find the. greatest number that will divide 80 ,118 and 60 leaving remainder8 ,10 and 12 respectively
Answers
Answer:
Step-by-step explanation:
80 remainer 8 is a number that evenly divides 72.
118 remainder 10 is a number that evenly divides 108.
60 remainder 12 is a number that evenly divides 48.
However, since the remainders are 8, 10, and 12 we know that x must be greater than 12.
From here the solution is GCD of 72, 108, and 48.
Looking at 72′s factors: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
We can start at a small range, since 12 and smaller can't be our number due to the largest remainder being 12. 36 and 72 are too large to be factors of 48 so those are also eliminated. That leaves us with 18 and 24. 18 is not a factor of 48 so 24 must be the number we are looking for. But just to be safe, 108/24=4.5. Uh oh. Either I did my math wrong or there is no solution to this problem.
Answer !!
Assuming an integer solution, the exactly divisible numbers are
80 -8 = 72
118 - 10 = 108
60 -12 = 48
The greatest number must be larger than 12 since it leaves a remainder of 12
The only factor of 48 that is greater than 12 is 16, which does not divide into 72 or 108