Tom wrote down two positive integers consisting of the following digits: 1,2,3,4,5, and 6. Each of the digits appeared in only one of the two numbers, and only

Question

once. When TI'm added up these numbers, he obtained 750. Whay positive integers did Tom write?

 

Answer 1

Using the variables a,b,c,d,e,f to represent digits 1 to 6 (only once for each value), we have:

    [abc] + [def] = [750]

 

Since the decimal number system is a positional number system, that means:

    100a + 10b + c + 100d + 10e + f = 700 + 50 + 0

 

Since each digit is greater than 0, c+f = 10   [it can't be 0 and it can't be 20]  Since each digit is used only once, they can't each be 5.  That leaves only 6 and 4 (in either order).

 

Once 6 and 4 are used, the list is reduced to 1,2,3,5 for a,b,d,e.   (b+e+1)*10 = 50    This happens when b+e=4 and that could only be 1 and 3 (in either order).

 

The list for a,d is now reduced to 2,5 (and these also could be either order).

 

So, we have: 

    (2 or 5)(1 or 3)(4 or 6)

 + (2 or 5)(1 or 3)(4 or 6)

   ----------------------------

        7          5         0

 

The integers that could satisfy the requirements are:

   214+536,   216+534,    234+516,    236+514,

  514+236,   516+235,    534+216,   536+214                

Answer 2

Hi Batoul.

 

Because the final digit is 0,  each number must end with final digits that add up to 10.  

 

Since cannot repeat, 5 won't work.  Out of 1, 2, 3, 4, 6.  Only 4 and 6 will add to 10.  So one number has 6 for its finsl digit, the other has 4.

 

Only 1, 2, 3 and 5 remain

 

If start adding

 

     4

     6

Get 0 and must carry 1 over to the 2nd column.

 

We need the 2nd column to add to 5, so the carried 1 + (   )  +  (  )  must = 5

 

With the digits left (1, 2, 3, 5)  only 3 and 1 will work.

 

So now we have 

 

   14

   36

 

Need the leading digits to add to 7.

 

Use 5 for one and 2 for the other.

 

514

236