e(11)=10 ? How? I thought, as per pattern ,it repeats after every 9, so e(11)=9 ,as I thought
e(11)=10
2^5/11=
32/11=10
..ans=10
e(11)=10 ? How? I thought, as per pattern ,it repeats after every 9, so e(11)=9 ,as I thought ?
joyz SaysCan somebody please explain how the last non zero digit no. of 25! in the above method is calculated. How have we arrived at 2^5*R(5)*R(0) and how is it coming out to be 2*2 ?
R(n!) i.e. right-most non-zero digit of n! is found out like this:
Step 1: Express n as 5a+b
Say, n = 37 = 35 + 2 = 5*7 + 2 => (a, b) = (7, 2)
Step 2:
R(n!) = R(2^a) * R(a!) * R(b!)
So, for 37 => R(2^7) * R(7!) * R(2!)
Note: We can repeat process for R(a!) if a is on the higher side
Here, R(37!) = R(128 ) * R(5040) * R(2)
= 8*4*2 = 64 => 4
Flynn Sayse(11)=10 ? How? I thought, as per pattern ,it repeats after every 9, so e(11)=9 ,as I thought
Euler's number for any number is equal to number of co-primes it has which are less that it.
And if n = (p1)^a * (p2)^b * ..... where p1, p2 are prime numbers
then E(n) = n * * * ...
A shortcut for prime numbers: E(p) = p-1
So, E(11) = 11-1 = 10
E(61) = 60, E(101) = 100, E(5) = 4, E(7) = 6 etc
FIND THE MAX VALUE SUCH OF n SUCH THAT 77! IS DIVISIBLE BY 720^n ....
niveditasharma SaysFIND THE MAX VALUE SUCH OF n SUCH THAT 77! IS DIVISIBLE BY 720^n ....
16*45
77!
power of 2 -- 38+19+9+4+2+1 -- 73/4 -- 18
power of 3 -- 25+8+2 -- 35 -- 17
power of 5 -- 15+3 -- 18
17 is the limiting value
n = 17
niveditasharma SaysFIND THE MAX VALUE SUCH OF n SUCH THAT 77! IS DIVISIBLE BY 720^n ....
720=9*8*10
Answer is either the number of 5's or no of 9 whichever is minimum( we can neglect 8 as power of two three times for 77! will be high)
77/5+77/25 = 18 is the no of 5s ------eq1
77/3+77/9+77/27=35 is the number of 3s
so 35/2=17 is the number of 9s----------eq2
if we compare eq 1 & 2, 17 is less ,
so ans is 17
Find two natural numbers whose sum is 85 and whose lcm is 102??
Got the answer
Factors of 102 are 2, 3, 17.
85 = 5*17 = (2+3)*17
The two numbers are
2*17 = 34
3*17 = 51
cataspirnt SaysFind two natural numbers whose sum is 85 and whose lcm is 102??
102 = 2*3*17
17X + 6Y = 85 ( no solution coz 17 is a factor of 85 and 6 is not a factor 17)
34X + 3Y = 85 (no solution)
51X + 2Y = 85 (X = 1 and Y = 17)
so , numbers are 51 and 34
1. which of the following is not a perfect square ?
a) 1,00,856
b) 3,25,137
c) 9,45,729
d) all of these
e) none of these
2. If 2
3.A test has 80 questions. There is one mark for a correct answer, while there is a negitive penalty of - 1/2 for a wrong answer and - 1/4 for an unattempted question. what is the number of questions answered correctly if the student has scored a total of 34.5 marks .
4. In above question 3 , if it is known that he has left 10 questions unanswered, the number of correct answers are ?
:
pls provide detailed explanation ....
1)what is the highest power of 3 available in the expression 58!-38!?
2) What is the remainder when 2(8!)-21(6!) divides 14(7!)+14(13!)?
a) 1 b)7! c)8! d)9!
3)A set S is formed by including some of the First one thousand naural numbers.S contains the max no. of numbers such that they satisfy the following questions.
1.No number of the set S is prime.
2.When the numbers of the set S are selected two at a time , we always see co prime numbers.
what is the number of elements in the set S?
a)11 b)12 c)13 d)7
1. which of the following is not a perfect square ?
a) 1,00,856
b) 3,25,137
c) 9,45,729
d) all of these
e) none of these
2. If 2
3.A test has 80 questions. There is one mark for a correct answer, while there is a negitive penalty of - 1/2 for a wrong answer and - 1/4 for an unattempted question. what is the number of questions answered correctly if the student has scored a total of 34.5 marks .
