Sprin BREAK MATH 2

AHH!! We only have 1 week of Sprin BREAK LEFT!! This thought is making me so sad....although I do miss some of my friends at school =/ Well anyways during last few day, I had a great opportunity to sleep in and have some great time with some of those GREAT People :D However, I did get bored playing so decided to do some math over the break. I didn't do too much of them but did enough to keep my brain functioning. There was this one particular problem I really liked. This question was very challenging yet very fascinating and fun.

Question:A positive integer is to be placed in each box. The product of any four adjacent integers is alway 120. Wht is the value of x?



When I saw this questions at first I was like, "Whoa... how am I suppose to do this?" However, like I said in previous posts before, if you read and think carefully, you would be able to come up with your own way of solving this problem.

So this was how I solved it. First we look at numbers 2 and 4. When 2 and 4 multiplies, the product would be 8. Meaning, in between 2 and 4 must be two digits that would add up to 15. (120/8) So now let's solve the column beside number 4. We now know the two columns on the left of 4 is 15. So we multiply 15 with 4 and come up with 60. Next you would divide 120 with 60 and the product would equal to two. Next we also know that next two columns adjacent to number 2 equal to 15. Then we may be able to solve the colum right side of the variable x. 2x15= 30, 120/30= 4 The column on the right side of the variable is 4. Next let's figure out the column that's on the RIGHT side of the numer 4. 15x4= 60, 120/60= 2. so the number is 2.

Finally we have everything we need to solve for the "x". We know three numbers adjacent to the variable. 4, 2 and 3. So we muliply all those and divide 120 using the product.

4x2x3= 24, 120/24= 5 x= 5