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In this, the options are sufficient to get almost to the answer. Only a small amount of calculations are required in the end.


We know that 15600 is a perfect cube who's cube root lies between 20 (\(20^3 = 8000\) and 30 (\(30^3 = 27000\)).


We also know that since 15600 ends in 2 zeroes, there has to be 2 fives in these 3 numbers (as a 0 is a product of a 2 and a 5), therefore one of the numbers is 25.


So the 3 numbers could be (a) 23, 24, 25 or (b) 24, 25, 26 or (c) 25, 26, 27

Now a glance at the options tells us that the sum of squares is an odd number. (All the options are odd)



If 2 numbers are Odd and the 3rd is even, then the resulting number will be even (\(O^2 + O^2 + E^2 = O + O + E = E + E = E)\), so (a) and (c) are ruled out and the numbers are therefore 24, 25 and 26.


Sum of squares \(24^2 + 25^2 + 26^2 = 576 + 625 + 676 = 1877\)



Option D

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I did the factors of 15600=2*2*2*2*3*5*5*13

Next I analysed that 13 is biggest prime and should be multiply by smallest prime keeping in mind that we have all 3 numbers as consecutive integers.

Also we know that there should numbers that have last digit and 5 and multiple of 2 to get a hundred at the end.

So from factors-->2*2*2*2*3*5*5*13
I took out 13*2=26
5*5=25
2*2*2*3=24

So we got 2 consecutive integers-->24,25,26

Next is sum of squares of this number=576+625+676=1877

Therefore D.
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The product of three consecutive integers is 15600.

One of the numbers will end with '5' as unit digit as we need '0' in unit digit because then number 15600 lies between [\((20)^3\)= 8000 and \((30)^3\) = 27000].

=> With a hit n trial we will get those three integers as 24, 25, and 26.

=> \((24)^2 + (25)^2 + (26)^2\) = 576 + 625 + 676

=> 1877

Answer D
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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

We need to sum the squares of three consecutive integers. If those integers are a-1, a and a+1, we need to find

(a-1)^2 + a^2 + (a + 1)^2 = a^2 - 2a + 1 + a^2 + a^2 + 2a + 1 = 3a^2 + 2

So the right answer must be 2 greater than some multiple of 3 (i.e. it must give a remainder of 2 when we divide it by 3). If you sum the digits of any positive integer at all, and divide that sum by 3, you'll always find the remainder you'd get when you divide the original number by 3. Here, if we add the digits of each answer choice, only answer D gives us a remainder of 2 when we divide by 3, so it is the only answer that could be correct
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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
\((n -1)n(n+1) = 15600\)

\(24*25*26 = 15600\)

So, \(n = 25\)

Sum of the squares of the 3 numbers is -

\(24^2 + 25^2 + 26^2 = 1877\). Answer must be (D)
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