A book has 213 problemsfrom 12 years, of competitions. At first there were 25 problems per year, then 16 problems per year and finally 20 problems per year. The number of years when there were 25 problems was
Solution
The equation shall be: 25a+16b+20c = 213 and a+b+c = 12
Since the final total is 213, the only possibility of a sum ending with 3 is when the resultant of 16 ends with 8 (48 or 128) and the resultant of 25 ends with 5, thereby giving 3 as the unit digit of the sum.
Case 1: When 16b is 48
This means b = 3
Hence, 25a + 20c = 165
Now, a+c = 9 (as b = 3)
These two equations are inconsistent since one of the values is coming out a negative integer.
Case 2: When 16b = 128
This means b = 8
Hence, 25a + 20c = 85
And a+c=4 (as b=8)
Hence, a=1 and c=3
By this, we know that the number of months with 25 problems is 1.
