The number $555,555$ can decompose, as the product of two factors of three digits, in how many ways?

Question

The number $555,555$ can decompose, as the product of two factors of three digits, in how many ways?

I've seen the answer to the question, and there is only one way: Since $555, 555 = 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 37$, the only way to combine the factors to achieve expressing it as a product of two three-digit numbers is $(3 \cdot 7 \cdot 37) (5 \cdot 11 \cdot 13)$. Regardless of this, I struggle to understand how the answer was formulated. Can someone show me the procedure?

Sorry if the question is poorly phrased, it is a rough translation of the original problem in Spanish.

Answer 1

It is intelligent brute force. The largest a three digit number can be is $999$ so you need to find a factor of $555,555$ that is between $556$ and $999$. The other will also be in that range so you are done. Next note that $3 \cdot 5 \cdot 7=105$ which is too small by itself and too large multiplied by any of the other factors, so two of $3,5,7$ have to be in one factor and one in the other. $11\cdot 13 \cdot 37 \gt 999$ so again two of those need to be in one factor and one in the other. We are down to $18$ combinations to try, three singletons from $3,5,7$ times all the one or two combinations from $11,13,37$. I missed Will Jagy's point that you need three factors in each set, so that decreases the number to try to $9$.

Answer 2

Nothing wrong with the approach Will gives in the comments. Here's another way. Obviously $555,555=555\times1001$, but $1001=7\times11\times13$ is a little too large. The way to make it a little smaller is to swap the factor 7 with the factor 5 in 555, which gives you your solution, $(3\times7\times37)(5\times11\times13)$.

Answer 3

The factors $7,11,13$ can't be used together, since $(7)(11)(13)=1001$.

So one of the groups, group $1$ say, must have exactly two of the factors $7,11,13$.

Hence the factors $3$ and $5$ can't both be in group $1$, else the product of the factors in group $1$ would be at least $(3)(5)(7)(11) > (7)(11)(13)$.

Similarly, the factor $37$ can't be in group $1$, else the product of the factors in group $1$ would be at least $(37)(7)(11) > (7)(11)(13)$.

Label the other group as group $2$.

Thus, group $2$ contains

• $37$
• Exactly one of $7,11,13$.
• At least one of $3,5$.

But since $(37)(27) = 999$, the factors in group $2$ other than $37$ must have a product which is at most $27$.

It follows that neither of the factors $11$ or $13$ is in group $2$, since $(11)(3) > 27$, and $(13)(3) > 27$.

So the factor $7$ must be in group $2$, and the factors $11,13$ must be in group $1$.

Since the factor $7$ is in group $2$, the factor $5$ can't be in group $2$, since $(7)(5) > 27$.

Hence, the factor $5$ must be in group $1$, and the factor $3$ must be group $2$.

Thus, group $2$ has the factors $37,7,3$, and group $1$ has the factors $11,13,5$.

Answer 4

The prime factor are $ 3 · 5 · 7 · 11 · 13 · 37$ so there are $2^6=64$ factors and $32$ complement pairs. Just list them all but don't bother with those that are less than $555555/999=556$

Toss out, $1,3,5,7,11,13,37,3*5,3*7,3*11,3*13,3*37,3*5*7,3*5*11,3*5*13, 3*5*37,3*7*11,3*7*13$ (that's 18 that are too small).

$3*7*37=777$ and its compliment is $5*11*13=715$. (That's 1 that is acceptible)

We can continuing tossing out $3*11*13$ and $3*11*37$and $3*13*37$ are too high so we toss them. (That's 21 that are unaccptible)

$3*5*7*11$ is too high so there wonvt be any more factors that are multiples of $3$ in range. Hence no other complements which aren't multiples of $3$ will be in range either.

Of the 9 we haven'haven't considered: $3*5*7*13,3*5*7*37,3*5*11*13,3*5*11*37,3*5*13*37,3*7*11*13,3*7*11*37,3*7*13*37,3*11*13*37$ are all too big.

So that was exhaustive.

Answer 5

We can write $$555,555=5\times111,111$$ and notice that $111=37\times3$, so we have $$555,555=5\times37\times3003$$ and since $1001=7\times11\times13$, the prime factorization is $$555,555=3\times5\times7\times11\times13\times37.$$

If we multiply each of the three combinations $11\times13$, $11\times37$ and $13\times37$ by $3$, we see that only $3\times(13\times37)$ exceeds $1000$ which is a four-digit number.

Hence $37$ must pair with two other one-digit numbers, and $11\times13$ must pair with either $3$ or $5$ since $7\times11\times13>1000$.

If $11\times13$ pairs with $3$, then the other product must be $$37\times(5\times7)>1000$$ which is not a three-digit number.

Therefore, the only possible combination for $555,555=P_1\times P_2$ is $$P_1=5\times11\times13,\quad P_2=3\times7\times37.$$

Answer 6

I think the procedure might be based on Fermat method.

We can calculate that the number $555555$ can be expressed as a difference of two squares:

$746^2 - 31^2=556516-961$

Number $31$ is the distance from $746$ both ways which gives $715$ and $777$ accordingly. From here the smallest prime factor of:

$777$ is number $3$ which equals to $3\cdot259$ and for

$715$ is number $5$ which equals to $5\cdot143$

Further factorisation results in final solution where $143=11\cdot13$ and $259=7\cdot37$ therefore:

$(5\cdot11\cdot13)(3\cdot7\cdot37)$ can satisfy equation.

Answer 7

If $555{,}555=ab$ with $a,b\lt1000$, then we must also have $a,b\gt555$ (e.g., if $b\lt1000$, then $a=555{,}555/b\gt555{,}555/1000\gt555$). Now since $555{,}555=3\cdot5\cdot7\cdot11\cdot13\cdot37$, we may assume, without loss of generality, that $a=37k$. From $a\gt555=15\cdot37$, we see that $k\gt15$, and from $a\lt1000$, we see that $k\le\lfloor1000/37\rfloor=27$. The only product $k$ of the primes $3$, $5$, $7$, $11$, and $13$ that falls in the interval $15\lt k\le27$ is $k=3\cdot7=21$. Thus $a=37\cdot21=777$, $b=5\cdot11\cdot13=715$ is the only factorization of $555{,}555$ into two three-digit numbers.