Elitmus
Exam
Numerical Ability
Number System
The sum of a non zero number, its square, its cube and its four power is 7380. If five power of the number is added to 7380 and divided by the number, then quotient will be?
Read Solution (Total 10)
-
- x + x^2 + x^3 + x^4= 7380
and x + x^2 + x^3 + x^4 + x^5 = x(1 +x + x^2 + x^3 + x^4)
when it is divided by x the quotient will be 1 + x + x^2 + x^3 + x^4 = 1+ 7380
=7381 Ans. - 10 years agoHelpfull: Yes(30) No(1)
- LET NON ZERO NUMBER =X
X+X^2+X^3+X^4=7380
BY SIMPLFY
X=9
SO A/C TO GIVEN QUESTION "If five power of the number is added to 7380 and divided by the number, then quotient will be?"
SO
(X^5+7380)/X = ((9^5+7380)/9)
=7381
SO QUOTIENT = 7381 - 10 years agoHelpfull: Yes(18) No(0)
- answer may be 7381
- 10 years agoHelpfull: Yes(3) No(0)
- the non zero number is 9
9+9^2+9^3+9^4=7380
9^5+7380=66429
66429/9=7381. - 10 years agoHelpfull: Yes(3) No(0)
- Let no.= x
x+x^2+x^3+x^4=7380
x[1+x+x^2+x^3]=7380
x^3+x^2+x+1=7380/x.......(i)
also (x^5+7380)/x
=x^4+7380/x
using (i)
x^4+x^3+x^2+x+1
7380+1
=7381
=x^4+x - 10 years agoHelpfull: Yes(3) No(0)
- ans: 7381
- 10 years agoHelpfull: Yes(2) No(0)
- number is 9 ,and ans is 7381, 9+9^2+9^3+9^4=7380 and 9^5+7380=66429 so 66429/9 =7381
- 10 years agoHelpfull: Yes(1) No(0)
- Number is 9 and ans is 7381,
9 + 9^2 + 9^3 + 9^4 =7380
and 9^5 +7380 = 66429 so 66429/9 =7381......... - 10 years agoHelpfull: Yes(1) No(0)
- Number is 9 and ans is 7381,
9 + 9^2 + 9^3 + 9^4 =7380
and 9^5 +7380 = 66429 so 66429/9 =7381......... - 10 years agoHelpfull: Yes(1) No(0)
- let number be a;
a+a^2+a^3+a^4=7380
a(1+a)(1+a^2)=7380;
factor of 7380=2*2*3*3*5*41
so if we take a=9;
then it satisfies the above equation.
(9^5+7380)/9=(9^4+4*5*41)
=7381
- 9 years agoHelpfull: Yes(0) No(0)
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