Hi,
I m stuck on the below question:
The remainder when 2^2 + 22^2 + 222^2 +.....(222....49twos)^2 is divided by 9 is:
a)2 b) 5 c) 6 d) 7
Its from Arun sharma LOD 3. Also how much time shd i give per Qstn while solving LOD3?
Hi,
I m stuck on the below question:
The remainder when 2^2 + 22^2 + 222^2 +.....(222....49twos)^2 is divided by 9 is:
a)2 b) 5 c) 6 d) 7
Its from Arun sharma LOD 3. Also how much time shd i give per Qstn while solving LOD3?
My take is 6.
Remainder by 9 is nothing but sum of digits.
Now, sum of digits are:
2, 4, 6, 8, 1, 3, 5, 7, 9 and then it repeats..
So, remainder of squares will be:
4, 7 (same as -2), 0, 1, 1, 0, -2, 4, 0 => Total = 4
Now, as we have 49 numbers, this will repeat 5 times.
Then we will have 4 more numbers.
So, final remainder = 4*5 + 4 + (-1) + 0 + 1 = 24 => 6
u will get this type of serries of zero
5+10+15+20+30+35+40+45+50+12
Help me with that last 12
Hi,
Please help me wid the below qstn's solution
The sum of the 1st 20 terms and 1st 50 terms of an AP is 420 and 2550. Find the 11th term of a GP whose 1st term is same as the AP and common ratio of the GP is equal to common diff. of the AP.
a) 560 b) 512 c) 1024 d) 1120 e) 3072.
My ans is cuming 1024 but the ans is given as 3072 :(.
Hi,
Please help me wid the below qstn's solution
The sum of the 1st 20 terms and 1st 50 terms of an AP is 420 and 2550. Find the 11th term of a GP whose 1st term is same as the AP and common ratio of the GP is equal to common diff. of the AP.
a) 560 b) 512 c) 1024 d) 1120 e) 3072.
My ans is cuming 1024 but the ans is given as 3072 :(.
How are you getting 1024???I got 2048
Please Help me with the following questions
1)Find the number of zeroes in 100^1*99^2*98^3....*1^100
Ans provided is 1124
2)Find the number of zeroes in 1^1!*2^2!*3^3!.....10^10!
Ans provided is 5!+10!
spectramind07 SaysHow are you getting 1024???I got 2048 :shocked:
Should be 2048 only...
Please Help me with the following questions
1)Find the number of zeroes in 100^1*99^2*98^3....*1^100
Ans provided is 1124
2)Find the number of zeroes in 1^1!*2^2!*3^3!.....10^10!
Ans provided is 5!+10!
1) My take is 1124
First let's calculate number of 5s in each term.
100, 95, 90, ..., 5 will give one 5 each.
100, 75, 50 and 25 will gice one more 5 each.
So, number of 5s as per first case = 1 + 6 + 11 + ... + 96 ....(100^1, 95^6, ...., 5^96)
= 970 ... sum of terms in AP.
And with second case, we will get 1+26+51+76 = 154 more 5s.
Now, each 5 when multiplied by 2, will end in 0.
So, total 0s in the end = 970+154 = 1124.
2) My take is 5!+10!
Here, only 2 terms will give 0.
5^(5!) and 10^(10!)
5^(5!) will give 5! zeros and 10^(10!) will give 10! zeros in the end.
So, total 0s = 5!+10!.
Thanks ThinkAce that completely makes sense 😃
Guys please help me the below sums:
1. How many triangles and quadrilaterals are possible for a set of 25 points on a plane, of which 7 are collinear?
2.There are 10 subjects in the school day,but students have only 5 periods in a day.In how many ways can we form a time table for the day for students?
Thanks
Guys please help me the below sums:
1. How many triangles and quadrilaterals are possible for a set of 25 points on a plane, of which 7 are collinear?
2.There are 10 subjects in the school day,but students have only 5 periods in a day.In how many ways can we form a time table for the day for students?
Thanks
1) My take for triangles is 25C3 - 7C3
Total number of groups of 3 points = 25C3
So, we can make maximum 25C3 triangles.
But as 7 poins are colinear, they cannot make any triangle between them.
So, we need to remove 7C3 triangles.
So, total (25C3-7C3).
Similarly, for quadrilaterals, it should come out to be 25C4-18*7C3.
Here, we need to remove cases when 3 poins on line are selected and one point out of other 18 poins is selected.
Note: Assuming no limit on max. angle in case of quadrilaterals.
2) My take is 10P5.
We have 10 subject which need to permuted in 5 ways.
