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AMC 10 Math Olympiad USA Math Olympiad

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

Check the Answer


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 \(13\)

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Algebra AMC 10 Math Olympiad USA Math Olympiad

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

Check the Answer


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 \(|S|=|A \cup B|=|A|+|B|-|A \cap B|=4008-|A \cap B|=4008-286=3722\)

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