A 4-digit zeroless number N=ABCD has a certain specific feature:
The last 3 digits of a product ABCD*DCBA are zeros.
List all such numbers, explaining your reasoning.
A number ending with 3 zeros is a multiple of 1000=2^3*5^3. Since there are no zeros, nether number contains both 2 and 5 as factors. Therefore one is a multiple of 5^3=125, lets let it be ABCD
So ABCD has one of the four forms:
A125
A375
A625
A875
Its reversal must be a multiple of 8
5210 = 2mod8 so the numbers could be 5216 and reversal 6125
5730 = 2mod8 so the numbers could be 5736 and reversal 6375
5260 = 4mod8 so the numbers could be 5264 and reversal 4625
5780 = 4mod8 so the numbers could be 5784 and reversal 4875
These are the eight possible values of N.
|
|
Posted by Jer
on 2020-06-04 12:56:02 |
Analytic solution
