Divisibility - Understanding the Rule : AMC 8, 2016 Problem 24

Divisibility is the study of finding remainder when a number is divided by another number.

The digits $1$, $2$, $3$, $4$, and $5$ are each used once to write a five-digit number $PQRST$. The three-digit number $PQR$ is divisible by $4$, the three-digit number $QRS$ is divisible by $5$, and the three-digit number $RST$ is divisible by $3$. What is $P$?

$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$

AMC 8, 2016 Problem number 24

Divisibility

6 out of 10

Mathematical Circles

Knowledge Graph

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Use some hints

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See this image:

So we have $5$ blank space to fill with the numbers $1,2,3,4,5$ following the given restriction.

See the given conditions care fully :

$1.\quad PQR$ is divisible by $4$,

$2.\quad QRS$ is divisible by $5$,

$3. \quad RST$ is divisible by $3$

Can you make a definite choice for any of the places ???

$1.\quad PQR$ is divisible by $4 \Rightarrow$ $QR$ is divisible by $4$

$2.\quad QRS$ is divisible by $5 \Rightarrow $, $S$ is either $5$ or $0$. [But we don't have $0$ so $S$ must be $5$]

$3. \quad RST$ is divisible by $3 \Rightarrow $ $R+S+T$ is divisible by 3.

Try to find out other choices, other blanks !!

$QR$ is divisible by $4$ so $R$ needs to be $2$ or $4$.

Let $R=2$

$\Rightarrow R+S+T=2+5+T=7+T$

As, $R+S+T$ is divisible by 3 then possible values of $T$ are $2, 5, 8$

We have already used $2$ and $5$. And $8$ is not in the given set of numbers. (Contradiction)

Therefore $R \ne 2 \Rightarrow R = 4$

Now the things are easy, try to drive it from here !!!!!!!!!

Again, $RST$ is divisible by $3$

$\Rightarrow R+S+T$ is divisible by $3$

$\Rightarrow 4+5+T$ is divisible by $3$

Then the possibilities for $T$ are $3, 6$

$6$ is not in the given number, hence $T=3$

We have two numbers left and two blank places left.

if $Q=1$

then $QR=14$ is not divisible by $4$

Then $Q=2$ [$24$ is divisible by $4$]

And certainly $P=1$

The number is :

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