9th task: "Coding information, volume and transmission of information"
Difficulty level - basic,
The maximum score is 1,
Estimated completion time is 5 minutes.

Unified State Exam in Informatics 2017 task 9 FIPI option 1 (Krylov S.S., Churkina T.E.):

What is the minimum amount of memory (in KB) you need to reserve so that you can save any bitmap of size 160 x 160 pixels, provided that the image can use 256 different colors?

Answer:25

✍ Show solution:

• We use the formula for finding the volume:
• Let's calculate each factor in the formula, trying to bring the numbers to powers of two:
• M x N:

160 * 160 \u003d 20 * 2³ * 20 * 2³ \u003d 400 * 2 6 \u003d \u003d 25 * 2 4 * 2 6

Finding the coding depth i:

256 \u003d 2 8 i.e. 8 bits per pixel (from the formula number of colors \u003d 2 i)

Find the volume:

I \u003d 25 * 2 4 * 2 6 * 2 3 \u003d 25 * 2 13 - total bits for the whole image

We translate into Kbytes:

(25 * 2 13) / 2 13 \u003d 25 KB

Unified State Exam in Informatics task 9.2 (source: 9.1 option 11, K. Polyakov):

Figure size 128 on the 256 pixels occupies in memory 24 kB (excluding compression). number of colors in the image palette.

Answer:64

✍ Show solution:

• where M * N Is the total number of pixels. We find this value using powers of two for convenience:

128 * 256 = 2 7 * 2 8 = 2 15

In the above formula i Is the color depth, on which the number of colors in the palette depends:

Number of colors \u003d 2 i

Find i from the same formula:

i \u003d I / (M * N)

Let's take into account that 24 kB must be translated into bits... We get:

2 3 * 3 * 2 10 * 2 3: i \u003d (2 3 * 3 * 2 10 * 2 3) / 2 15 \u003d \u003d 3 * 2 16/2 15 \u003d 6 bits

Now let's find the number of colors in the palette:

2 6 = 64 color options in the color palette

Topic: Image encoding

Unified State Exam in Informatics task 9.3 (source: 9.1 option 24, K. Polyakov):

After converting the raster 256-color graphic file in 4-color format its size decreased by 18 kB. What was the sizesource file in KB?

Answer:24

✍ Show solution:

• According to the formula for the size of the image file, we have:

where N - the total number of pixels,
and i

• i can be found by knowing the number of colors in the palette:

number of colors \u003d 2 i

before conversion: i \u003d 8 (2 8 \u003d 256) after conversion: i \u003d 2 (2 2 \u003d 4)

Let us compose a system of equations based on the available information, take for x number of pixels (resolution):

I \u003d x * 8 I - 18 \u003d x * 2

Let us express x in the first equation:

x \u003d I / 8

I (file size):

I - 18 \u003d I / 4 4I - I \u003d 72 3I \u003d 72 I \u003d 24

Topic: Image encoding

Unified State Exam in Informatics task 9.4 (source: 9.1 option 28, K. Polyakov, S. Loginova):

The color image was digitized and saved as a file without using data compression. Received file size - 42 MB 2 times less and the color depth was increased by 4 times more than the original parameters. No data compression was performed. Please indicate file size in MBobtained during re-digitization.

Answer:42

✍ Show solution:

• According to the formula for the size of the image file, we have:

where N
and i

• In this kind of tasks, it is necessary to take into account that reducing the resolution by 2 times means reducing by 2 times the pixels separately in width and height. Those. overall N decreases 4 times!
• Let's compose a system of equations based on the available information, in which the first equation will correspond to the data before the file transformation, and the second equation - after:

42 \u003d N * i I \u003d N / 4 * 4i

Let us express i in the first equation:

i \u003d 42 / N

Substitute into the second equation and find I (file size):

\\ [I \u003d \\ frac (N) (4) * 4 * \\ frac (42) (N) \\]

After the abbreviations, we get:

I \u003d 42

Topic: Image encoding and bit rate

Unified State Exam in Informatics task 9.5 (source: 9.1 option 30, K. Polyakov, S. Loginova):

The image was digitized and saved as a bitmap file. The resulting file was transferred to cities via communication channel for 72 seconds... Then the same image was digitized again with a resolution of 2 times more and color coding depth in 3 times less than the first time. No data compression was performed. The resulting file was transferred to city \u200b\u200bB, the bandwidth of the communication channel with the city B c 3 times higher than the communication channel with the city A.
B?

