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TestFunda Answers  Higher Maths
the series will end up (1)^3+(2)^3.... when calcupated we get 2 

TestFunda Answers  Higher Maths


TestFunda Answers  Higher Maths
What is the last non zero digit in 100!? please explain its method also 


TestFunda Answers  Higher Maths


TestFunda Answers  Higher Maths


TestFunda Answers  Higher Maths
Since you only want the last digit only concentrate on the last digits which would be multiplied. This question in eeffective becomes, last digit of (9!)^{10}. if you know the answer of 9! then awesome or else, break it down. You would know, 7! = 5040 So, last digit = 4. Last digit of 8! = 2 (4*8=32) Last digit of 9! = 8 (9*2=18) Hence, last digit of 100! = last digit of 8^{10 }= last digit of (2^{10})^{3} 2^{10 }= 1024 4^{3 }= 64 Last nonzero digit of 100! is 4 

TestFunda Answers  Higher Maths
Since you only want the last digit only concentrate on the last digits which would be multiplied. This question in eeffective becomes, last digit of (9!) ^{10}. if you know the answer of 9! then awesome or else, break it down. You would know, 7! = 5040 So, last digit = 4. Last digit of 8! = 2 (4*8=32) Last digit of 9! = 8 (9*2=18) Hence, last digit of 100! = last digit of 8 ^{10 }= last digit of (2 ^{10}) ^{3} 2 ^{10 }= 1024 4 ^{3 }= 64 Last nonzero digit of 100! is 4 

TestFunda Answers  Higher Maths
what is the remainder when 97! is divided by 101? 


TestFunda Answers  Higher Maths
2x+4x=240; x=40; then two whl=2*40=80 4 whl =4*40=160; ans: 160/4= ie 40 Read more at http://www.m4maths.com/2639Therearesome2wheelersand4wheelersparkedtotalnumberofwheelspresentis240thenhowmany4.html#F39eXTbVYGBrwWuI.99 

TestFunda Answers  Higher Maths
First Read the question carefully, According to the given question all the numbers will be decided by alok and first 2 would be put by bhanu whereas alok will put the 3rd one.So for getting maximum alok will tell 7 as the first Numbertwo cases arise1) if bhanu will put this number as Z then 9, 9 will be the next 2 number. and 9+x+yz = 202) if bhanu will put this number X, then alok will tell 4 or 5 as next number.if bhanu will put 4 or 5 as Z then alok will put 9 as Y. So N will be9+x+yz= 9+7+94 = 21 OR 9+7+95 = 20If bhanu will put 4 or 5 as Y then alok will put 0 as Z9+x+yz = 9+7+40 = 20 OR 9+7+50 = 21So if both are playing with there best effort then 9+x+yz will be 2020 is the correct answer


TestFunda Answers  Higher Maths
A and B walk up an escalator (moving staircase). The escalator moves at a constant speed. A takes 6 steps for every 4 steps of B . A gets to the top of the escalator after having taken 50 steps while B (because of his slower pace lets the escalator do a little more of the work) takes only 40 steps to reach the top. If the escalator were turned off, how many steps would they have to take to walk up?  Options  


TestFunda Answers  Higher Maths
Let speed of escalator be E and total steps be X. Then i) 50 + E. 50/Sa = X where Sa is speed of A. Similarly , ii ) 40 + E.40/Sb = X where Sb is speed of b in steps per second. Also , iii) Sa/Sb = 6/4. Subtrating (ii) from (i) and substituting (iii) , we get X=100 

TestFunda Answers  Higher Maths
Wilson's Theorem is about primes, namely, for integer n>1, (n1)! = 1 mod n iff n is prime. Since 101 is prime, we can use it, along with the fact that 100*99*98*97! = 100! Then taking mod 101 and using Wilson's Theorem, 100*99*98*97! = 1 mod 101 (1)(2)(3)97! = 1 mod 101 so that, multiplying both sides by 1, 6*97! = 1 mod 101 Now, we'd like to find the inverse mod 101 of 6. To do this, use the fact that 6*17 = 102 = 1 mod 101, to multiply both sides by 17, yielding 17*6*97! = 97!mod 101 = 17 mod 101 So it's 17 

TestFunda Answers  Higher Maths
The number will be of the form 44k+13.Put k=0, rem is 13put k=1, rem is 24put k=2, rem=2put k=3, rem=13put k=4,rem=24...so cycle followingtherefore sum of distinct rems=2+24+13=39 

TestFunda Answers  Higher Maths
The approach taken is that the limiting factor in both 35^6 and the 21^12 is 7.On calculating we will see that 42! has highest power of 7 as 6. And 76! has highest power of 7 as 11 but 77 will have highest power of 7 as 12.so we need to discard 77. Therefore total number of value of n possible =7642+1=35
Last edited by vimbloid; 14Nov14 at 11:12 AM.


TestFunda Answers  Higher Maths
100<x<200100/7=rem(2)...ie 98 divivisible by 7...so in 100 to 200 range lowest no divisible by 100=98+7=105200/7=rem(4)...i.e. 196 is the largest within 200 to be divisible by 7...so now do highestlowest/7=196105/7=91/7=13 

TestFunda Answers  Higher Maths
Let 'n' be the total number of participants Therefore, No. of men= n2 & No. of women= 2 No. of chess matches betwee men= 2 x [(n2)C2] = (n2)(n3) {Factor 2 comes from the info that "every participant played 2 games with every other participant"} AND No. of chess matches betwee men and women = [2 x (n2)] + [2 x (n2)] = 4 x (n2) Ac. to question: No. of chess matches betwee men = 66 + No. of chess matches betwee men and women Or (n2)(n3) = 66 + 4 x (n2) Solving for n, we get n=13 

TestFunda Answers  Higher Maths
2520= 7 x 5 x 2x2x2 x 3x3 . Here, 7 is the highest prime. So counting the number of 7's in 50! we get: 50/7 + 50/49 =7 + 1 =8. 

TestFunda Answers  Higher Maths
Out of a pck of 52 card one is lost , from the remaining of the pack two cards are drawn and are found to be spade. find the chances is the spade.  Options  1)  11/50  2)  11/49  3)  10/49  4)  10/50 
 Additional Notes: Answer is giving option a. but i did not get concept, please explaine. 


TestFunda Answers  Higher Maths
Three people Kishore, Rakesh and Ajeet start their journey from Bangalore to Delhi. They leave Bangalore at 7 am, 11 am and 4 pm respectively. In how many hours after Ajeet starts, will they meet if their speeds are 25km/hr, 30km/hr and 40 km/hr respectively? A ) 24 hrs B ) 20 hrs C ) 15 hrs D ) 12 hrs SOLUTION Let they meet after x hours. Distance travelled by them should be the same So 40x = 30(x + 5) 10x = 150 x = 15 hours The correct
Three people Kishore, Rakesh and Ajeet start their journey from Bangalore to Delhi. Theyleave Bangalore at 7 am, 11 am and 4 pm respectively. In how many hours after Ajeet starts, willthey meet if their speeds are 25km/hr, 30km/hr and 40 km/hr respectively?  Options  1)  24hrs  2)  20hrs  3)  15hrs  4)  12hrs 


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