Question

how many same numbers may there exist by 4 digits multiplied by a 5th different number will result the first number inversed order
abcd*e=dcba
a,b,c,d and e are natural numbers

Answer 1

i have got one

2178*4
--------
8712

there may exist ?

Answer 2

@Hero
@Preetha
@agent0smith

Answer 3

@ganeshie8
@mathmath333
@mathmale
@Directrix

Answer 4

@TheSmartOne

Answer 5

Does this count:

0009*1000=9000

Answer 6

every digits need being different from zero till 9
- this is why i noted there abcd*e - different letters for different digits

Answer 7

@TheSmartOne ty

Answer 8

nice to meet you dear my friend

Answer 9

@Directrix

Answer 10

*

Answer 11

Let the number (in general) be pqrs =\(1000p+100q+10r+s\)
the number be multiplied by an integer k, then we want
\(k(1000p+100q+10r+s)=1000s+100r+10q+p\)
Solve for k
\(\large k=\frac{1000s+100r+10q+p}{1000p+100q+10r+s}\)
Set p=2, q=1, r=7, s=8 and evaluate k.
Hint: k is the square of the first prime number.

Answer 12

Oh, you want other ones, give me a sec!

Answer 13

@jhonyy9

Here's what I've got for all 4 digit numbers (i.e. first digit is 1-9):

k=1 for all palindromes, example 1*1001=1001, 1*2442=2442, etc.

If k≠1, which is probably what you want, you have hit half of the lot:
{2178, 1089} are the non-palindrome solutions ( k≠1 )

Someone brilliant can perhaps give the mathematical solution.

If all digits including k must be different, your solution is the only one.

Answer 14

ty @mathmale yes sure how i wrote above all digits must be different so than this is sure that my solution is the only one ? ty.

Answer 15

Exactly. You've obtained the one and only one solution where all five digits are different...by exhaustive search. :)

Answer 16

@BOOKISHFAIRY73

Answer 17

@bookishfairy73