Number Property

Most people feel that the topic of number properties is hard or at least a little tricky. The reason is that no matter how much effort you put into it, you will still come across new concepts every time. There will be some concepts you don’t know and will need to “figure out” during the actual test. Say you have N consecutive integers (starting from any integer). What can you say about their sum? What can you say about their product? Say N = 3 The numbers are 5, 6, 7 (any three consecutive numbers) Their sum is 5 + 6 + 7 = 18 Their product is 5*6*7 = 210 Note that both the sum and the product are divisible by 3 (i.e. N). Say N = 5 The numbers are 2, 3, 4, 5, 6 (any five consecutive numbers) Their sum is 2 + 3 + 4 + 5 + 6 = 20 Their product is 2*3*4*5*6 = 720 Again, note that both the sum and the product are divisible by 5 (i.e. N) Say N = 4 The numbers are 3, 4, 5, 6 (any five consecutive numbers) Their sum is 3 + 4 + 5 + 6 = 18 Their product is 3*4*5*6 = 360 Now note that the sum is not divisible by 4, but the product is divisible by 4. If N is odd then the sum of N consecutive integers is divisible by N, but this is not so if N is even. Why is this so? Let’s try to generalize – if we have N consecutive numbers, they will be written in the form: (Multiple of N), (Multiple of N) +1, (Multiple of N) + 2, … , (Multiple of N) + (N-2), (Multiple of N) + (N-1) In our examples above, when N = 3, the numbers we picked were 5, 6, 7. They would be written in the form: (Multiple of 3) + 2 = 5 (Multiple of 3)       = 6 (Multiple of 3) + 1 = 7 In our examples above, when N = 4, the numbers we picked were 3, 4, 5, 6. They would be written in the form: (Multiple of 4) + 3 = 3 (Multiple of 4)        = 4 (Multiple of 4) + 1 = 5 (Multiple of 4) + 2 = 6 etc. What happens in case of odd integers? We have a multiple of N and an even number of other integers. The other integers are 1, 2, 3, … (N-2) and (N-1) more than a multiple of N. Note that these extras will always add up in pairs to give the sum of N: 1 + (N – 1) = N 2 + (N – 2) = N 3 + (N – 3) = N … So when you add up all the integers, you will get a multiple of N. What happens in case of even integers? You have a multiple of N and an odd number of other integers. The other integers are 1, 2, 3, … (N-2) and (N-1) more than a multiple of N. Note that these extras will add up to give integers of N but one will be leftover: 1 + (N – 1) = N 2 + (N – 2) = N 3 + (N – 3) = N … The middle number will not have a pair to add up with to give N. So when you add up all the integers, the sum will not be a multiple of N. For example, let’s reconsider the previous example in which we had four consecutive integers: (Multiple of 4)      = 4 (Multiple of 4) + 1 = 5 (Multiple of 4) + 2 = 6 (Multiple of 4) + 3 = 3 1 and 3 add up to give 4 but we still have a 2 extra. So the sum of four consecutive integers will not be a multiple of 4. Let’s now consider the product of N consecutive integers. In any N consecutive integers, there will be a multiple of N. Hence, the product will always be a multiple of N. Now take a quick look at the question that brought this concept into focus: Which of the following must be true? 1) The sum of N consecutive integers is always divisible by N. 2) If N is even then the sum of N consecutive integers is divisible by N. 3) If N is odd then the sum of N consecutive integers is divisible by N. 4) The Product of K consecutive integers is divisible by K. 5) The product of K consecutive integers is divisible by K! (A) 1, 4, 5 (B) 3, 4, 5 (C) 4 and 5 (D) 1, 2, 3, 4 (E) only 4 Let’s start with the first three statements this question gives us. We can see that out of Statements 1, 2 and 3, only Statement 3 will be true for all acceptable values of N. Therefore, all the answer choices that include Statements 1 and 2 are out, i.e. options A and D are out. The answer choices that don’t have Statement 3 are also out, i.e. options C and E are out. This leaves us with only answer choice B, and therefore, B is our answer. This question is a direct application of what we learned above so it doesn’t add much value to our learning as such, but it does have an interesting point. By establishing that B is the answer, we are saying that Statement 5 must be true. 5) The product of K consecutive integers is divisible by K!