sorry maam i was bit intoxicated that time. this questions needs a little thought. i guess this is one of rarest of rare cases. For the number to be divided by 99, it must be divisible by 11 and 9. 11 ka divisibility to main bhuul gaya, actually kabhi samjha hi nahi class mein carefully. In case of 999, it must be divisible by 9, 111.... in short, mujhse nahi hoga, its a very good question, i'll have to think about it(when i can). 123123123... i guess if written as 123*10(power)100+123*10(power)97+.. but it wont do the trick. 99=100-1,999=1000-1 can be used but how i dont know. Ye question to main hi solve karunga maam, promise raha mohit ka...
I could get the answer for 99 only. i am in the thought process for 999 but i not so sure. for 999, one thought that came to my mind is that when a number is divisible by 1000 and the quotient got is 'x' then the number minus quotient is divisible by 999, but this works only in case of small numbers. when 123123... is divided by 99, the remainder is 33. 123123... is divisible by 11 but not by 9, so i subtracted 11, then again it is not divisible by 9, so if i subtract 22 more, number is divisible by 9 and 11 both. so remainder is 33.bahut dimag lagaya isme nut i know there's some trick involved in it else u wouldn't have given it. abhi mat batana maam. sorry maam i got the answer in such a long time, and after 2 failures, i'll try not to repeat it again.
the ans are 312 n 33...wen 1000 is divided by 999 it gives the remainder 1..similarly if 10^6 divided by 999 gives remainder 1 n so on...so we can write the no as 123(10^297+10^294+10^291+.......+10^0)...so applying remainder theorem we get 123*100...so nw we can get de ans...n in case of 99...wen 100 divided by 99 gives remainder 1 n wen divided by 1000 gives 10..so for even powers of 10 99 gives remainder 1 n for 0dd powers it gives remainder 10..so the no becomes after applying remainder theorem...123(50*10 + 50)..
9 comments:
hi mohit...u r applying divisibility rule fr 9....bt d question is fr 99 nd 999...think toda more..
sorry maam i was bit intoxicated that time. this questions needs a little thought. i guess this is one of rarest of rare cases.
For the number to be divided by 99, it must be divisible by 11 and 9. 11 ka divisibility to main bhuul gaya, actually kabhi samjha hi nahi class mein carefully.
In case of 999, it must be divisible by 9, 111....
in short, mujhse nahi hoga, its a very good question, i'll have to think about it(when i can).
123123123... i guess if written as
123*10(power)100+123*10(power)97+..
but it wont do the trick.
99=100-1,999=1000-1 can be used but how i dont know.
Ye question to main hi solve karunga maam, promise raha mohit ka...
I could get the answer for 99 only. i am in the thought process for 999 but i not so sure. for 999, one thought that came to my mind is that when a number is divisible by 1000 and the quotient got is 'x' then the number minus quotient is divisible by 999, but this works only in case of small numbers.
when 123123... is divided by 99, the remainder is 33.
123123... is divisible by 11 but not by 9, so i subtracted 11, then again it is not divisible by 9, so if i subtract 22 more, number is divisible by 9 and 11 both. so remainder is 33.bahut dimag lagaya isme nut i know there's some trick involved in it else u wouldn't have given it. abhi mat batana maam.
sorry maam i got the answer in such a long time, and after 2 failures, i'll try not to repeat it again.
well mohit ans is correct bt m still nt gettin d way u've solved it....cn u make n anthr try???
the ans are 312 n 33...wen 1000 is divided by 999 it gives the remainder 1..similarly if 10^6 divided by 999 gives remainder 1 n so on...so we can write the no as 123(10^297+10^294+10^291+.......+10^0)...so applying remainder theorem we get 123*100...so nw we can get de ans...n in case of 99...wen 100 divided by 99 gives remainder 1 n wen divided by 1000 gives 10..so for even powers of 10 99 gives remainder 1 n for 0dd powers it gives remainder 10..so the no becomes after applying remainder theorem...123(50*10 + 50)..
sorry 4 being late...hmmm u r correct...gd explantion.....wel do me a favor ....write d solution 4 2nd one on d main pag..
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