Problem Solving

Vincen wrote: ↑

Wed Jul 22, 2020 12:54 pm

How many digits are there in the product \(2^{23}\cdot 5^{24}\cdot 7^3?\)

A. 24
B. 25
C. 26
D. 27
E. 28

[spoiler]OA=D[/spoiler]

Source: Veritas Prep

Looking at \(2^{23}\cdot 5^{24}\cdot 7^3,\) we find that there are 2s and 5s; a 2 and a 5 make a 10. Since there are 23 2s and 24 5s, the product of 23 2s and 23 5s would make 23 10s or

\(2^{23}\cdot 5^{24}\cdot 7^3\) = \(10^{23}\cdot 5\cdot 7^3\)

\(10^{23}\) would give 23 0s.

Again, \(10^{23}\cdot 2,115 = 2,115\times10^{23}\)

So, the number \(2^{23}\cdot 5^{24}\cdot 7^3\) would have 4 + 23 = 27 digits.

Correct answer: D

Hope this helps!

-Jay
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Vincen wrote: ↑

Wed Jul 22, 2020 12:54 pm

How many digits are there in the product \(2^{23}\cdot 5^{24}\cdot 7^3?\)

A. 24
B. 25
C. 26
D. 27
E. 28

[spoiler]OA=D[/spoiler]

Solution:

Since 2^23 x 5^24 x 7^3 = (2 x 5)^23 x 5 x 7^3 = 10^23 x 1,715, we see that the product is the number 1,715 followed by 23 zeros. Therefore, there are 4 + 23 = 27 digits in the product.

Answer: D