several yrs. older, divides each of a different set of 5 consecutive integers by her age, the sum of the five remainders she gets is also 32. What is the sum of the ages of Alvin and Susan?

Let either one of their ages be A.
Let the 5 consecutive integers be I1,I2,I3,I4,I5
Let the remainders when these integers are divided
by A be R1,R2,R3,R4,R5

If none of the consecutive integers is A or
or a multiple of A, then the 5 remainders will 
themselves be 5 consecutive integers.

Let's see if there exist 5 consecutive integers 
which the remainders could be, so as to have the sum 32.






No, that is not possible, for n doesn't come out
to an integer.

That means one of 5 consectutive integers must
be a multiple of A, say kA

Whichever one of the 5 is kA, The remainder left when
dividing by A will of course be 0.  

Any of the 5 remainders of members of the 5 that are 
less than kA, will be consecutive integers, the largest 
of which must be A-1, then the one below that will be
A-2, and so on.

Any of the 5 remainders that are greater than the multiple
of A will be consecutive integers 1,2,3, and so on

Then we have 5 cases to consider, that is, when kA comes
1st, 2nd, 3rd, 4th, and 5th.

         I1    I2    I3    I4    I5     R1   R2   R3   R4  R5   Sum  
case 1.  kA   kA+1  kA+2  kA+3  kA+4     0    1    2    3   4    10
case 2. kA-1   kA   kA+1  kA+2  kA+3    A-1   0    1    2   3   A+5
case 3. kA-2  kA-1   kA   kA+1  kA+2    A-2  A-1   0    1   2    2A
case 4. kA-3  kA-2  kA-1   kA   kA+1    A-3  A-2  A-1   0   1  3A-5    
case 5. kA-4  kA-3  kA-2  kA-1   kA     A-4  A-3  A-2  A-1  0  4A-10
 
We set each of those sums on the far right equal to 32 and see which,
if any, have an integer solution:

Case 1. This is out because 10 does not equal 32.

Case 2.  A+5 = 32
           A = 27

That's a whole number, so one of their ages can be 27.

Case 3.  2A = 32 
          A = 16

That's a whole number, so one of their ages can be 16.

Case 4. 3A-5 = 32
        2A   = 37
         A   = 27/2

That is not a whole number, so Case 4 is out.

Case 5. 4A-10= 32
        4A   = 42  
         A   = 21/2

That is not a whole number, so Case 5 is out.

So Alvin is 16 and Susan is 27, making the sum of their ages 43.

Now there is no way to tell whether Susan's 5 consecutive
integers were 

26,27,28,29,30, or 53,54,55,56,57, or 80,81,82,83,84 or etc.

There is also no way to tell whether Alvin's 5 consecutive
integers were 

14,15,16,17,18, or 30,31,32,33,34, or 46,47,48,49,50 or etc.
 
However he must be 16 and she must be 27.

Edwin