AMC 8, 2023, Problem Marathon

[Arisha Roy]

Problem 21

Alina writes the numbers \(1,2, \ldots, 9\) on separate cards, one number per card. She wishes to divide the cards into 3 groups of 3 cards so that the sum of the numbers in each group will be the same. In how many ways can this be done?

[Arisha Roy]

Solution:

Sum of the numbers of the whole group will be,
\(1+2 \cdots+9=\frac{9(10)}{2}=45\)
After dividing the cards in 3 groups, the sum of the numbers in each group,
\(\frac{9\times10}{2}=45\)

To make the counting easier, first we will make the possible groups in which there is 9. 2 groups can be posible i.e. \((9,2,4)\) and \((9,1,5)\).
We will repeat the same process for 8 using the remaining elements in the list.
Possible groups are \((8,2,5)\), \(8,1,6\), \(8,3,4\).
Again we will repeat the same process for 7 using the remaining elements in the list.
Possible groups are \((7,2,6)\), \(7,3,5\).
Again we will repeat the same process for 6 using the remaining elements in the list.
Possible groups are \((6,4,5)\).
We will not get any new set.
So, possible sets will be \((9,1,5)(8,3,4)(7,2,6)\) and \((9,2,4)(8,1,6)(7,3,5)\)
Answer:  2

• This reply was modified 1 year, 5 months ago by Arisha Roy.
• This reply was modified 1 year, 5 months ago by Arisha Roy.

[Arisha Roy]

Problem 22

In a sequence of positive integers, each term after the second is the product of the previous two terms. The sixth term in the sequence is 4000. What is the first term?

[Arisha Roy]

Solution:

Let the 1st term is \(a\) and 2nd term is \(b\).

The 3rd term will be \(ab\).

The 4th term will be \(ab^2\).

The 5th term will be \(a^2b^3\).

The 6th term will be \(a^3b^5\).

So, the series will be, \(a\), \(b\), \(ab\), \(ab^2\), \(a^2b^3\), \(a^3b^5\).

So, \(a^3b^5 = 4000\)

\(\Rightarrow a^3b^5 = 5^32^5\)

\(\Rightarrow a=5\)

First term will be \(5\)

 

• This reply was modified 1 year, 5 months ago by Arisha Roy.

[Arisha Roy]

Problem 24

Isosceles \(\triangle A B C\) has equal side lengths \(A B\) and \(B C\). In the figure below, segments are drawn parallel to \(\overline{A C}\) so that the shaded portions of \(\triangle A B C\) have the same area. The heights of the two unshaded portions are 11 and 5 units, respectively. What is the height of \(h\) of \(\triangle A B C\) ?

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