product of three successive numbers is 1860744. Their sum will be

M4maths Previous Puzzles

17 Jan 2013 sum and product

product of three successive numbers is 1860744. Their sum will be

OPtion

1) 432
2) 357
3) 135
4) 789
5) 246
6) 369
7) 468
8) 123
9) 531
10) none of these

Solution

1860744 = 2*2*2*3*31*41*61
on arranging these factors we get 122*123*124

Correct Option: 6 Best Solution (12)

Dr.Dipin Singh   12 years AGO

1860744 = 122*123*124

122+123+124=369 .. option 6

Like? Yes (6) | No (2)  

Manideep   12 years AGO

product of three successive numbers is 1860744.
Last unit digit number to get 4 should be multiple of 2,3,4
so,122*123*124=1860744
their sum is 369

Like? Yes (4) | No (1)  

Garima   12 years AGO

Numbers are 122,123 and 124.
122+123+124= 369

Like? Yes (3) | No (1)  

juhi yadav   12 years AGO

the three successive numbers let suppose X,X+1,X+2..
then product is 1860744 that is
X(X+1)(X+2)=1860744
X^3+3X^2+2X-1860744=0
then X=122
then the sum is 122+123+124=369
so ans is 6)369

Like? Yes (3) | No (2)  

Vinay Sipani    12 years AGO

x(x + 1)(x + 2) = 1860744 = (x+2) P 3
factorizing 1860744 ,
1860744 = 2 * 2 * 2 * 3 * 31 * 41 * 61
= (31 * 4) * (41 * 3) * (61 * 2)
= 124 * 123 * 122

=> 122 + 123 + 124 = 369

OR

(x-1)x(x+1) = 1860744
x^3 - x = 1860744
x^3 = (1860744 + x)
The nearest cube greater than 1860744 is 123^3
=> x = 123
=> sum = 3x = 369

Like? Yes (3) | No (3)  

kirti komal   12 years AGO

by factorization we get 1860744=122*123*124.
their sum = 122+123+124 = 369
so, ans.6)369

Like? Yes (2) | No (2)  

Devendra Marghade   12 years AGO

If three successive numbers are x,x+1,x+2
then x(x+1)(x+2)=x^3+3x^2+2x=1860744
Solving for x, we get x=122,So x+1=123,x+2=124
Therefore sum=122+123+124=369

Like? Yes (2) | No (2)  

Muhammad Nadeem Haider Naseem   12 years AGO

let x, x+1, x+2 be three consecutive nos. then x+x+1+x+2=369
3x+3=369
3x=366
x=122, 123,124 so 122*123*124=1860744

Like? Yes (2) | No (2)  

Balakumaran N.A   12 years AGO

Numbers are 122,123,124.

Like? Yes (2) | No (2)  

vivek   12 years AGO

the factors are 2 2 2 3 31 41 61
the successive numbers are 122 123 124
the sum is 369

Like? Yes (2) | No (3)  

shilpesh   12 years AGO

let first unknown num=x
x*(x+1)*(x+2)=1860744
(x^2+x)*(x+2)=1860744
x^3+3x^2+2x=1860744
solving this equation
x=122
3x+3=369

Like? Yes (2) | No (3)  

saikumar   12 years AGO

let x-1 , x , x+1 be the three numbers
product of those numbers : (x-1)x(x+1)=1860744
x^3-x=1860744

[:. 123^3-123=1860744]
so from the above statement we can conclude that
x=123
the sum of three number is: 122+123+124=369

Like? Yes (2) | No (3)