sum of all the numbers that can be formed using digits 1,2,3,4 without repetition

i did not get the solution given in the topic test. so please give an alternate approach if possible

Hi Aditya,When 1 is at thousands position, the other 3 numbers can be arranged in 3! ways so there will be 6 such numbers. For each number value of 1 will be 1000. Similarly when other numbers occupy thousands place, there will be 6 numbers and each time value will be 1000.Similarly when any occupies 100's position, its value will be 100 and son on.So required answer is 1*1000*6 + 2*1000*6 + 3*1000*6 +4*1000*6 + 1*100*6 +2*100*6 ....... I hope you get the patternAfter taking the terms common it reduces to (1+2+3+4)*1111*6 = 10*6*1111In general if a,b,c,d...n are the numbers then general formula for sum is (a+b+c+d...+n)*1111...n times *(n-1)!.Hope that clears your doubt. If not please revert back with your queryThanks.

is there any shortcut formula to find remainders ?

There is shortcut as such but there are some thoerums which will help you solve the questions faster.These are- Wilson's Theorum, Fermet's Theorum, Chinese remainder theorum etc. We have videos on every type of remainder question and all the theorums on remainders. You can subscribe to that if you want or you can also google the above mentioned theorums.Thanks

2n+1C12n+1C1 + 2n+1C22n+1C2 + ... 2n+1Cn−12n+1Cn−1 + 2n+1Cn2n+1Cn = 63 ......... How to solve this equation?

Sum is 63 so N should not be above 4. There are just 3 values to be checked. i.e 2,3 and 4. On checking we can see that n=3 satisfies this equation.Now if you want to know that why I said it won't be above 4, because if you put n=4, you will get 9c1, 9c2, 9c3 and 9c4. 9c4 it self is greater than 63. So N has to be below 4.Thanks

there is no formula? we jus use hit n trial?

N is a natural number. 2N is having 28 factors, 3N is having 30 factors. How many factors 6N will have. options 36, 35, 32, 40, non of these.

Since 2N and 3N have different number of factors, so number N must have 2's in it. On factorising of 28 (14*2, 7*4, 28*1) we can observe that N must be 2^5*3^3.So 6N would be 2^6*3^4 . Hence the number of factors will be 7*5 = 35. Option B

N is a natural number. 2N is having 28 factors, 3N is having 30 factors. How many factors 6N will have. options 36, 35, 32, 40, non of these.

Since 2N and 3N have different number of factors, so number N must have 2's in it. On factorising of 28 (14*2, 7*4, 28*1) we can observe that N must be 2^5*3^3.So 6N would be 2^6*3^4 . Hence the number of factors will be 7*5 = 35. Option B

How do I solve para jumbled

Hi Rahini,We have a blog post where we have covered this topic in detail. You can find it here https://cracku.in/blog/solve-parajumbles-cat/. It will surely help you, please go through it.Thanks

How do I solve para jumbled

Hi Rahini,We have a blog post where we have covered this topic in detail. You can find it here https://cracku.in/blog/solve-parajumbles-cat/. It will surely help you, please go through it.Thanks

How do I solve para jumbled

Hi Rahini,We have a blog post where we have covered this topic in detail. You can find it here https://cracku.in/blog/solve-parajumbles-cat/. It will surely help you, please go through it.Thanks

when the cat 2016 notification will be out?....any idea?....

Hi KuppaCAT 16 notification will by out in the last week of July or first week of August.

akhilesh singh sir as you have put the bracket in the equation their was no bracket in the question in my test paper but the ans is even only thank you ((((713+6)12+1)11+4)7)((((713+6)12+1)11+4)7)

Hi Sam,I put the brackets to remove any sort of confusion.Happy to have helped you.Feel free to post your doubts here, we will try to revert at the earliest.Thanks