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umm ... must be less than 2500
those r five digit numbers..
I don't think this is possible, unless you count 00000 as a 5 digit number.
why do u think so?
i know the answer..21978 and 87912..but i am looking for the process
hm I stand corrected. haha
(Ax10000+Bx1000+Cx100+Dx10+E) × 4 = Ex10000+Dx1000+Cx100+Bx10+A
well, not so tough problem..
if a number is divisible by 4, its last 2 digits must be divisible by 4 too.. so, after multiplication of 4, clearly, the last 2 digits have to b divisible by 4 the number must be less than 25000, so, after multiplication of 4, the last 2 digits can be 12 or 32 but, in case of 32, after multiplication by 4, the product will start with 9, which opposes the condition.. so, first 2 digits are 21 so, after multiplication with 4, first digit of the product will be 8, so, last digit of the original number is 8 so, we get the condition: (21008+100x+10y)*4=80012+1000x+100y simplifying: y= (67+5x)/16 as x and y, both r integers with domain from 0 to 9, the condition will only be satisfied in case of x=9, y=7 so, the number is 21978
this is the longest reply i have typed in openstudy^^ :D
nice answer !!
thanx, @experimentX, at least u have honored my labor to type it :D
oh ...! i had bookmarked this problem for a while .. thanks to you!
it's solved now :D
yup.. i solved similar prob before.. 4 digit, 5 digit is a little more complicated..
@gingerkid101, this is possible except 00000 :D