# A Numbers Problem

davegage
Posers and Puzzles 12 Apr '05 03:15
1. 12 Apr '05 03:15
This problem is intended to be solved without the use of a calculator (ie, it's only a good challenge if you don't use a calculator):

Find a 2-digit number and a 3-digit number such that when you place the 2-digit number to the left of the 3-digit number, the 5-digit number that is created is exactly 9 times the product of the original 2-digit and 3-digit numbers.

As far as I can tell, the answer is unique.
2. 12 Apr '05 05:11
Let m = 10a + b
Let n = 100z + 10y + z

Then 9mn = 1000m + n

or

m = n / (9n - 1000) >= 10

So n = 112, m = 14 is the only solution
and the number is 14112
3. 12 Apr '05 06:54
Originally posted by THUDandBLUNDER
Let m = 10a + b
Let n = 100z + 10y + z

Then 9mn = 1000m + n

or

m = n / (9n - 1000) >= 10

So n = 112, m = 14 is the only solution
and the number is [b]14112

[/b]
yes...14112 is correct.

Out of curiosity: how did you find the answer (m,n) = (14,112) and also determine it is unique? did you substitute in and try to reason out the values of a,b,x,y,z (one of your z's should be an x or something) directly?

4. 12 Apr '05 09:042 edits
Originally posted by davegage
yes...14112 is correct.

Out of curiosity: how did you find the answer (m,n) = (14,112) and also determine it is unique? did you substitute in and try to reason out the values of a,b,x,y,z (one of your z's should be an x or something) directly?

m = n / (9n - 1000) >= 10
gives
89n =< 10000
n =< 112

and

9n - 1000 > 0
gives
n > 111
5. 12 Apr '05 16:13
Originally posted by THUDandBLUNDER
m = n / (9n - 1000) >= 10
gives
89n =< 10000
n =< 112

and

9n - 1000 > 0
gives
n > 111
nice...