Question

Find two five-digit numbers such that
one is 4 times the other, and the digits
of the first are the reverse of the digits of the second.
In other words, ABCDE × 4 = EDCBA?

Answer 1

umm ... must be less than 2500

Answer 2

those r five digit numbers..

Answer 3

I don't think this is possible, unless you count 00000 as a 5 digit number.

Answer 4

why do u think so?

Answer 5

i know the answer..21978 and 87912..but i am looking for the process

Answer 6

hm I stand corrected. haha

Answer 7

(Ax10000+Bx1000+Cx100+Dx10+E) × 4 = Ex10000+Dx1000+Cx100+Bx10+A

Answer 8

well, not so tough problem..

Answer 9

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

Answer 10

this is the longest reply i have typed in openstudy^^ :D

Answer 11

nice answer !!

Answer 12

thanx, @experimentX, at least u have honored my labor to type it :D

Answer 13

oh ...! i had bookmarked this problem for a while .. thanks to you!

Answer 14

it's solved now :D

Answer 15

yup.. i solved similar prob before.. 4 digit, 5 digit is a little more complicated..

Answer 16

@gingerkid101, this is possible except 00000 :D

Answer 17

what happened??