Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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?

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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
what happened??

Not the answer you are looking for?

Search for more explanations.

Ask your own question