Find all natural two digit numbers such that when you substract by seven times the sum of its digit from the number you get a prime number.
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Tags: number theory
MathArt4
27.02.2021 19:51
Say the number is expressed in the form $10a+b$. We want to find all numbers such that $10a+b-7a-7b=3a-6b$ is prime. Note that we can factor out $3$ from this expression to get $3(a-2b)$. The only way for this to be prime is if $a-2b=1$. Checking all the values of $b$ from $0$ to $9$, the only values of $b$ that work are $0,1,2,3,$ and $4$. The corresponding $a$ values for this are $1,3,5,7,$ and $9$. This means the answer is $\boxed{10,31,52,73,94}$.
EdiBerisha
27.02.2021 19:56
MathArt4 wrote:
Say the number is expressed in the form $10a+b$. We want to find all numbers such that $10a+b-7a-7b=3a-6b$ is prime. Note that we can factor out $3$ from this expression to get $3(a-2b)$. The only way for this to be prime is if $a-2b=1$. Checking all the values of $b$ from $1$ to $9$, the only values of $b$ that work are $1,2,3,$ and $4$. The corresponding $a$ values for this are $3,5,7,$ and $9$. This means the answer is $\boxed{31,52,73,94}$.
You forgot b=0 Which means a=2b+1 => a=2(0)+1 = 1 Meaning there is also 10 as a solution
MathArt4
27.02.2021 19:57
EdiBerisha wrote: MathArt4 wrote:
Say the number is expressed in the form $10a+b$. We want to find all numbers such that $10a+b-7a-7b=3a-6b$ is prime. Note that we can factor out $3$ from this expression to get $3(a-2b)$. The only way for this to be prime is if $a-2b=1$. Checking all the values of $b$ from $1$ to $9$, the only values of $b$ that work are $1,2,3,$ and $4$. The corresponding $a$ values for this are $3,5,7,$ and $9$. This means the answer is $\boxed{31,52,73,94}$.
b=0? Whoops, I missed that case, it should be fixed now, thanks!
Rijadinho
28.04.2023 00:01
We can express the number as $10a+b$ $10a+b-7(a+b)=10a+b-7a-7b=3a-6b=3(a-2b)$ Since $3(a-2b)$ is prime, $a-2b=1$ Therefore, $(a,b)=(1,0),(3,1),(5,2),(7,3),(9,4)$ Giving us $10,31,52,73,94$ as the only solutions $\blacksquare$