So, 10P5.
Or you also say that, first one 10 ways, second one in 9 ways and so on.
So, 10*9*8*7*6.. same as 10P5.
hey! could anyone solve this for me?! in an examination, there are 4 sections ; each section of 45 marks. Find the no. of ways in which a student can qualify if the qualifying marks is 90.
1) My take for triangles is 25C3 - 7C3
Total number of groups of 3 points = 25C3
So, we can make maximum 25C3 triangles.
But as 7 poins are colinear, they cannot make any triangle between them.
So, we need to remove 7C3 triangles.
So, total (25C3-7C3).
Similarly, for quadrilaterals, it should come out to be 25C4-18*7C3.
Here, we need to remove cases when 3 poins on line are selected and one point out of other 18 poins is selected.
Note: Assuming no limit on max. angle in case of quadrilaterals.
2) My take is 10P5.
We have 10 subject which need to permuted in 5 ways.
So, 10P5.
Or you also say that, first one 10 ways, second one in 9 ways and so on.
So, 10*9*8*7*6.. same as 10P5.
In the first problem wont it be 25C3/3 and 7C3/3 to remove any repeat combinations.
For e.g. in case we name the points A,B,C,D,E...and so on we will have ABC,BCA and CAB as three triangles in 25C3 which are actually the same.
In the first problem wont it be 25C3/3 and 7C3/3 to remove any repeat combinations.
For e.g. in case we name the points A,B,C,D,E...and so on we will have ABC,BCA and CAB as three triangles in 25C3 which are actually the same.
No need to divide by 3. Here, we are taking combinations and not permutataions. Had I taken permutations, it would have been needed to divide by 3!.
Hi puys,
Can ny one pl solve dis qn ?
In an examination the maximum marks for each of the 3 papers are 50 each.Maximum marks for the fourth paper are 100. Find the number of ways in which the candidate can score 60% marks in the aggregate?
Chapter:Functions
LOD 3
Q.44Which of the following is equal to /?
Please need a quick reply...HELP.
3*34*199*3/236 any quick method to find remainder?
Hi puys,
Can ny one pl solve dis qn ?
In an examination the maximum marks for each of the 3 papers are 50 each.Maximum marks for the fourth paper are 100. Find the number of ways in which the candidate can score 60% marks in the aggregate?
It should be (103C3 - 3*52C3)
Total marks = 250
60% makrs = 150
So, one has to lose 100 marks to get 150.
Let, a, b, c and d be marks lost in 4 subjects.
So, a+b+c+d = 100 => 103C3 ways
But, a, b and c cannot be more than 50.
So, we need to remove those cases.
Let a > 50
And a = 51 + a'
So, a'+51+b+c+d=100 => a'+b+c+d = 49 => 52C3 ways
Now, b and c can also br greater than 50.
So, total cases to beremoved = 3*52C3
So, final ways = 103C3 - 3*52C3
himanshuarora Says3*34*199*3/236 any quick method to find remainder?
It should be 6.
236 = 4*49
Remainder by 4 = (-1)*(-2)*(-1)*(-1) = 2
Remainder by 59 for 3*34*199*3
=> (3*34) * 199 * 3
=> (102) * (22) * 3 ......199 = 59*3+22
=> (-16) * 66 ..............102 = 59*2-16
=> (-16) * 7
=> -112
=> 6
So, we need to find a number, 4x+2 = 59y+6 => 6 ... at x=1 and y=0
So, remainder is 6.
How many numbers smaller than 2*10^8 and are divisible by 3 can be written by means of the digits 0,1 and 2? (Exclude single digit and double digit numbers)
a) 4369
b) 4353
c) 4373
d) 4351
How many numbers smaller than 2*10^8 and are divisible by 3 can be written by means of the digits 0,1 and 2? (Exclude single digit and double digit numbers)
a) 4369
b) 4353
c) 4373
d) 4351
I am getting 4371. Not sure why it is not in options.
First we need to find out how many numbers we can form.
It is like forming numbers in ternary system.
So, we can take numbers till 8-digit numbers => 3^8 numbers from 0 to 3^8 - 1.
And then, for 9 digit numbers, we can take only 1 as first digit and last digits can be filled in 3^8 ways.
So, we will have total 2*3^8 numbers.
Out of these, (1/3)rd will be divisible by 3.
So, numbers divisible by 3 = 2*3^7 = 4374
But we need to remove 1 and 2 digit numbers.
So, we need to remove 0, 12 and 21. ... (We have taken numbers from 0)
So, total cases = 4374 - 3 = 4371