=> 11111111.......1001 times/143 i hope this is clear now to make a trend consider single digit no '1' and divide by 143, 1/143, further take 1, two times, then 3 times and further, u find a similar trend after six times so get repetition up to 995 times & we left with 5 so reminder will be 100 => 1/143 => rem=1 => 11/143=> rem=11 111/143=> r=111 1111/143=>r=110 11111/143=> r= 100 111111/143=>r=0
thus for every 6th r= 0 1001/6=> r=5 thus Rem=100, now multiply common divisor '7' now 100*7=700
The best I can thk of solving this problem is by square root of any number method given in T.I.M.E material. If u have time material u can look into it.As its quite difficult to post it here.
N is 111111111(nine 1's) This is a typical pattern: 11^2 = 121 111^2=12321 and so on
EagleMenace Says
The best I can thk of solving this problem is by square root of any number method given in T.I.M.E material. If u have time material u can look into it.As its quite difficult to post it here.
Hi
As pointed out by Individualist, this is a typical pattern like square(11) = 121 square(111) =12321 so square(111111111) = 12345678987654321
Q: Find the 4th digit from right in the expression 4002^11? (please post the soln. also)
4002^11 = (4000 + 2)^11
We need to consider only last two terms in the binomial expansion of (4000 + 2)^11, as rest of the terms have exponent of 10 more than 6 and hence they won't contibute to the last four digits. Last two terms are:
11*(2^10)*4000 + 2^11 = 1024*44000 + 2048 => last 4 digits will be 8048
2. In a particular country .all the numbers are represented by alphabets a,b,c. 15 is written as abc. 6 is written as bc. 60 is writtenas bcbc.
what wud be 17?
3. N is a 1001 digit number consisting of 1001 sevens. What is the remainder when N is divided by 1001?
4 .Suppose N is an integer such tha the sum of the digits of N is 2, and 10^9
hw did u get 776???
2. we convert to base 3 and get the answer as abb
we know (15)in base10 = (1111) in base 2
so when base decreases...no of digits in the number increases..here 15 became 1111
now we take 15 and abc into consideration (15)base 10 = (abc )base X no of digits has increased ,hence base X is smaller than 10(must hv decreased)
we take (1111) base 2 and (abc)base x into consideration no of digits decreased ,hence base x is greates than 2(must hv increased)
hence X lies in between 2 and 10 ..hit and trial we get 3... it fits... convert 17 to base 3 ,we get abb in these type of questions i think the main problem is to identify..the chapter..from which the questions are being asked..here it was base conversion
I have an interesting question , but it is difficult to categorize under a specific thread. Since its more like a puzzle involving numbers, i am posting it in this thread:
There is a six digit number and if the last digit of this number is moved in front of the number, the resulting number is 5 times the originAL number. Find the original number.
I have an interesting question , but it is difficult to categorize under a specific thread. Since its more like a puzzle involving numbers, i am posting it in this thread:
There is a six digit number and if the last digit of this number is moved in front of the number, the resulting number is 5 times the originAL number. Find the original number.
Hi, The no is 142857.
let the no be abcdef. 5* abcdef = fabcde if a>2 then fabcde will be 7 didgit. Hence a=1. so, 5 * 1bcdef= f1bcde Now, 5( 100000+ 10000*b+1000*c+100*d+10*e+f) = 100000*f+10000+1000*b +100*c+10*d+e Simplifying, 49(10000+1000*b+100*c+10*d+e) = 99995*f or,7(10000+1000*b+100*c+10*d+e) =14285*f Since,14285 is not divisible by 7 and RHS has to be a factor of 7=> f should be 7. so, (10000+1000*b+100*c+10*d+e) = 14285 or, 1bcde= 14285, hence, b=4,c=2,d=8,e=5.
If the lcm of (10!)(18!) and (12!)(17!) is expressed as a!b!/c! where a, b are two digit positive numbers and c is a one digit positive number. Then abc equals ? a) 54 b) 648 c) 216 d) 324 e) none of these
If the lcm of (10!)(18!) and (12!)(17!) is expressed as a!b!/c! where a, b are two digit positive numbers and c is a one digit positive number. Then abc equals ? a) 54 b) 648 c) 216 d) 324 e) none of these
If the lcm of (10!)(18!) and (12!)(17!) is expressed as a!b!/c! where a, b are two digit positive numbers and c is a one digit positive number. Then abc equals ? a) 54 b) 648 c) 216 d) 324 e) none of these
* 12!* 3 = (18!*12!)/ 3!* 12!* 3 = (18!*12!)/ 3!