Hi puys,
51^3 when divided by 7 leaves a remainder of 1....
Thus,
1375 by 3 leaves a remainder of 1.
thus ,
we can write
((51^1374)*51) by7
((51^3)^45
*51 by 7
reduced to
51 by 7
therefore remainder is 2:)
Thanx chillfactor bhai 😃
A DS Questions
A.is x > y:
1.Three times x is equal to twice y
2. twice of x is 7 more than y
it can be answered by both statements together.Can neone show how to approach these kind of questions .
B. is the %age increase in the perimeter of a rectangle less than 15%?
1.the area of the rectangle is increased by 14%
2.the length of the rectangle is increased by 20% and breadth is increased by 10%
answer is either statements can answer.Plz elaborate.
A.is x > y:
1.Three times x is equal to twice y
2. twice of x is 7 more than y
it can be answered by both statements together.Can neone show how to approach these kind of questions .
B. is the %age increase in the perimeter of a rectangle less than 15%?
1.the area of the rectangle is increased by 14%
2.the length of the rectangle is increased by 20% and breadth is increased by 10%
answer is either statements can answer.Plz elaborate.
6) A triangular number is defined as a number which has the property of being expressed as a sum of consecutive natural numbers starting with 1. how many triangular numbers less than 1000, have the property that they are difference of squares of two consecutive natural numbers?=> 22 such numbers
a) 20 b)21 c)22 d)23
Triangular numbers are of form n(n + 1)/2.
Any number that can be written as difference of squares of two consecutive natural numbers are odd numbers
=> n(n + 1)/2 should be odd
=> n should be of form 4k - 2 or 4k - 3
Chill Bhai how do you take these values of n,why can't it be say 4k+2 or 4k+3, i know we will miss 1 here,but how do u construct these linear eq's depending on the condition.Hit and trial or is there a method?
Help me out with this....
What will be the last three digits of he number N, where
N = 10! + 11! + 12! + .... + 49! + 50!
a) 500
b) 400
c) 300
d) 200
e) 100
Detailed explaination would be appreciated 😃
Help me out with this....
What will be the last three digits of he number N, where
N = 10! + 11! + 12! + .... + 49! + 50!
a) 500
b) 400
c) 300
d) 200
e) 100
Detailed explaination would be appreciated :)
Is it 200? Detailed explanation if answer is correct.
yaa its correct!!!
hw did u get it????
yaa its correct!!!e
hw did u get it????
The last three digits will be the addition of factorials from 10! to 14! since from the 15th factorial onwards, the last three digits, due to the presence of 3 5s in the factorial is 000. Therefore 10!+11!+12!+13!+14!
10!=2^8 x 3^4 x 5^2 x 7
We need to get the last non zero digit of 10!
The last non zero digit of 10! is the factorial minus the 5s and the corresponding 2s which add multiply to multiples of 10.
Therefore last non zero digit of 10!= 2^6 x 3^4^7= 8
Now the addition of the last non zero digits of 10! to 14! gives the last non zero digit of the addition.
Help me out with this....
What will be the last three digits of he number N, where
N = 10! + 11! + 12! + .... + 49! + 50!
a) 500
b) 400
c) 300
d) 200
e) 100
Detailed explaination would be appreciated :)
As we want the last three digits of N. So, we consider only the last three digits of each of the factorial here:
=> 10! = 3628800. So, we'll take only the last 3 digits = 800
11! = 800*11 = 8800 = 800
12! = 12*800 = 9600 = 600
13! = 13*600 = 7800 = 800
14! = 14*800 = 11200 = 200
15! = 15*200 = 3000 = 000 So, from 15! onwards, the last three digits would be zeroes. => sum of 800+800+600+800+200+000 = 3200. => last three digits = 200.
As we want the last three digits of N. So, we consider only the last three digits of each of the factorial here:
=> 10! = 3628800. So, we'll take only the last 3 digits = 800
11! = 800*11 = 8800 = 800
12! = 12*800 = 9600 = 600
13! = 13*600 = 7800 = 800
14! = 14*800 = 11200 = 200
15! = 15*200 = 3000 = 000 So, from 15! onwards, the last three digits would be zeroes. => sum of 800+800+600+800+200+000 = 3200. => last three digits = 200.
i started preparing for CAT a few days back.. i have CL notes n stuff.
but should i purchase arun sharma book, cz d notes don't have such concepts taught in depth????
would it be helpful???? o jus d notes are enough????
e
The last three digits will be the addition of factorials from 10! to 14! since from the 15th factorial onwards, the last three digits, due to the presence of 3 5s in the factorial is 000. Therefore 10!+11!+12!+13!+14!
10!=2^8 x 3^4 x 5^2 x 7
We need to get the last non zero digit of 10!
The last non zero digit of 10! is the factorial minus the 5s and the corresponding 2s which add multiply to multiples of 10.
Therefore last non zero digit of 10!= 2^6 x 3^4^7= 8
Now the addition of the last non zero digits of 10! to 14! gives the last non zero digit of the addition.
Arun bhai, could you explain how did you express 10! in that form??? sorry bhai poor in quants...... in fact yet to start......
chandrakant.k SaysArun bhai, could you explain how did you express 10! in that form??? sorry bhai poor in quants...... in fact yet to start......
10!=1*2*...*10
4=2^2; 6=2*3; 8=2^3; 9=3^2; 10=2*5
thus there are 8 2's , 4 3's 1 7 and 2 5's
10 is formed by 2*5,so remove both 2's and 5's
hence v r lft wid 6 2's 4 3's and 1 7
thus..lst digit is 2^6*3^4*7
If its 1!+2!+...+1000! then the ten's digit is 0.
shouldn't the tenth place digit be 1..?
From 10! onwards no number contributes to 10's digit.
If its 5!+6!+7!+.....+1000! then the ten's digit is 8.
How did you find it is 8??
shouldn't the tenth place digit be 1..?
How did you find it is 8??
Just consider last two digits:-
01 + 02 + 06 + 24 + 20 + 20 + 40 + 20 + 80 (1! to 9!)
So, last two digits will be 13
In case of 5! on wards, it will be 80 (just add last 5 terms in the above expression)
Just consider last two digits:-
01 + 02 + 06 + 24 + 20 + 20 + 40 + 20 + 80 (1! to 9!)
So, last two digits will be 13
In case of 5! on wards, it will be 80 (just add last 5 terms in the above expression)
yeah. this is exactly what I'm saying.
from 1!to 9! ten's digit will be 1 and not 0 as said in the above thread.
Hello Everyone,
I have a doubt. Is the five-digit number pqr86 a perfect square or not???
also tel me what are the properties that we should keep in our mind while solving such kind of qsns.!
Thank you!
Hello Everyone,
I have a doubt. Is the five-digit number pqr86 a perfect square or not???
also tel me what are the properties that we should keep in our mind while solving such kind of qsns.!
Thank you!
It is not......an eve number is a perfect square only if it is divisible by 4.....
Divisibility check by 4 states that a number is divisible by 4 if its last two digits are divisible by 4.....
Now 86 is not divisible by 4...hence pqr86 is not divisibe by 4.....
This one is from testfunda.. Please explain
The integer N is given by N = 2^M 5^386, where M is an integer. If N is known to have 388 digits, how many values can M take?
OPTIONS
1)1
2)2
3)3
4)4
5)None of these
This one is from testfunda.. Please explain
The integer N is given by N = 2^M 5^386, where M is an integer. If N is known to have 388 digits, how many values can M take?
OPTIONS
1) 1
2) 2
3) 3
4) 4
5) None of these
Option 3???
spectramind07 SaysOption 3???
Yes... explaination????
Thanku