adiG Sayswhat is the correct answer...
answer is 1!
answer is 1!
well that wasnt really a long time to find the series of remainders...it would take around 1 min...
then to find 11 cube and figuring it out , another 1 min...so not a really time taking thing...
hope it helps
please ignore
Good One----->
Let f be a factor of 120, then the number of positive integral solutions of xyz = f is
(a) 160 (b) 240 (c) 320 (d) 480 (e) none of the foregoing
All the best
Let f be a factor of 120, then the number of positive integral solutions of xyz = f is
(a) 160 (b) 240 (c) 320 (d) 480 (e) none of the foregoing
All the best
jain_ashu SaysCan smbdy pls try this???
N is a number such that the ratio of sum of its digit to product of its digits is 3: 40. If N is divisible by 37 and N is the smallest such number, how many factors does N have?
a. 8
b. 16
c. 20
d. 30
Very Nice Question...
Is the Answer 30...Getting It For N = 5328
5328 = 37*144 = 37* 2^4 * 3^2
Number of factors = 2*5*3 = 30....
Very Nice Question...
Is the Answer 30...Getting It For N = 5328
5328 = 37*144 = 37* 2^4 * 3^2
Number of factors = 2*5*3 = 30....
Don't have OA but it seems correct. Thanx
how many zeros will occur at the end of 1^1*2^2*3^3*................. 77^77...........
i know its easy but try
how many zeros will occur at the end of 1^1*2^2*3^3*................. 77^77...........
i know its easy but try
750 is the answer
how many zeros will occur at the end of 1^1*2^2*3^3*................. 77^77...........
i know its easy but try
10^10 have 10..zeros
20^20 have 20
so.....upto 70^70..have 70.
total 280.
new counting 5 in other terms.
5^5 have 5 fives
15^15 have 15.
25... have 50
35 have 35
45 have 45
.
.
.
75 have 150
total 470.
and number of 2 are more than 470 so total number of zero are
280+470 = 750
how many zeros will occur at the end of 1^1*2^2*3^3*................. 77^77...........
i know its easy but try
Lo main aagayaaaa....!!Yaar simple si baat hai...
10 banta hai like this
10=2*5
Finding higest powers of 5 we can find the no of zeroea an expression will have...
so highest power 5 is= 5^((5+10+15+20+25+75)+(25+50+75))
threfore total zeroes will be =590
Yaar simple si baat hai...
10 banta hai like this
10=2*5
Finding higest powers of 5 we can find the no of zeroea an expression will have...
so highest power 5 is= 5^((5+10+15+20+25+75)+(25+50+75))
Soryy puyss...
I am correcting myself here...
highest power 5 in above expression= 5^(5+10+15+20++30+35+40+45++55+60+65+70+)
Power in square brackets r due to 25^25,50^50,75^75
Ans=750
Good One----->
Let f be a factor of 120, then the number of positive integral solutions of xyz = f is
(a) 160 (b) 240 (c) 320 (d) 480 (e) none of the foregoing
All the best
This Is not the Method To calculate...don't go by this ever in CAT....It is preferrable to leave such kinda question.....Though a very bad method....Searching For better One.....
120 = 2^3 * 3 * 5
Total factors = 16...
Factors = 1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120
Case 1: (x,y,z) = (1,1,1) = 1 way
Case 2: (x,y,z) = (1,1,2) = 3 ways
Case 3: (x,y,z) = (1,1,3) = 3 ways
Similarly, Number of ways..rest Of the numbers can be written as
Case 4: (x,y,z) = 6
Case 5: (x,y,z) = 3 ways
Case 6: (x,y,z) = 9
Case 8: (x,y,z) = 10
Case 10: (x,y,z) = 9
Case 12: (x,y,z) = 18
Case 15: (x,y,z) = 9
Case 20: (x,y,z) = 18
Case 24: (x,y,z) = 30
Case 30: (x,y,z) = 21
Case 40: (x,y,z) = 30
Case 60: (x,y,z) = 60
Case 120: (x,y,z) = 90
Total Ways: 320..Approach was Lengthy..But Final Answer is 320....miscalculated it last time...counted 6 more ways...Finally got 320.....
...................skip
viveknitw Says300! is divisible by (24!)^n. what is the max. possible integral value of n?
divide by 23...which is the least possible prime number in 24!....n u 'll get answer!!!
viveknitw Says300! is divisible by (24!)^n. what is the max. possible integral value of n?
one method to solve this kind of questions is to find the factors of denomiantor (here 24). and then take the greatest number....in this case it will be 3.
hey guys just got across this questions !!!!!
please help me out !
find the simplified value of the expression
(7+1)(7^2+1)(7^4+1)(7^8+1).......(7^1024+1)
(1) (7^2096-1)/6
(2) (7^2048-1)/6
(3) (7^4096-1)/6
(4)7^2049
(5) None of these
hey guys just got across this questions !!!!!
please help me out !
find the simplified value of the expression
(7+1)(7^2+1)(7^4+1)(7^8+1).......(7^1024+1)
(1) (7^2096-1)/6
(2) (7^2048-1)/6
(3) (7^4096-1)/6
(4)7^2049
(5) None of these
Its. b. (7^2048-1)/6
(7+1)(7^2+1)(7^4+1)(7^8+1).......(7^1024+1)...Multiply Divide this equation by (7-1).....You will get terms of a^2 - b^2....
let's take two cases:-
xyz = 120 ==>
x = 120/yz
factors of 120 = 16
solutions will be = n(n+1)/2 = 136
xyz = 60 ==>
x = 60/yz
factors of 60 = 12
solution = 78
likewise i did of all, getting total = 420
correct me if smthing is wrong.
xyz = f
xyzk = 120 = 2^3*3*5
number of solutions are number of ways of distributing 2,3,5
3,5 can be distributed in 4*4 ways
2^3 can be distributed in (4c1 + 4c2*2 + 4c3)
Total solutions are thus 16*20 = 320
Its. b. (7^2048-1)/6
(7+1)(7^2+1)(7^4+1)(7^8+1).......(7^1024+1)...Multiply Divide this equation by (7-1).....You will get terms of a^2 - b^2....
thanks a lot dude !!!!
Himanshu1989 Saysthanks a lot dude !!!!
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