Latest Elitmus Aptitude Question SOLUTION: Q. What is the last term of this eq after adding 1^4+2^4+3^4...........+99^4

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Q. What is the last term of this eq after adding
1^4+2^4+3^4...........+99^4

Asked In Elitmus nishant singh bisht (10 years ago)

Unsolved Read Solution (12)

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Ankit Goyal  (10 years ago)

The last digit of any number raised to power 4 is either 1,6,5,0
In the given series if you will find the power of the number you will get series of unit digits as:

1^4 : unit digit = 1
2^4 : unit digit = 6
3^4 : unit digit = 1 ...

over all series of unit digits will be like this

1,6,1,6,5,6,1,6,1,0 (1 - 10)
1,6,1,6,5,6,1,6,1,0 (11 - 20)
...
same for 21-30, 31-40, ....., 81-90, 91-99

adding the unit digits we get sum = 33 * 10 = 330

so unit digit = 0

Is this solution Helpfull? Yes (25) | No (9)  

RAKESH  (10 years ago)

1st consider
1^4+2^4+3^4+...+9^4+10^4
unit digit will be 1,6,1,6,5,6,1,6,1,0 (1^4 to 10^4)
divide 1 to 99 is divided into 10 groups each containing 10 numbers & last one 9 numbers i.e (1-10,11-20,21-30,...,81-90,91-99)
unit digit will be same for all 10 groups
sum of unit digit for 1^4 to 10^4=1+6+1+6+5+6+1+6+1+0=33
so sum of unit digit for 1^4 to 99^4=33*10=330
so last term of exp=0
ans 0

Is this solution Helpfull? Yes (24) | No   

saikumar  (10 years ago)

Hiii @ Ankit Your approach is correct


The last digit of any number raised to power 4 is either 1,6,5,0
In the given series if you will find the power of the number you will get series of unit digits as:

1^4 : unit digit = 1
2^4 : unit digit = 6
3^4 : unit digit = 1 ...

over all series of unit digits will be like this

1,6,1,6,5,6,1,6,1,0 (1 - 10)
1,6,1,6,5,6,1,6,1,0 (11 - 20)
...
same for 21-30, 31-40, ....., 81-90, 91-99

adding the unit digits we get sum = 33 * 10 = 330 // here is the mistake numbers from 1 to 90 so there are 9 such groups

so unit digit = 7
33*9 // units digit 7

Is this solution Helpfull? Yes (3) | No (10)  

saikumar  (10 years ago)

But the question in yesterday exam is 90 ...!!!!

@ ankit

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Abhishek Singh  (10 years ago)

i think ans ladt term is 7

Is this solution Helpfull? Yes (2) | No (6)  

Hanumesh GH  (10 years ago)

There are 10 numbers for each last digit. Eg:01,11,21,31,41,51,61,71,81,91. When they are rised to power 4, they all lead to same last digit. Hence when we sum up them we get zero for each last digit's. Hence last digit for given sum is zero

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Mudit Vaish  (10 years ago)

1^4=1 2^4=16 3^4=81 4^4=....6 5^+4=625 6^4=...6 ...
up to 9... last digit of sum is 1+6+1+6+5+6+1+6+1=33*10=330
so last digit is 0

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Ankit Goyal  (10 years ago)

@SAIKUMAR
QUESTION is 1 to 99

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ashes  (10 years ago)

last digit of any expression is the remainder when divided by 10.
let divide the expression by 10 and find the remainder.
1^4+2^4+......(1-9)
1^4+2^4+......(11-20)
.
.
.
.
1^4+2^4.........(91-99)
after addition of all terms
10*(1^4)+10*(2^4)+........10*(9^4)
above expression is divided by 10. hence remainder is zero.

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vishal rathore  (10 years ago)

last digets of each term are as , 1,6,1,6,1,6 now lets make group of (1,6) and add these two it comes 7 now if it is added 49 times then we will get last digit of (1^4+2^4+.....98^4) now 99^4 has last digit = 1 and last digit of 7*49 is 3 hence answer is 4 :P

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vishal rathore  (10 years ago)

oh ho.. big mistake :P , last term are as 1 ,6,1,6,5,6,1,6,1,0 now in the question please add 100^4 because it do not affect the last term okk ? now sum all the above last terms 1+6+1+6+5+6+1+6+1+0 = 3 as last digit now multilpy it with 100 we get 300 means 0 as last digit this is the answer 0

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aditya lagudu  (10 years ago)

i think answer is 3 the series will repeat after 10

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