When two digit numbers in base $5$ are multiplied the result is $4103_5$. What are the numbers in base $5$?

Question

When two digit numbers in base $5$ are multiplied the result is $4103_5$. What are the numbers in base $5$?

Well given by two digit numbers in base $5$ I tried out the multiplication and tried to simplify.

$(ab_5)(cd_5)=4103_5$ or $(5a+b)(5c+d)=4*5^3+1*5^2+0*5^1+3$

Then get that $5^2ac+5ad+5cb+bd=5^3*4+5^2+3$

I notice that $bd=3$ so b=1,3 and $d=1,3$. Now this is where I'm stuck I don't see a clear way to get what the numbers should be without a lot of trial and error.

Answer 1

Notice that $\;4103_5 = x^2 - 4x + 3\;$ where $\;x = 5^2.\;$ Now factor it as $\;(x-3)(x-1).$

Answer 2

Maybe this is cheating, but ... $4103_5 = 528 = 2^4 \cdot 3 \cdot 11$.

The pair of numbers greater than or equal to $5$ and less than $25$ that we can make with these factors is $22=42_5$ and $24=44_5$.

Answer 3

Some divisibility tests \begin{array}{ccl} \text{n base 10} & \text{n base 5} & \text{Divisibility tests} \\ 4_{10} & 4_5 & \text{$4_5|N \iff$ sum of digits is divisible by 4} \\ 3_{10} & 3_5 & \text{$3_5|10_5A+b \iff 3_5|A-b$} \\ 5_{10} & 10_5 & \text{$10_5|N \iff$ N ends in 0} \\ 7_{10} & 12_5 & \text{$12_5|10_5A+b \iff 12_5|A-4b$} \\ 11_{10} & 21_5 & \text{$21_5|10_5A+b \iff 21_5|A-2b$} \\ \end{array}

Since $4+1+0+3$ is clearly a multiple of $4$, then $4_5 \mid 4103_5$.

We find $4103_5 = 1012_5 \times 4_5$

Since $1+0+1+2$ is clearly a multiple of $4$, then $4_5 \mid 1012_5$.

We find $4103_5 = 113_5 \times 4_5 \times 4_5$

Testing $113_5$ for divisibility by $3_5$: \begin{array}{c} 1 & 1 & \require{cancel}\cancel{3} \\ - & 3 \\ - & - & \\ & 3 \end{array}

So $3_5 \mid 113_5$

We find $4103_5 = 21_5 \times 3_5 \times 4_5 \times 4_5$

So $4103_5 = 21_5 \times 3_5 \times 2_5^4$

Examining the factors, it seems that the two two-digit numbers must be

$(2_5)^3 \times 3_5 = 44_5$

and

$2_5 \times 21_5 = 42_5$.

Shortcut using hindsight

Note $(43_5)^2 = 4104_5$. Hence $4103 = 43^2-1 =(43-1)(43+1)=42 \times 44$

Answer 4

Continuing along your route, where: $$4103=(10a+b)(10c+d)$$ (notice all values are represented in base $5$ here)

All values will be in base $5$ throughout.

Notice that $40^2=3100$, which is less than $4103$, so $a$ and $c$ are both $4$: $$4103=(40+b)(40+d)=3100+40(b+d)+bd$$ Now, notice that $\left|{b-d}\right|=2$ in order for the last digit in the product to be a $3$. Since the order of factors is optional, and we know the first digits are the same, we will set $d=b+2$: $$4103=3100+40(2b+2)+b(b+2)$$ $$4103=3100+(130b+130)+(b^2+2b)$$ $$4103=3230+132b+b^2$$ $$0=b^2+132b-323$$ $$0=(b-2)(b+134)$$ $$ b=\{2,-134\}$$ Since the digit must be between $0$ and $4$ inclusive, only $2$ can be valid. For $d$: $$d=b+2=2+2=4$$ Finally, since $a=4$; $b=2$; $c=4$; $d=4$, the factors are $\{42_5, 44_5\}$

Answer 5

So $n_1 = 5a + b; n_2 = 5c + d$ and $(5a+b)(5c+d) = 4103_5 = 4*5^3 + 1*5^2 + 3$

$(5a + b)(5c+d) = ac*5^2 + 5(bc + ad) + bd = 4*5^3 + 1*5^2 + 3 = 21*5^2 + 0 + 3$

So $ac \approx 4*5$ which is a very large value considering $a,c \le 4$.

Note $40_5*100_5 = 4000_5 < 4103_5$ so as $(5a+c) < 100_5$ we now $(5b+d) > 40_5$ and vice versa. So $a = c = 4$.

$16*5^2 + 5(4b + 4d) + db = 4*5^3 + 1*5^2 + 3 = 21*5^2 +3$

$20(b+d) + db = 5^3 + 3$

So $20(b+d) \approx 125$ so $b + d \approx 6$.

$bd \equiv 3\mod 5$ and $b,d \le 4$ so $bd \le 13$. so $bd=3,8,13$.

$13$ is prime so that is out. $bd = 3\implies b,d = 1,3$ and $b + d = 4\not \approx 6$. $bd = 8\implies b,d =2,4$ and $b + d = 6$.

So if an such number exist they are $42_5$ and $44_5$.

And $(4*5 + 2)(4*5+4) = 16*5^2 + 5(2*4 + 4*4) + 8 =$

$3*5^3 + 5^2 + 5(24) + 5 + 3 = $

$3*5^3 + 5^2 + 5(25 - 1) + 5 + 3=$

$3*5^3 + 5^2 + 5^3 +3 =$

$4*5^3 + 5^2 + 3 = 4103_5$.

And $42_5*44_5 = 4103_5$ or in base 10 $22*24 = 528$.