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What is the sum of the multiples of 7 from 84 to 140?

This topic has 3 expert replies and 1 member reply

What is the sum of the multiples of 7 from 84 to 140?

Post Mon Sep 25, 2017 5:59 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    What is the sum of the multiples of 7 from 84 to 140, inclusive?

    A) 896
    B) 963
    C) 1008
    D) 1792
    E) 2016

    The OA is C.

    Is there any way to solve it without listing the numbers?

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    Post Mon Sep 25, 2017 6:17 pm
    Quote:
    What is the sum of the multiples of 7 from 84 to 140, inclusive?

    A)896
    B)963
    c)1008
    D)1792
    E)2016

    In other words, 84 + 91 + 98 + . . . 140 = ?

    Since the values are EQUALLY SPACED, we can use the rule: SUM = [(FIRST + LAST)/2][# of values]

    NUMBER of values
    Here's a nice rule: If x and y are multiples of k, then the number of multiples of k from x to y inclusive = [(y-x)/k] + 1
    So, for example, the NUMBER multiples of 3 from 6 to 21 inclusive = [(21 - 6)/3] + 1 = [15/3] + 1 = 6

    So, the NUMBER multiples of 7 from 84 to 140 inclusive = [(140 - 84)/7] + 1
    = [56/7] + 1
    = 9

    ------------------------------------

    Now apply the formula:
    SUM = [(FIRST + LAST)/2][# of values]
    = [(84 + 140)/2][9]
    = [224/2][9]
    = [112][9]
    = 1008
    = C

    Cheers,
    Brent

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    Post Tue Sep 26, 2017 5:16 pm
    We want 7*12 + 7*13 + ... * 7*20

    To find this, we could add the first 20 multiples of 7, then subtract the ones we don't want (the first 11 multiples of 7).

    That gives us:

    (7*1 + ... + 7*20) - (7*1 + ... + 7*11)

    From there, we could write this as:

    7*(1 + 2 + ... + 20) - 7*(1 + 2 + ... + 11)

    The two sums in parentheses are just triangular numbers, and the summation is a cinch. Smile

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    Post Tue Sep 26, 2017 9:48 pm
    Vincen wrote:
    What is the sum of the multiples of 7 from 84 to 140, inclusive?

    A) 896
    B) 963
    C) 1008
    D) 1792
    E) 2016

    The OA is C.

    Is there any way to solve it without listing the numbers?
    Sum of the multiples of 7 from 84 to 140, inclusive = 84 + 91 + ... + 140.

    Number of multiples of 7 from 84 to 140 = 140/7 - 84/7 + 1 = 20 - 12 + 1 = 9.

    Now, write these numbers twice, once in ascending order and then in descending order:

    S = 84 + 91 + .... + 133 + 140
    S = 140 + 133 + .... + 91 + 84

    The sum of each pair is 224. As the total number of pairs are 9.

    2S = 224*9 or S = 112*9 = 1008.

    Thus, the correct answer is C.

    Post Fri Oct 13, 2017 10:28 am
    Quote:
    What is the sum of the multiples of 7 from 84 to 140, inclusive?
    Hi Vincen,
    Let's take a look at your question.

    First of all we need to find the number of terms between 84 to 140 inclusive.
    Since we are talking about multiples of 7, lets find out the term numbers of 84 and 140 in the sequene of multiples of 7.
    $$\frac{84}{7}=12$$
    84 is 12th term.

    $$\frac{140}{7}=20$$
    140 is 20th term.

    Total number of terms = n = 20 - 12 + 1 = 9

    Sum of n terms in geometric series$$ =\frac{n}{2}\left(a_1+a_n\right)$$
    $$=\frac{9}{2}\left(84+140\right)$$
    $$=\frac{9}{2}\left(224\right)$$
    $$=\frac{9}{2}\left(84+140\right)$$
    $$=9\left(112\right)$$
    $$=1008$$

    Therefore, Option C is correct.
    Hope this helps.

    I am available, if you'd like any follow up.

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