Answer:32

✍ Show solution:

• By the formula for the file transfer rate, we have:

where I - file size, and t - time

• According to the formula for the size of the image file, we have:

where N - the total number of pixels or resolution,
and i - color depth (the number of bits allocated for 1 pixel)

• For this task, it is necessary to clarify that the resolution actually has two factors (pixels in width * pixels in height). Therefore, when the resolution is doubled, both numbers will increase, i.e. N will increase in 4 times instead of two.
• Let's change the formula for getting the file size for the city B:

\\ [I \u003d \\ frac (2 * N * i) (3) \\]

• For cities A and B, replace the volume values \u200b\u200bin the formula to get the speed:

\\ [V \u003d \\ frac (N * i) (72) \\]

\\ [3 * V \u003d \\ frac (\\ frac (4 * N * i) (3)) (t) \\]

\\ [t * 3 * V \u003d \\ frac (4 * N * i) (3) \\]

• Substitute the speed value from the formula for city A into the formula for city B:

\\ [\\ frac (t * 3 * N * i) (72) \u003d \\ frac (4 * N * i) (3) \\]

• Let us express t:

t \u003d 4 * 72 / (3 * 3) \u003d 32 seconds

Topic: Image encoding

Unified State Exam in Informatics task 9.6 (source: 9.1 option 33, K. Polyakov):

The camera takes pictures of the size 1024 x 768 pixels. One frame is stored for 900 kB.
Find the best possible number of colors in the image palette.

Answer:512

✍ Show solution:

• The number of colors depends on the color coding depth, which is measured in bits. For storing a frame, i.e. total pixels allocated 900 KB. Let's translate into bits:

900 KB \u003d 2 2 * 225 * 2 10 * 2 3 \u003d 225 * 2 15

Let's calculate the total number of pixels (from a given size):

1024 * 768 = 2 10 * 3 * 2 8

Let's determine the amount of memory required to store not the total number of pixels, but one pixel ([memory for the frame] / [number of pixels]):

\\ [\\ frac (225 * 2 ^ (15)) (3 * 2 ^ (18)) \u003d \\ frac (75) (8) \\ approx 9 \\]

9 bits per pixel

9 bit is i - color coding depth. Number of colors \u003d 2 i:

2 9 = 512

Topic: Image encoding

9 task. Demo version of the Unified State Exam 2018 informatics:

An automatic camera produces bitmap images in size 640 × 480 pixels. At the same time, the size of the image file cannot exceed 320 KB, data packing is not performed.
What maximum number of colors can be used in palette?

Answer: 256

✍ Show solution:

• According to the formula for the size of the image file, we have:

where N i - color coding depth (number of bits allocated for 1 pixel)

• Let's see what from the formula has already been given to us:

I\u003d 320 KB, N \u003d 640 * 420 \u003d 307200 \u003d 75 * 2 12 total pixels, i - ?

The number of colors in the image depends on the setting iwhich is unknown. Let's remember the formula:

number of colors \u003d 2 i

Since the color depth is measured in bits, it is necessary to convert the volume from Kilobytes to bits:

320 KB \u003d 320 * 2 10 * 2 3 bits \u003d 320 * 2 13 bits

Find i:

\\ [i \u003d \\ frac (I) (N) \u003d \\ frac (320 * 2 ^ (13)) (75 * 2 ^ (12)) \\ approx 8.5 bits \\]

Find the number of colors:

2 i \u003d 2 8 \u003d 256

9_21: : Unified State Exam in Informatics task 9.21 (source: K. Polyakov, 9.1 option 58):

For storage in the information system, documents are scanned with permission 300 ppi... No image compression methods are used. Average scanned document size is 5 MB... In order to save money, it was decided to switch to a permit 150 ppi and a color system containing 16 colors... The average size of a document scanned with the modified settings is 512 kB.

Define number of colors in the palette before optimization.

