geekygirl Sayswhy will they be repeated 50 times ?
See total 300 digits
121 = 3 digit
212 = 3 digit
121 repeated 50 times =>150
212...........................=> 150
so toal 300 digits.
See total 300 digits
121 = 3 digit
212 = 3 digit
121 repeated 50 times =>150
212...........................=> 150
so toal 300 digits.
My take666
Take three digits at a time as a single number. Add them up
121+212+121+....212
we will have 121 repeated 50 times and 212 repeated 50 times
sum = 121*50+212*50 = 16650
Now find the remainder when 16650 is divided by 999 = 666
geekygirl Sayswhy will they be repeated 50 times ?
We have 300 digits that have been separated into groups of 3 digits each. In this way we will have 100 groups of 3 digits each.
We have two types of groups( 121 and 212 ) so 50 sets for each group.
Hope its clear now
We have 300 digits that have been separated into groups of 3 digits each. In this way we will have 100 groups of 3 digits each.
We have two types of groups( 121 and 212 ) so 50 sets for each group.
Hope its clear now
how can i check remainder for 12121212121...300 digits when divided by111?
sanu8080 SaysFind the remainder when 1212..1212(300 digits) is divided by 999 ?
i ll go with rem=333,
1212121212..../111*9
now,121212...is divisible by 9.
also 121212...is divisible by 111.
so rem should be 111n+0=9m+0 which is minimum in the case of 333.
may b i think i shud factorize a bit more to 111&9
jaiswalicsi Sayshow can i check remainder for 12121212121...300 digits when divided by111?
take set of 3 numbers from left, divide by 111 and add remainder.
121+121+...................(100 times)
10+................(100 times)
10*100 = 1000 remainder by 111 again
1000% 111 so remainder 1.
find this thread now.....
I think it will be useful person like me who are not very good in quant...
will post some questions from tomorrow...
1-Find the arithmetic mean of the AP with 41 terms whose 1st term is 2.5 and common difference is 0.75.
2-Find the 15th term of the sequence 20,15,10.........
1-Find the arithmetic mean of the AP with 41 terms whose 1st term is 2.5 and common difference is 0.75.
2-Find the 15th term of the sequence 20,15,10.........
Solutions :-
1) A.P nth term = a + (n-1)*d , where a is first term and d is common difference
Arithmetic mean = (first term + Last term)/2
Given a = 2.5
d = 0.75
Last term = t(41) = a + 40 d = 2.5 + 40*0.75 = 2.5 + 30 = 32.5
Mean = (2.5+32.5)/2 = 35/2 = 17.5
2) Here a = 20 and d= -5
t(15) = 20 + 14*(-5) = 20 - 70 = -50
take set of 3 numbers from left, divide by 111 and add remainder.
121+121+...................(100 times)(pls explain this,u ve taken 121121....instead of121212)
10+................(100 times)
10*100 = 1000 remainder by 111 again
1000% 111 so remainder 1.[/QUOTE
Solutions :-
2) Here a = 20 and d= -5
t(15) = 20 + 14*(-5) = 20 - 70 = -50
why d= -5 (is it because its reducing?)
i got only 5 and my answer was 90 which was offcource incorrect
1-Divide 124 into 4 parts which are in AP such that the product of 1st and 4th part is 128 less than the product of the 2nd and 3rd part.
Sol-The basic formulae is same here... but why those 4 numbers cant be a,2a,3a and 4a.
As in book its a-3, a-1, a+1 , a+3
2-The 63rd and 6th terms of an AP are -77 and 37. find the 17th term.
2. T63 = a + 62d = -77
T6 = a + 5d = 37
Hence a = 47 and d = -2,
So, T17 = a + 16d = 47 - 32 = 15.
1. Good question! You cant be sure that the four parts are multiples of each other. Hence your assumption is wrong.
To generalize the problem, we can assume the four parts to be a-3d, a-d, a+d, a+ 3d (we assume this because the sum eliminates extra variable)
Hence sum of 4 parts = 4a = 124. or a = 31
also (a-d)(a+d) - (a-3d)(a+3d) = 128
or a^2 - d^2 - a^2 + 9d^2 = 128
or 8d^2 = 128
or d^2 = 16
or d = 4 or -4
Hence your 4 parts are : 19, 27, 35 and 43.
Cheers!
why d= -5 (is it because its reducing?)
i got only 5 and my answer was 90 which was offcource incorrect
d = tn - tn-1 = t2 - t1 = 15 - 20 = -5
1-Divide 124 into 4 parts which are in AP such that the product of 1st and 4th part is 128 less than the product of the 2nd and 3rd part.
Sol-The basic formulae is same here... but why those 4 numbers cant be a,2a,3a and 4a.
As in book its a-3, a-1, a+1 , a+3
2-The 63rd and 6th terms of an AP are -77 and 37. find the 17th term.
1) Those 5 numbers cannot be a,2a,3a,4a because a,2a,3a,4a has d=a , but we don't know what the difference is. So we need to take a term with both the variables a and d. So, generally if there are 4 terms given we take
a-3d,a-d,a+d,a+3d.
As per the given condition
(a-3d)*(a+3d) = (a-d)*(a+d) - 128
a^2 - 9d^2 = a^2 - d^2 - 128
8d^2 = 128
d^2 = 16
d = +- 4
2) t63 = a+62d = -77
t6 = a+5d = 37
Solving above equations we get
57d = -114
d = -2
Substitute value od d in a+5d=37, we get a = 47
So t17 = a +16d = 47 - 32 = 15
Nice threat anna.
Subscribed.
jaiswalicsi Sayshow can i check remainder for 12121212121...300 digits when divided by111?
My take
0111 = 37 * 3
we know that 121212 is divisible by 37
write the given number as 121212*10^24+121212*10^18+....121212
this thing is divisible by 37
also the number is divisible by 3
by C.R.T
37a = 3b
remainder = 37*3 = 111 = 0
12121212....300 digits is fully divisible by 111
rocking thread hyd rocker
hey thanx yr
if length of a rectangle is increased of 40% and breadth is
decreased by 40%.wht is the net effect on the area?
pliz tell me the trick??also
if length of a rectangle is increased of 40% and breadth is
decreased by 40%.wht is the net effect on the area?
pliz tell me the trick??also
Length = l
Breadth = b
Area of rectangle = l*b
When changes are made according to the given conditions
Length = 1.4 l
Breadth = 0.6 b
Area = 1.4 l * 0.6 b = 0.84 lb
So the net effect on area is that it decreases by 16%.
Note :- When ever there is n% loss(decrease) and at the same time n% profit (increase), then the net effect will be n^2/100 %.