Theresa C. answered • 09/25/21
Certified Math and Science Teacher with Tutoring Experience
Ok, so your note at the bottom is exactly what you need to find, a number less than 85 which is divisible by 5, but not divisible by 2, 3, or 4. So, I started by listing all the numbers less than 85 that are divisible by 5. So that gives me 5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,and 80 (remember it has to be LESS than 85). You can cross out all the even numbers (all of them will be divisible by 2). So that leaves us with 5,15,25,35,45,55,65 and 75. Now, remember that if a number is divisible by 4, it is also divisible by 2, so removing all the even numbers actually removes both of those exclusions. Lastly, we have to remove the numbers divisible by 3. Numbers that are divisible by 3 have the rule that the individual digits sum to give you a number also divisible by 3. So as an example: 1+5 = 6, 6 is divisible by 3, which means 15 (and 51) are divisible by 3. So eliminating those numbers leaves us with 5,25,35,55, and 65.
These are the possible numbers. Looking at the last limitation is that if she breaks the students into the groups of 2,3, or 4, she will have 1 piece left over. So using process of elimination, 5 cannot be the answer, because you have a remainder of 2 when dividing by 3. 35 cannot be the number, because when dividing by 3, you have a remainder of 2, 55 cannot be the number because when dividing by 4 you have a remainder of 3, and 65 cannot be the answer because when dividing by 3, you have a remainder of 2.
I purposely skipped 25, because that is the answer. When dividing by 2, she would have 12 students with 1 left over. When dividing by 3, she would have 8 students with 1 left over. When dividing by 4, she would have 6 students with 1 candy remaining, and when dividing by 5, it divides evenly.
Sorry to be long winded, but tried to explain each step and how we complete a problem of this type. Hope this helps!
