Quote:
Originally Posted by great_math  There is a 3 digit number abc which satisfies the condition 49a + 7b + c = 286 Find all possible values of 100a+10b+c |
Since a, b, and c must be one digit integers, it is certainly possible to write c= 286- 49a- 7b and see which values of a and b give a one digit value for c.
for example, the largest that b can be is 9 and then c= 286- 63- 49a= 223- 49a. For what value of a is c= 223- 49a= 9? That is the same as 49a= 214 and a= 214/49= 4.36.... That is, a cannot be smaller than 5.
If a= 5, c= 286- 7b- 5(49)= 41- 7b. In order that that be less than 9, 7b must be larger than 41-9= 32 and b must be larger than 4. 41- 5(7)= 41-35= 6. One answer is 556. 41-6(7)= 41-42 is negative so there is no other solution with a= 5.
If a= 6, c= 286- 7b- 6(49)= -8- 7b which can't be positive!
The only such number is 556.