Categories

## Greatest Common Divisor | AMC-10A, 2018 | Problem 22

Try this beautiful problem from Algebra based on Greatest Common Divisor from AMC 10A, 2018, Problem 22.

## Greatest Common Divisor – AMC-10A, 2018- Problem 22

Let $a, b, c,$ and $d$ be positive integers such that $\gcd(a, b)=24$, $\gcd(b, c)=36$, $\gcd(c, d)=54$, and $70<\gcd(d, a)<100$. Which of the following must be a divisor of $a$?

• $5$
• $7$
• $13$

### Key Concepts

Number theory

Gcd

Divisior

Answer: $13$

AMC-10A (2018) Problem 22

Pre College Mathematics

## Try with Hints

TO find the divisor of $a$ at first we have to find the value of $a$.can you find the value of $a$?

Given that $\gcd(a, b)=24$, $\gcd(b, c)=36$, $\gcd(c, d)=54$, and $70<\gcd(d, a)<100$

so we can say $a=24 \times$ some integer and $b=24 \times$ some another integer (according to gcd rules)

similarly for the others c & d…..

now if we can find out the value of $\gcd(d, a)$ then we may use the condition $70<\gcd(d, a)<100$

Can you now finish the problem ……….

so we may say that $gcd(a, b)$ is $2^3 * 3$ and the $gcd$ of $(c, d)$ is $2 * 3^3$. However, the $gcd$ of $(b, c) = 2^2 * 3^2$ (meaning both are divisible by 36). Therefore, $a$ is only divisible by $3^1$ (and no higher power of 3), while $d$ is divisible by only $2^1$ (and no higher power of 2).

can you finish the problem……..

so we can say that $gcd$ of $(a, d)$ can be expressed in the form $2 \times 3 \times$ some positve integer and now $k$ is a number not divisible by $2$ or $3$. so from the given numbes it will be $13$ because $2 \times 3 \times k$ must lie $70<\gcd (d, a)<100$. so the required ans is $13$

Categories

## Arithmetic Progression | AMC-10B, 2004 | Problem 21

Try this beautiful problem from Algebra based on Arithmetic Progression.

## Arithmetic Progression – AMC-10B, 2004- Problem 21

Let $1$; $4$; $\ldots$ and $9$; $16$; $\ldots$ be two arithmetic progressions. The set $S$ is the union of the first $2004$ terms of each sequence. How many distinct numbers are in $S$?

• $3478$
• $3722$
• $3378$

### Key Concepts

algebra

AP

Divisior

Answer: $3722$

AMC-10B (2004) Problem 21

Pre College Mathematics

## Try with Hints

There are two AP series …..

Let A=$\{1,4,7,10,13……..\}$ and B=$\{9,16,23,30…..\}$.Now we have to find out a set $S$ which is the union of first $2004$ terms of each sequence.so if we construct a set form i.e $A=\{3K+1,where 0\leq k <2004\}$ and B=$\{7m+9, where 0\leq m< 2004\}$.Now in A and B total elements=4008.

Now $S=A \cup B$ and $|S|=|A \cup B|=|A|+|B|-|A \cap B|=4008-|A \cap B|$

Now we have to find out $|A \cap B|$

Can you now finish the problem ……….

To find out $|A \cap B|$ :

Given set A=$\{1,4,7,10,13……..\}$ and B=$\{9,16,23,30…..\}$.Clearly in the set A $1$ is the first term and common difference $3$.and second set i.e B first term is $9$ and common difference is $7$.

Now $|A \cap B|$ means there are some terms of $B$ which are also in $A$.Therefore $7m+9 \in A$ $\Rightarrow$ $1\leq 7m+9 \leq 3\cdot 2003 + 1$, and $7m+9\equiv 1\pmod 3$”.

The first condition gives $0\leq m\leq 857$, the second one gives $m\equiv 1\pmod 3$.

Therefore $m$= $\{1,4,7,\dots,856\}$, and number of digits in $m$= $858/3 = 286$.

can you finish the problem……..

$S=A \cup B$ and $|S|=|A \cup B|=|A|+|B|-|A \cap B|=4008-|A \cap B|=4008-286=3722$