M4maths Previous Puzzles

13 Nov 2009 Successive Number

The sum of squares of three successive integers is 27650. Sum of these integers is

OPtion

1) 297
2) 256
3) 198
4) 234
5) 304
6) 288
7) 312
8) 252
9) 228
10) 346

Solution

(a-1)^2 + a^2 + (a+1)^2 = 27650 , solve the equation

Correct Option: 6 Best Solution (4)

Dimple   15 years AGO

For last digit to be zero of square of 3 consecutive integers, these integers should have last digits as 3,4,5 or 5,6,7.
Last digit of sum should be 2 or 8.

For sum to be 27***, numbers should be of range of 90+.

considering this and checking options, 288 satisfies all conditions and gives desired result.

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Dr.Dipin Singh   15 years AGO

95^2+96^2+97^2= 9025 +9216+9409= 27650

95+96+97=288

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Pargat Pal Singh   15 years AGO

95*95+96*96+97*97 = 9025+ 9216+9409= 27650

Hence sum of integers = 95+96+97=288

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Dolly   15 years AGO

Let the numbers be (x-1), x, (x+1)

Then (x-1)^2+ x^2+ (x+1)^2= 27650

3x^2+2= 27650
3x^2= 27648
x^2= 9216
x= 96
Hence numbers are
95,96,97
sum = 3x = 3*96=288.... option 6

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