Given two non-negative integers $m>n$, let's say that $m$ ends in $n$ if we can get $n$ by erasing some digits (from left to right) in the decimal representation of $m$. For example, 329 ends in 29, and also in 9. Determine how many three-digit numbers end in the product of their digits.
Problem
Source: Central American Olympiad 2007, Problem 5
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dyaka
12.06.2007 08:57
I recommend checking my casework.
Consider the one-digit ends: For a*b*c=c, a=b=1, so the possible numbers abc are 110, 111, 112, ..., 119; as well as 100.
Consider the two-digit ends: For a*b*c=bc, we have a=(10b+c)/(b*c)=10/c+1/b=int. 1/b is never an integer so 10/c must not be; c cannot be 1, 2 or 5. Examining the other cases:
c=3 - 1/b= 2/3,5/3,... (b not int)
c=4 - 1/b= 2/4,6/4,... (b=2 in 1/2) 324 works
c=6 - 1/b= 2/6,8/6,... (b=3 in 1/3) 236 works
c=7 - 1/b= 4/7,13/7,... (b not int)
c=8 - 1/b= 6/8, 14/8, ... (b not int)
c=9 - 1/b= 8/9, 17/9, ... (b not int)
The only possible numbers are 100, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 236, 324 - 13 in total.
dengmi
12.06.2007 23:45
You missed something.
Actually, there's 101.
You observed that 111-119 are possible, but you ignored most of the ones ending in 0.
There's 9 third digit choices.
There's 10 second digit choices.
There's only 1 third digit choice since that's 0.
Thus, there's 99 one-digit numbers' products that are the ends of the three-digit numbers.
You found 2 two-digit numbers where the products are the two-digit end numbers.
Thus, we have 101.
eLeM
08.08.2007 13:28
Jutaro wrote: Given two non-negative integers $ m>n$, let's say that $ m$ ends in $ n$ if we can get $ n$ by erasing some digits (from left to right) in the decimal representation of $ m$. For example, 329 ends in 29, and also in 9. Determine how many three-digit numbers end in the product of their digits. Let $ a,b,c$ be digits of a number. If $ n$ is one-digit number then we have: $ abc=c$ or $ c(ab-1)=0$. If $ c=0$ then we 90 numbers that satisfy conditions. If $ ab=1$ then where are 9 such numbers. If $ n$ is two-digit number then we have: $ c(ab-1)=10b$ or $ k(ab-1)=10$ where $ c=kb$. After checking three cases $ k=1, k=2$ and $ k=5$ we get 4 solutions: $ 236, 324, 612$ and $ 315$. All in all, there are 103 such numbers.
michito
05.11.2007 04:48
Hello. I'm new here and I'm from Chile. This forum will give me the possibility to practise English and to learn MATH. First it's very easy to notice that the product of the digits of all 3-digits numbers that ends in 0 is cero. (0) The number abc can ends in bc or c. That's why that now we have 2 possibilities: (1) c=a*b*c (c>0) --> ab=1 --> a=b=1 (because they must be positive integers) (2) bc=a*b*c --> 10*b + c=a*b*c --> 10*b = c*(a*b -1) so now he have that c can be 1,2,5,10,b,2*b,5*b,10*b . But c<=9 so the possibilities now are 1,2,5,b,2b,5b. c=1 --> ab-1=10b --> b(a-10)=1 Impossible (a<=9) c=2 --> ab-1=5b --> b(a-5)=1 --> a=6, b=1 c=5 --> ab-1=2b --> b(a-2)=1 --> a=3, b=1 c=b --> ab-1=10 --> ab=11 (impossible, they are integer and <=9) c=2b --> ab-1=5 --> ab=6 --> a=2,3,6 and b=3,2,1 --> c=6,4,2 (in that order) c=5b --> ab-1=2 --> ab=3 --> a=3, b=1, c=5 So the numbers are: (0) :100,110,120...200,210,...300,......990 (1): 111, 112, 113,....,119 (2) 236, 315, 324, 612. Finally from (0) we get 90 numbers, from (1) 9 numbers and from (2) 4. So there are 90+9+4=103 numbers.