The number 2160 can be expressed as 2a ×3b×5. Find the value of a and b

Answer 1

To express the number 2160 in the form \(2^a \times 3^b \times 5\), we need to perform its prime factorization.

Starting with 2160, we repeatedly divide by prime factors:

1. Divide by 2:
\[
2160 \div 2 = 1080
\]
2. Divide by 2 again:
\[
1080 \div 2 = 540
\]
3. Divide by 2 again:
\[
540 \div 2 = 270
\]
4. Divide by 2 again:
\[
270 \div 2 = 135
\]
Now 135 is not divisible by 2.

Next, we divide by 3:
5. Divide by 3:
\[
135 \div 3 = 45
\]
6. Divide by 3 again:
\[
45 \div 3 = 15
\]
7. Divide by 3 again:
\[
15 \div 3 = 5
\]
Now 5 is not divisible by 3.

Finally, we divide by 5:
8. Divide by 5:
\[
5 \div 5 = 1
\]

Now, we have reached 1 and we have fully factored 2160 as:
\[
2^4 \times 3^3 \times 5
\]

Thus, comparing this form with \(2^a \times 3^b \times 5\), we find:
- \(a = 4\)
- \(b = 3\)

So, the values of \(a\) and \(b\) are:
\[
a = 4, \quad b = 3
\]