q + some digit on the end = 13q

D

DAXE

New member
Joined
Sep 9, 2019
Messages
5
Some positive integer q was appended with a digit at the end, thus obtaining a number 13 times greater than the number q.
Calculate all numbers q having this property.
Please help me in answer, calculations and proof it.
 
When you append a digit to a number, you are effectively moving each digit one place to the left (multiplying it by 10), then adding the new digit. So if the original number is q and you append the digit d, the new number is 10q + d.

Use that to write an equation, and solve. Please show us any work you do and where you are stuck, so we can talk about what to do next.
 
That sure is a nice problem you've got there. But what are your thoughts? What have you tried? Please re-read the Read Before Posting thread that's stickied at the top of each sub-forum and comply with the rules found therein. In particular, please share with us any and all work you've done on this problem, even the parts you know for sure are wrong. Thank you.

Hint by means of two examples:
  • 1369 = 1360 + 9 = 10(136) + 9
  • 253 = 250 + 3 = 10(25) + 3
 
I think these are only three right solutions:
1 & 3 -> 13/1=13
2 & 6 -> 26/2=13
3 & 9 -> 39/3=13
next larger numbers do not meet this rule.
 
Correct. The equation is 10q + d = 13q, so that d = 3q. Since d must be a single digit, q can't be anything more than 3.
 
You must log in or register to reply here.
Top