4. In above question 3 , if it is known that he has left 10 questions unanswered, the number of correct answers are ?:
1. After trying to find square root, none are perfect squares.
2. 2 7
-1 5
Draw a straight number line and try to find left and right extremes after for x+y and x-y
3. p+q+r =80
p-q/2-r/4=34.5
Eliminate r,
p= 43+ (q+3)/5, so for max p=> q=0, hence p=43.
4. Put r= 10, and solve for p and q, p=48.
hello!
i needed help with a pattern in number system.
the question types are:
1)find the maximum value of n such that 50! is perfectly divisible by 2520^n?
here the largest prime factor(i.e 7) was taken as restricting factor..
2)) find the maximum value of n such that 50! is perfectly divisible by 12600^n?
here instead of the biggest prime no.(i.e 7), 3 was taken as restricting prime factor.
so,
can somebody plz help me with this concept? i know that the longer way would be to calculate it for all the prime factors and see which one will be the smallest...but thus can be really lengthy (like in case of 504^n if we take 2 as the prime factor).. so can sumbody plz tell dat how is the limiting factor considered?
thanks..
what ll be the remaider when 1223334444....100 digits /7 ?
Gopal, Hari and Karthik started a business with investments of Rs 8000, Rs 12000 and Rs 16000 respectively. Hari and Karthik left the business after X months. Out of the ... If X is an integer, find the ratio of Gopal's, Hari's and Karthik's share?
a. 27:21:28
b. 24:21:28
c. 30:27:36
d. 32:30:40
a. 27:21:28
b. 24:21:28
c. 30:27:36
d. 32:30:40
Gopal, Hari and Karthik started a business with investments of Rs 8000, Rs 12000 and Rs 16000 respectively. Hari and Karthik left the business after X months. Out of the ... If X is an integer, find the ratio of Gopal's, Hari's and Karthik's share?
a. 27:21:28
b. 24:21:28
c. 30:27:36
d. 32:30:40 :banghead:
gopal = 12*8000
hari = 7*12000
karthik = 7*16000
ratio = 24:21:28
pls provde detailed solution for these questions
1)How many four-digit numbers can be formed using the digits 1, 2, 4, 6, 7 and 9 such that each number is divisible by 3 but not by 9? (Repetition of digits is not allowed) ?
2)What is the remainder when 1! + 2! + 3! + ... + 100! is divided by 11!?
3)Find the reminders when 10^ 2 + 11^ 2 + 12^ 2 + . . . . 28 ^2 is divided by 19
4)The average of 25 distinct positive integers is 25. What is the greatest possible value of any one of these 25 numbers?
5)What is the remainder when 25 x (1331)2196 is divisible by 13?
:
pls provde detailed solution for these questions
1)How many four-digit numbers can be formed using the digits 1, 2, 4, 6, 7 and 9 such that each number is divisible by 3 but not by 9? (Repetition of digits is not allowed) ?
2)What is the remainder when 1! + 2! + 3! + ... + 100! is divided by 11!?
3)Find the reminders when 10^ 2 + 11^ 2 + 12^ 2 + . . . . 28 ^2 is divided by 19
4)The average of 25 distinct positive integers is 25. What is the greatest possible value of any one of these 25 numbers?
I think for the 4 th question answer is
4)325
explanation:consider numbers from 1 to 24,325
sum of numers from 1 to 24 is 300 so add 325 to dat 300+325=625
average=625/25=25
pls provde detailed solution for these questions
1)How many four-digit numbers can be formed using the digits 1, 2, 4, 6, 7 and 9 such that each number is divisible by 3 but not by 9? (Repetition of digits is not allowed) ?
2)What is the remainder when 1! + 2! + 3! + ... + 100! is divided by 11!?
3)Find the reminders when 10^ 2 + 11^ 2 + 12^ 2 + . . . . 28 ^2 is divided by 19
4)The average of 25 distinct positive integers is 25. What is the greatest possible value of any one of these 25 numbers?
Is the answer for Question 3 : 18??
2) What is the remainder when 2(8!)-21(6!) divides 14(7!)+14(13!)?
a) 1 b)7! c)8! d)9!
My Answer:
The remainder will be 1 !