Make Zero Without Zero!

Number Theory Level 1

Magician Mike Michael claims to know two positive whole numbers that multiply to 1000, neither of which contain the digit 0.

What is the sum of these 2 numbers?

1001 133 110 254

9 solutions

Discussions for this problem are now closed

Shourya Pandey
May 10, 2014

Suppose $xy=1000$ , where $x$ and $y$ are positive integers. This means that both these numbers are factors of $1000= 2^3 * 5^3$ . So both of them will have only $2$ and $5$ as its prime factors. If one of the numbers has $2$ as its prime factor, them it should not have $5$ as its prime factor, or else it would become divisible by $10$ and thus end in a zero. A similar argument shows that if $5$ is a factor of one of $x$ or $y$ , then $2$ shouldn't be its factor.

What this means is that one of these two numbers is completely made of $2$ s and the other is completely made of $5$ s. It follows that the numbers are none other than $2^3 = 8$ and $5^3 = 125$ .

So the two numbers are $8 , 125$ and the sum is $8+ 125 = 133$ .

Imran Raja
May 7, 2014

Prime factorization of 1000 is $2^3 \cdot 5^3$ . This is equivalent to $8 \cdot 125$ , so the answer is 125+8=133.

Can you quickly explain why that is the only possible solution?

Calvin Lin Staff - 7 years, 1 month ago

@Calvin Lin Here is a quick proof. Note that $1000=2^3\times5^3.$ Assume that the pair of numbers we are looking for $(x,y)$ has $x$ and $y$ of the form $2^m5^n$ for positive $m$ and $n.$ But this will end in a $0$ because any even multiplied by a multiple of $5$ is a multiple of $10,$ meaning it ends in a $0.$ So $x$ and $y$ are of the form $2^m$ and $5^n$ (assume WLOG those are respective). To have $x$ and $y$ multiply to $1000,$ $m$ and $n$ must both be $3.$ So the $(x,y)$ pairing we are looking for is $(2^3,5^3)=(8,125),$ and $x+y=\boxed{133}.$