Answer:1024

Show solution:

• According to the formula for the size of the image file, we have:

where N Is the total number of pixels or resolution, and i - color coding depth (number of bits allocated for 1 pixel).

• Since, according to the task, we have a resolution expressed in pixels per inch, in fact this means:

I \u003d ppi value 2 * N * i

• Color formula:

number of colors \u003d 2 i

• Let's see what the formula has already been given to us before the economical option and with the economical option:

Waste option: I\u003d 5 MB \u003d 5 * 2 23 bits, N - ?, i -? 300 ppi Economical option: I\u003d 512 KB \u003d 2 9 * 2 13 bits \u003d 2 22 bits, N - ?, i \u003d 4 bits (2 4 \u003d 16) 150 ppi

Since in the economy mode we know all the components of the formula, except for the resolution (N), we will find the resolution:

N \u003d I / (i * 150 * 150 ppi) N \u003d 2 22 / (4 * 22500)

We substitute all known values, including the found N, into the formula for the uneconomical mode:

I \u003d N * 300 * 300 ppi * i 5 * 2 23 \u003d (2 22 * \u200b\u200b300 * 300 * i) / (22500 * 4);

Let us express i and calculate its value:

i \u003d (5 * 2 23 * 22500 * 4) / (2 22 * \u200b\u200b300 * 300) \u003d 9000/900 \u003d 10 bits

According to the formula for finding the number of colors in the palette, we have:

2 10 = 1024

Topic: Sound coding

Unified State Exam in Informatics 2017 task 9 FIPI option 15 (Krylov S.S., Churkina T.E.):

In the studio with a four-channel ( quad) sound recordings with 32 -bit resolution for 30 seconds a sound file was recorded. No data compression was performed. The file size is known to be 7500 KB.

From what sampling rate (in kHz) were you recording?Enter only a number as an answer; you do not need to specify units of measurement.

Answer:16

✍ Show solution:

• Using the formula for the volume of a sound file, we get:

I \u003d β * t * ƒ * S

• From the task we have:

I\u003d 7500 KB β \u003d 32 bits t\u003d 30 seconds S\u003d 4 channels

ƒ - sampling rate - unknown, we express it from the formula:

\\ [ƒ \u003d \\ frac (I) (S * B * t) \u003d \\ frac (7500 * 2 ^ (10) * 2 ^ 2 bits) (2 ^ 7 * 30) Hz \u003d \\ frac (750 * 2 ^ 6 ) (1000) KHz \u003d 2 ^ 4 \u003d 16 \\]

2 4 = 16 kHz

Unified State Exam in Informatics task 9.9 (source: 9.2 option 36, K. Polyakov):

The piece of music was digitized and recorded as a file without using data compression. The resulting file was transferred to the city AND through the communication channel. Then the same piece of music was digitized again with resolution in 2 3 times less than the first time. No data compression was performed. The received file was transferred to the city B behind 15 seconds; bandwidth of the communication channel with the city B in 4 times higher than the communication channel with the city AND.

How many seconds did the file transfer to the city take A? In the answer, write down only an integer, you do not need to write a unit of measurement.

Answer:90

✍ Show solution:

• To solve, you need a formula for finding the data transfer rate of the formula:
• Let's also recall the formula for the volume of an audio file:

I \u003d β * ƒ * t * s

where:
I - volume
β - coding depth
ƒ - sampling frequency
t - time
S - number of channels (if not specified, then mono)

• We will write out separately, all data relating to the city B (about AND practically nothing is known):

city \u200b\u200bB: β - 2 times higher ƒ - 3 times less t - 15 seconds throughput (speed V) - 4 times higher

Based on the previous point, for city A we get the opposite values:

cities: β B / 2 ƒ B * 3 I B / 2 V B / 4 t B / 2, t B * 3, t B * 4 -?

