How many integers between 1 and 2009 inclusive can be expressed as a sum of five consecutive positive multiples of 7?

newsss Aug 12, 2023

#1**0 **

Let the five consecutive multiples of 7 be 7n, 7(n+1), 7(n+2), 7(n+3), and 7(n+4). Then their sum is 35n+105=7(5n+15), so the integers which can be expressed as the sum of five consecutive positive multiples of 7 are just the multiples of 7 which are of the form 5n+15 for some integer n. The smallest such multiple is 7⋅2=14 and the largest is 7⋅287=2009, so there are 281 such integers.

Guest Aug 12, 2023

#2**0 **

7 (n + n+1 + n+2 + n+3 + n +4)=35n + 70

35n + 70 <= 2009, solve for n

35n =2009 - 70=1939

n <= 1939 / 35 =55.40

**n ==55 - number of 5 consecutive multiples of 7 between 1 and 2009.**

Check : 7 + 14 + 21 + 28 + 35 = 105 - This is the smallest sum.

385 + 392 + 399 + 406 + 413 =1995 - this is the largest sum

The difference between any two consecutive sums of 5 multiples of 7 is =5 x 7 = 35.**[1995 - 105] / 35 + 1 =55**

Guest Aug 12, 2023

#3**0 **

When you have a sequence of consecutive integers, the sum of the sequence can be found by multiplying the average of the first and last term by the number of terms. In this case, you have a sequence of five consecutive positive multiples of 7. The average of the first and last term would be (7 + 7*4) / 2 = 21.

So, the sum of the five consecutive positive multiples of 7 is:

Sum = Average * Number of Terms = 21 * 5 = 105.

Now, let's consider the possible values of the sum that can be obtained using this sequence. The minimum sum you can get is 105, which is when you use the sequence {7, 14, 21, 28, 35}. The maximum sum you can get is 105 + 4 * 7 = 133, which is when you use the sequence {28, 35, 42, 49, 56}.

Any integer between 105 and 133 (inclusive) can be expressed as the sum of five consecutive positive multiples of 7 using different sequences. Since both 105 and 133 are inclusive, there are a total of 133 - 105 + 1 = 29 integers between 1 and 2009 (inclusive) that can be expressed as the sum of five consecutive positive multiples of 7.

JohnChong Aug 16, 2023