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07 Sep 2018, 00:31
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Difficulty:
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Question Stats:
66% (03:02) correct
34%
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wrong
based on 160
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If the product of three consecutive positive integers is 15600 then the sum of the squares of these integers is
A) 1777
B) 1785
C) 1875
D) 1877
E) 1897
A) 1777
B) 1785
C) 1875
D) 1877
E) 1897
26 Aug 2019, 04:37
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Hi,
If the product of three consecutive positive integers is 15600 then the sum of the squares of these integers is
A) 1777
B) 1785
C) 1875
D) 1877
E) 1897
Since, the product of 3 numbers is 15600, one of the numbers should be a multiple of 5. Since, 20^3 is 8000 and 30^3 is 27000, so one of the numbers should by 25. Join the dots and try to use POE (process of elimination) rather than trying to use algebra.
Now, make factors of 15600. You will get 2^4 x 3 x 5^2 x 13. Thus, one of the numbers should be a multiple of 13. So, it has to be 26.
Now, the 3 consecutive numbers could be either 24, 25, 26 or 25, 26, 27. We will pick 24, 25, 26 as 27 can not be part of the list. We have only one 3 in the factorization of 15600.
So, rather than using algebra, we used hit and trial to get the three consecutive numbers.
Now, you can either calculate the square the numbers and get the sum as 1877, which is going to be quite time consuming. Alternate way is to find the unit digit and then use ballparking.
Square of unit digit of 24 = 6
Square of unit digit of 25 = 5
Square of unit digit of 26 = 6
6 + 5 + 6 = 17
So, the unit digit of the resultant number should be 7. Eliminate B and C.
Now, find the square of 25 and multiply it with 3. You will get 1875. So, the answer should be close to 1875.
Thus, the answer is D.
Please give kudos if you liked the solution.
Deepti Singh
Master Trainer - GMAT
Manya - The Princeton Review
If the product of three consecutive positive integers is 15600 then the sum of the squares of these integers is
A) 1777
B) 1785
C) 1875
D) 1877
E) 1897
Since, the product of 3 numbers is 15600, one of the numbers should be a multiple of 5. Since, 20^3 is 8000 and 30^3 is 27000, so one of the numbers should by 25. Join the dots and try to use POE (process of elimination) rather than trying to use algebra.
Now, make factors of 15600. You will get 2^4 x 3 x 5^2 x 13. Thus, one of the numbers should be a multiple of 13. So, it has to be 26.
Now, the 3 consecutive numbers could be either 24, 25, 26 or 25, 26, 27. We will pick 24, 25, 26 as 27 can not be part of the list. We have only one 3 in the factorization of 15600.
So, rather than using algebra, we used hit and trial to get the three consecutive numbers.
Now, you can either calculate the square the numbers and get the sum as 1877, which is going to be quite time consuming. Alternate way is to find the unit digit and then use ballparking.
Square of unit digit of 24 = 6
Square of unit digit of 25 = 5
Square of unit digit of 26 = 6
6 + 5 + 6 = 17
So, the unit digit of the resultant number should be 7. Eliminate B and C.
Now, find the square of 25 and multiply it with 3. You will get 1875. So, the answer should be close to 1875.
Thus, the answer is D.
Please give kudos if you liked the solution.
Deepti Singh
Master Trainer - GMAT
Manya - The Princeton Review
07 Sep 2018, 04:33
Kudos
Bookmarks
AkshdeepS
If the product of three consecutive positive integers is 15600 then the sum of the squares of these integers is
A) 1777
B) 1785
C) 1875
D) 1877
E) 1897
A) 1777
B) 1785
C) 1875
D) 1877
E) 1897
\(15600=2*2*3*13*2*5*2*5\)
A immediate though should no 3 consecutive integers will be multiple of 5, so both 5s have to be together
so one number is 25 or a multiple of 25
let us work on remaining factors 2*2*2*2*3*13... 13*2 will give us 26 and 2*2*2*3 will give 24
so 3 integers = 24*25*26
sum of square = \(24^2+25^2+26^2=(25-1)^2+25^2+(25+1)^2=25^2-2*25+1+25^2+25^2+2*25+1=3*25^2+2=3*625+2=1875+2=1877\)
D
