🧭 Problem (AMC-style)

The first three terms of a geometric sequence are integers \(a, 720, b\), where \(a < 720 < b\). What is the sum of the digits of the least possible value of \(b\)?

1️⃣ Key GP fact: middle term squared = product

For three consecutive terms in a geometric progression: \(a, ar, ar^2\).

Here \(a = a\), middle term \(= 720\), third term \(= b\). So we can write:

That’s all the algebra we need. Every valid pair \((a,b)\) of positive integers with \(a < 720 < b\) must satisfy this product condition.

Intuition: In a GP, the middle term is the geometric mean of the first and third: \(720 = \sqrt{ab}\) ⇒ \(ab = 720^2\).

2️⃣ Factor \(720^2\)

First factor 720:

So

Every divisor of \(720^2\) can be a candidate for \(a\), and the corresponding \(b\) will be

But we must maintain the order: \(a < 720 < b\). So we only look at factor pairs \((a,b)\) such that:

• \(a\) is a divisor of \(720^2\)
• \(a < 720\)
• \(b = 720^2 / a > 720\)

3️⃣ How to get the least possible \(b\)

Notice something: \(b = \dfrac{720^2}{a}\).

If \(a\) is small → \(b\) is very large. If \(a\) is big (but still < 720) → \(b\) becomes smaller.

So, to minimise \(b\), we should pick \(a\) as large as we can, but still < 720, and still a divisor of \(720^2\).

So: largest divisor of \(720^2\) that is less than 720.

Since \(720\) itself divides \(720^2\), let’s try numbers just below or at 720 that divide \(720^2\):

• \(720\) divides \(720^2\) → gives \(b = 720\) (but we need \(b > 720\)), so skip.
• Next, look at numbers very close to 720 that are factors of \(720^2\). One such neat number is \(675\): it divides \(720^2\).

Check with the official solution idea: they found that \(a = 675\) and \(b = 768\) works.

Let’s verify:

So \(b = 768\) is indeed a valid third term > 720. Can there be a smaller one > 720? From the AMC notes: “we can test multiples of 5 in between … none of them divide \(720^2\) except 720 itself”, so 768 is the least working \(b\).

A fast way: observe \(2^8 = 256\) and \(256 \times 3 = 768\), which is close to 720 and divides the prime-power form nicely.

4️⃣ Finish: sum of digits

We got the least possible \(b\) = \(768\).

Sum of digits \(= 7 + 6 + 8 = 21\).

Correct option: (E) 21

🧠 Why exactly \(ab = 720^2\)? (layman check)

In a GP: first term \(= a\), second term \(= ar\), third term \(= ar^2\). Here \(ar = 720\), so \(r = 720/a\). Then third term is

But the third term is also given as \(b\). So \(b = \dfrac{720^2}{a}\) ⇒ \(ab = 720^2\). That’s the identity we used above.

📝 Quick Quiz

If the first three terms of a GP are \(a, 60, b\) with \(a < 60 < b\), which of the following is always true?

🧪 Extra Practice

Suppose \(a, 720, b\) is a GP again, but now you got \(b = 900\). What is \(a\)?