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?
Stacey Warren - Expert brainly.com
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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.
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:
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..