Latest Elitmus Aptitude Question SOLUTION: Find numbers of n for which (100^n- (112^2)) divisible by 3?

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(#M40055121) ELITMUS Numerical Ability question Keep an EYE

Find numbers of n for which (100^n- (112^2)) divisible by 3?

Asked In Elitmus yash (10 years ago)

Unsolved Read Solution (24)

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Read Solution (24) :

Gurudeep  (10 years ago)

The min value of n have to be 3... So have to be => 3...

Is this solution Helpfull? Yes (15) | No (2)  

vikas kumar singh   (10 years ago)

least value be 3
expl=when n=3
x=1000000
112^2=12544
so y=1000000-12544 = 987456
divisiblity rule for 3 is 9+8+7+4+5+6=39
987456 is divisible be 3

Is this solution Helpfull? Yes (9) | No (3)  

SMRITI  (10 years ago)

Least value will be n=6
let x = ((100^n)-(112^2))
As 1oo^6 = 1000000000000
And 112^2 = 12544
therefore
x = 1000000000000 - 12544
=> x =999999987546
now to check whether number is divisible by 3 , check whether summation of its digits is divisible by 3 or not
as 9+9+9+9+9+9+9+8+7+5+4+6 is equal to 93
so, 999999987546 is divisible by 3.

like wise for each n>=6,
number will be divisible by 3
but least value will be 6.




After subtraction

Is this solution Helpfull? Yes (4) | No (2)  

Dipankar Dey  (10 years ago)

2,3,4,5,6,7,8,9,........................(from 2 all positive integer)

Is this solution Helpfull? Yes (3) | No (4)  

P.Nikhil Reddy  (10 years ago)

all positive numbers...1,2,3.........

Is this solution Helpfull? Yes (3) | No (8)  

maharaja  (10 years ago)

n value should be 3 and above.......as we subtract 1000000-12544...only then we get positive answer....

Is this solution Helpfull? Yes (3) | No   

vikash  (10 years ago)

nd 1 more conditon n

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

mritunjaysingh  (10 years ago)

n is greater than 6 (n>6)
options;-
a. 6
b.17
c.16
d.8

Is this solution Helpfull? Yes (1) | No   

Shrimant  (10 years ago)

If n=1 100^1/3=Reminder is 3
if 3/3=1
then 1-(112^2)/3=1-(1/3)
=1-1=0
IF n=2 then
(100^2-(112^2)=(1^2)-(1^2)=1-1=0

Then we can clearly say then N =Any Positive Integer number.....
Ans:=1,2,3,..............................................+Infinity

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javed iqbal  (10 years ago)

hey how to post a question here in e litmus page of m4 maths

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suryakant  (10 years ago)

Given n

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suryakant  (10 years ago)

Given n

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md shahjad khan  (10 years ago)

here question is 100^n-112^2 is positive and n

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vikash  (10 years ago)

this question is incomplete ....

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prasanthi  (10 years ago)

for all positive numbers from 1,2,3...............of n

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sravan kumar  (10 years ago)

n can be anything....because
consider it (100)^n/3-112^2/3 then
(99+1)^n/3 gives modules 1
112^2/3 gives modules 1
so 1^n-1 wil gives 0...
so n value can be any postive integer

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swati priya  (10 years ago)

the condition will satisfy for all n>=3

Is this solution Helpfull? Yes | No (1)  

love sharma  (9 years ago)

112^2=12544 100^3=1000000-12544=987456 add all digit 9+8+7+4+5+6=39 and 39 is divisible by 3 so n=3

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ghanshyam gupta  (9 years ago)

any no greater or equal to 1 here it is not given in question that number should positive.

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Nupur  (9 years ago)

It is not necessary for the result of subtraction to be +ve . Even -ve nos are divisible by 3
So for n = 1
100 - 12544 = 12444 is divisible by 3
Hence for n = 1,2 3, 4,.....

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Anand Kumar Sharma  (9 years ago)

for all values of n=0,1,2,3,...
because its not mentioned that the answer must be +ve

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Asmita  (9 years ago)

given number = 3
let n=3
then 100 pow 3 = 1000000
x=1000000
112 pow 3 = 12544
now
x= 1000000-12544
we get 987456
now divisibility rule of 3
9+8+7+4+5+6=39 and 39 vl be divisible by 3
so 987456 will be divisible by 3

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shilpa  (8 years ago)

n=3 becoz 112^2=12544 and 1000000-12544=987456 which is divisible to 3

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Adil khan  (7 years ago)

112^2/3 will give remainder 1 ad 100/3 remainder 1/3 so 1^n/3 always gives remainder 1 so
1-1=0 n=1,2,3... but we cant take 1 and 2 because i ll become negative thats y
n>=3

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