Let's explain the obtained data:

as coding depth ( β ) for the city B higher in 2 times, then for the city AND she will be lower in 2 times, respectively, and t decrease in 2 times:

t \u003d t / 2

as sampling rate (ƒ) for the city B less in 3 times, then for the city AND it will be higher in 3 times; I and t change proportionally, which means that with an increase in the sampling rate, not only the volume will increase, but also the time:

t \u003d t * 3

speed ( V) (bandwidth) for the city B higher in 4 times, so for the city AND it will be 4 times lower; times the speed is lower, the time is higher in 4 times ( t and V - inversely proportional dependence from the formula V \u003d I / t):

t \u003d t * 4

Thus, taking into account all indicators, the time for the city AND changes like this:

\\ [t_A \u003d \\ frac (15) (2) * 3 * 4 \\]

90 seconds

Topic: Sound coding

Unified State Exam in Informatics task 9.10 (source: 9.2 option 43, K. Polyakov):

The piece of music was recorded in stereo ( two-channel recording), digitized and saved as a file without using data compression. Received file size - 30 MB. Then the same piece of music was re-recorded in the format monoand digitized with permission in 2 times higher and a sampling rate of 1,5 times less than the first time. No data compression was performed.

Please indicate file size in MBobtained when rewriting. In the answer, write down only an integer, you do not need to write a unit of measurement.

Answer:20

✍ Show solution:

I \u003d β * ƒ * t * S

I - volume
β - coding depth
ƒ - sampling frequency
t - time
S -number of channels

• Let's write down separately, all data concerning the first state of the file, then the second state - after transformation:

1 state: S \u003d 2 channels I \u003d 30 MB 2 state: S \u003d 1 channel β \u003d 2 times higher ƒ \u003d 1.5 times lower I \u003d?

Since it was originally 2 communication channel ( S), and began to be used one communication channel, then the file has decreased in 2 times:

I \u003d I / 2

Coding depth ( β ) increased in 2 times, then the volume ( I) will increase by 2 times (proportional dependence):

I \u003d I * 2

Sampling frequency ( ƒ ) decreased in 1,5 times, so the volume ( I) will also decrease by 1,5 times:

I \u003d I / 1.5

Let's consider all the changes in the size of the converted file:

I \u003d 30 MB / 2 * 2 / 1.5 \u003d 20 MB

Topic: Audio coding and bit rate

Unified State Exam in Informatics task 9.11 (source: 9.2 option 72, K. Polyakov):

The piece of music was digitized and recorded as a file without using data compression. The resulting file was transferred to cities via communication channel for 100 seconds. Then the same piece of music was digitized again with permission 3 times higher and sampling rate 4 times lessthan the first time. No data compression was performed. The resulting file was transferred to city \u200b\u200bB behind 15 seconds.

How many times is the speed (bandwidth) of the city B more bandwidth to the city AND ?

Answer:5

✍ Show solution:

• Let's recall the formula for the volume of an audio file:

I \u003d β * ƒ * t * S

I - volume
β - coding depth
ƒ - sampling frequency
t - time

• Let us write out separately, all the data regarding the file transferred to the city ANDthen the converted file transferred to city B:

AND: t \u003d 100 s. B: β \u003d 3 times higher ƒ \u003d 4 times lower t \u003d 15 s.

✎ 1 solution:

The data transfer speed (bandwidth) depends on the file transfer time: the longer the time, the lower the speed. Those. how many times the transmission time increases, how many times the speed decreases and vice versa.

From the previous paragraph, we see that if we calculate how many times the time for transferring a file to the city will decrease or increase B (compared to city A), then we will understand how many times the data transfer rate to the city will increase or decrease B (inverse relationship).

Accordingly, let's imagine that the converted file is transferred to the city AND... File size changed to 3/4 times (coding depth (β) in 3 times higher, sampling rate (ƒ) in 4 times lower). Volume and time vary proportionally. This means that time will change in 3/4 times:

t A to convert. \u003d 100 seconds * 3/4 \u200b\u200b\u003d 75 seconds

Those. the converted file would be transferred to the city AND 75 seconds, and into the city B 15 seconds. Let's calculate how many times the transmission time has decreased:

75 / 15 = 5

Time transfer to town B decreased in 5 times, respectively, the speed increased by 5 time.

