how to check? Why did you accepted 13/2=6.5 and not 23/3 = 7.666 ? plz edify !
see actually i m testing for authenticity A/B here both shud be a less than the base system. for eg. in base system 10 all digits are frm 0 to 9 hence A/B = 2/3 or 4/5 is possible but 11/2 is not ...same thing
Now to let n divide 500,n has to be its factors Therefore 500=2^2 * 5^3 Hence no. of factors = 3*4 =12
But acc. to question,n cannot be 1 or 2.It cannot be 250 or 500 So we are left wid 4,5,10,20,25,50,100,125 Check by inserting it in the above equation. For n=4, a=125-3=122 (Not possible as number has to be odd)
Checking all,I am getting only one set for n=10,a=41(Not taking into account negative integers,Else Two) Wats the OA?
2(15!+25!+50!) is the answer. Can any of u please explain the approach?
wen we observe the base 10! is the smallest number there....so just take 10! common out of the bracket with the power which is raised to the whole bracket.......
now in the bracket we r left with (1+.....) now in the bracket before adding 1 we will have many zeros in the final term but wen we will add 1 in the end it will spoil all the zeros by adding to the last zero of the number.... so we r left wid the 10! and the same power........ 10! will have 2 zeros in the end and theses 2 zeros will get multipied by the power to give the answer....... i hope it is clear
What is the remainder when (1! + 2! + 3! + + 100!)^100 is divided by 10?
ans is 1.. after 5!..we'l start having zeros..so it will be like (1+2+6+24+60+-----X00000000000..24 zeros)^100 after 4,every term will be divisible by 10..as they have zeros..
so add it 1! till 4!=so its 33^100= 3^100=9^50/10 =(10-1)^50/10 =1 remainder
Guys any idea, how to solve problem like this in an efficient way...I am able to solve this by following some pattern what is the last non zero digit of 1786!
means like in no. 183400, 4 is the last non-zero digit.
!*81*82*83*84*85*86