Question

Find an integer n whose first digit is three, such that 3n/2 is obtained by removing the 3 at the beginning and putting it at the end.

Answer 1

\[n=3a,3/2n=a3\]

Answer 2

@TuringTest @dpaInc

Answer 3

The number is 3529411764705882

I started by looking for the last digit of n.
n = 3....k and 3n/2 = .....3
3*k/2 = 3 means k has to be 2.

n = 3 ... 2 and 3n/2 = .... 23

Step 1

.....?
x 3
-----
3
? = 1

Step 2
1
__________
2 | 3 ........ 2
|
v
_____
0 2
2
----
0
Step 3:
... ?1
x 3
-----
23

? = 4 carry 1

Step 4:
41
__________
2 | 3 ........?2 2 goes into ? 4 times with no remainder
| |
v |
08 |
8 v
-----
0 2
2
----
0
? = 8
so n = 3 ... 82
and 3n/2 = ...823

Step 5:
1
... ?41
x 3
------
823
? = 9 since 3 * 9 + 1 = 28 (carry 2)

and continue...
No joking, this took a while and I started to think I wasn't getting anywhere, but it did finally work out.

Answer 4

medal medal medal