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01 Mar 2017, 07:18
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D
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Difficulty:
65%
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Question Stats:
60% (02:15) correct
40%
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wrong
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Integer y equals the sum of all multiples of 21 between 210 and 441, inclusive. What is the greatest prime factor of y?
A. 7
B. 11
C. 19
D. 31
E. 37
A. 7
B. 11
C. 19
D. 31
E. 37
01 Mar 2017, 08:03
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Bunuel
there are (21-9=12) values that need to be added to find value of y
the sum, y = (210+441)*(12/2) [(First value+Last value)*(No. of terms/2)] in any arithmetic progression
the sum, y = 21*(10+21)*6 = 21*31*6
Biggest prime factor = 31
answer: Option D
02 Mar 2017, 18:21
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Bunuel
We can first determine the sum of all multiples of 21 between 210 and 441 inclusive using the following formula:
sum = average x quantity
The quantity is (441 - 210)/21 + 1 = 231/21 + 1 = 11 + 1 = 12.
When we have a set of evenly spaced integers, we can calculate the average by using the following formula:
Average = (smallest multiple in the set + largest multiple in the set)/2. Thus:
Sum = (210 + 441)/2 x 12
Sum = 651/2 x 12 = 651 x 6
We need to determine the largest prime factor of 651 x 6.
Since we know the prime factors of 6, let’s prime factorize 651.
651 = 3 x 217 = 3 x 7 x 31. Thus, 31 is the greatest prime factor of y.
Alternative solution:
We are given that y = 210 + 231 + 252 + … + 441. We can factor 21 from each number. That is:
y = 21(10 + 11 + 12 + … + 21)
We can use the formula sum = average x quantity to calculate 10 + 11 + 12 + … + 21 as follows:
10 + 11 + 12 + … + 21 = [(10 + 21)/2] x 12 = 31/2 x 12 = 31 x 6
Thus y = 21(31 x 6) = 3 x 7 x 31 x 2 x 3.
So the largest prime factor of y is 31.
Answer: D
