Homework help-Please explain your answer?

Question

If you divide a particular three-digit number by 2, 3, 4, 5, or 6 there would be a remainder of 1. If you divide3 the same three-digit number by 11, there would be no remainder. What is the smallest possible positive three-digit integer that will meet there conditions.

Answer 1

the answer is 121 i think. All you had to do is think of three digit numbers that are divisible by 11 and are in the 100's (just to start out...if none of those work you'd have to continue). so theres 110, 121, 132, 143, 154, and so on....now all the even ones can be thrown out since their divisible by 2, and all the ones that end with 5 can be thrown out since they are divisible by 5. lastly, all the ones where the three numbers add up to a number that can be divisble by three can be thrown out since their divisible by three (for example 123 would be thrown out if it was in the list since 1+2+3=6 and 6 is divisible by 3.) that leaves you with 121, 143, and 187. then you would just pick the smallest number

Answer 2

Well first off, you need to look at the clues given to you:
2, 3, 4, 5, 6 division leaves a remainder of one - so this tells me that the answer has to be an odd number because you cannot divide by an even number and have a remainder of 1 unless the number is an odd number
The next clue is the division by 11 leaves no remainder - so this must mean the the total number must be a multiple of 11
And with a little deduction I came up with 121 as the answer

121/6 = 20 1/6
121/5 = 24 1/5
121/4 = 30 1/4
121/3 = 40 1/3
121/2 = 60 1/2
121/11 = 11

Notice that all of the above answers have 1/x fraction which is the same as having a remainder of 1 while the last answer has no fraction thus no remainder

Hope this helps :)

Answer 3

Let's look for a number divisible by 4, 5 and 6.
We can do this because any number divisible
by 4,5 and 6 is automatically divisible by 2 and 3.
The LCM of 4 and 6 is 4*6/2 = 12
and the LCM of 12 and 5 is 60.
So the LCM of 4,5,6 is 60.
The smallest 3 digit multiple of 4,5,6 is therefore 120.
So 121 leaves a remainder of 1 when divided by 2,3,4,5
and 6. Since 121 = 11*11, the answer to your problem is 121.