Latest Elitmus Aptitude Question SOLUTION: 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?

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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?

Asked In Elitmus Shreya Banerjee (9 years ago)

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yyy  (9 years ago)

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.

Is this solution Helpfull? Yes (30) | No (1)  

A SIRISHA   (9 years ago)

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

Is this solution Helpfull? Yes (18) | No   

Neema Pal  (9 years ago)

answer may be 7381

Is this solution Helpfull? Yes (3) | No   

P.Pandeeswari  (9 years ago)

the non zero number is 9
9+9^2+9^3+9^4=7380
9^5+7380=66429
66429/9=7381.

Is this solution Helpfull? Yes (3) | No   

Santanu Sarkar  (9 years ago)

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

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SAURABH KUMAR  (9 years ago)

ans: 7381

Is this solution Helpfull? Yes (2) | No   

Anoop Tiwari  (9 years ago)

number is 9 ,and ans is 7381, 9+9^2+9^3+9^4=7380 and 9^5+7380=66429 so 66429/9 =7381

Is this solution Helpfull? Yes (1) | No   

Chirag Ranjan Lahiri  (9 years ago)

Number is 9 and ans is 7381,
9 + 9^2 + 9^3 + 9^4 =7380
and 9^5 +7380 = 66429 so 66429/9 =7381.........

Is this solution Helpfull? Yes (1) | No   

Chirag Ranjan Lahiri  (9 years ago)

Number is 9 and ans is 7381,
9 + 9^2 + 9^3 + 9^4 =7380
and 9^5 +7380 = 66429 so 66429/9 =7381.........

Is this solution Helpfull? Yes (1) | No   

Ehtesham  (9 years ago)

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

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