Question

If a= 5^99 x 2^98 and b= 5^72 x 4^38 show ab has 172 digits when written as a ordinary number

This is a proof question. Please show step by step.

Answer 1

Explanation:

ab

= (5⁹⁹ * 2⁹⁸) * (5⁷² * 4³⁸)

= 2⁹⁸ * 2⁷⁶ * 5⁹⁹ * 5⁷²

= 2¹⁷⁴ * 5¹⁷¹

= 2³ * (2¹⁷¹ * 5¹⁷¹).

The 2¹⁷¹ * 5¹⁷¹ will give 171 ending zeroes as 2 * 5 = 10.

And there is a 8 infront as 2³ = 8.

Hence altogether there are 171 + 1 = 172 digits.

Answer 2

The product of
`\(a = 5^(99) * 2^(98)\)` and
`\(b = 5^(72) * 4^(38)\)` is
`\(ab = 5^(171) * 2^(174)\)`, resulting in 120 digits when expressed as an ordinary number.

1. Express a and b in their prime factorization form:

`\[ a = 5^(99) * 2^(98) \] \[ b = 5^(72) * 4^(38) \]`

Note that
`\(4^(38)\)` can be expressed as
`\((2^2)^(38) = 2^(76)\)`.

2. Multiply a and b to find ab:

`\[ ab = (5^(99) * 2^(98)) * (5^(72) * 2^(76)) \]`

Combine the like terms:

`\[ ab = 5^(99 + 72) * 2^(98 + 76) \]`

Simplify the exponents:

`\[ ab = 5^(171) * 2^(174) \]`

3. To find the number of digits in ab, we can use the formula:

`\[ \text{Number of digits} = \lfloor \log_(10)(ab) \rfloor + 1 \]`

Using this formula:

`\[ \text{Number of digits} = \lfloor \log_(10)(5^(171) * 2^(174)) \rfloor + 1 \] \[ \text{Number of digits} = \lfloor 171 \cdot \log_(10)(5) + 174 \cdot \log_(10)(2) \rfloor + 1 \]`

4. Calculate the value:

`\[ \text{Number of digits} \approx \lfloor 171 \cdot 0.69897 + 174 \cdot 0.30103 \rfloor + 1 \] \[ \text{Number of digits} \approx \lfloor 119.52467 \rfloor + 1 \] \[ \text{Number of digits} = 119 + 1 \] \[ \text{Number of digits} = 120 \]`

Therefore, ab has 120 digits when written as an ordinary number.