Answer: 5

✎ 2 solution:

Let's write down separately all the data regarding the file transferred to the city AND: AND: t A \u003d 100 s. V A \u003d I / 100

Since increasing or decreasing in some times the resolution and sampling rate leads to a corresponding increase or decrease in the file size (proportional dependence), then we will write down the known data for the converted file transferred to the city B:

B: β \u003d 3 times higher ƒ \u003d 4 times lower t \u003d 15 s. I B \u003d (3/4) * I V B \u003d ((3/4) * I) / 15

Now we find the ratio of V B to V A:

\\ [\\ frac (V_B) (V_A) \u003d \\ frac (3 / _4 * I) (15) * \\ frac (100) (I) \u003d \\ frac (3 / _4 * 100) (15) \u003d \\ frac (15 ) (3) \u003d 5 \\]

(((3/4) * I) / 15) * (100 / I) \u003d (3/4 * 100) / 15 \u003d 15/3 \u003d 5

Topic: Sound coding

Unified State Exam in Informatics task 9.12 (source: 9.2 option 80, K. Polyakov):

Produced four-channel (quad) sound recording with sampling rate 32 kHz and 32-bit resolution. The recording lasts 2 minutes, its results are written to a file, no data compression is performed.

Determine the approximate size of the resulting file (in MB). Enter the integer closest to the file size as your answer, multiple of 10.

Answer:60

✍ Show solution:

• Let's recall the formula for the volume of an audio file:

I \u003d β * ƒ * t * S

I - volume
β - coding depth
ƒ - sampling frequency
t - time
S - number of channels

• For simplicity of calculations, we will not take into account the number of channels yet. Let's consider what data we have, and which of them need to be converted to other units of measurement:

β \u003d 32 bits ƒ \u003d 32kHz \u003d 32000Hz t \u003d 2 min \u003d 120 s

Let's substitute the data into the formula; take into account that the result must be obtained in MB, respectively, the product will be divided by 2 23 (2 3 (bytes) * 2 10 (Kbytes) * 2 10 (MB)):

(32 * 32000 * 120) / 2 23 = =(2 5 * 2 7 * 250 * 120) / 2 23 = = (250*120) / 2 11 = = 30000 / 2 11 = = (2 4 * 1875) / 2 11 = = 1875 / 128 ~ 14,6

I \u003d β * ƒ * t * S

I - volume β - coding depth \u003d 32 bits ƒ - sampling rate \u003d 48000 Hz t - time \u003d 5 min \u003d 300 s S - number of channels \u003d 2

Let's substitute the existing values \u200b\u200binto the formula:

I \u003d 48000 * 32 * 300 * 2

Since the values \u200b\u200bare large, numbers are needed 48000 and 300 express in powers of two:

48000 | 2 24000 | 2 12000 | 2 6000 | 2 = 375 * 2 7 3000 | 2 1500 | 2 750 | 2 375 | 2 - no longer divisible 187.5 300 | 2 \u003d 75 * 2 2 150 | 2 75 | 2 - no longer divisible 37.5

We get:

I \u003d 375 * 75 * 2 15

In the proposed answer options, we see that the result is everywhere in MB. Hence, it is necessary to divide the result we obtained by 2 23 (2 3 * 2 10 * 2 10):

I \u003d 375 * 75 * 2 15/2 23 \u003d 28125/2 8

Find an approximate number 28125 value to the power of two:

2 10 = 1024 1024 * 2 2048 * 2 4096 * 2 8192 * 2 16384 * 2 32768

We get:

2 10 * 2 5 = 2 15 = 32768 2 10 * 2 4 = 2 14 = 16384

Number 28125 lies between these values, so we take them:

2 15 / 2 8 = 2 7 = 128 2 14 / 2 8 = 2 6 = 64

We choose the answer, the value of which is between these two numbers: option Substitute the existing values \u200b\u200bin the formula. For convenience, we will use powers of two:

ƒ \u003d 4 kHz \u003d 4 * 1000 Hz ~ 2 2 * 2 10 B \u003d 64 bits \u003d 2 6/2 23 MB t \u003d 1 min \u003d 60 s \u003d 15 * 2 2 s S \u003d 2

Let's substitute the values \u200b\u200binto the formula for the volume of the sound file:

I \u003d 2 6 * 2 2 * 2 10 * 15 * 2 2 * 2 1/2 23 \u003d 15/4 ~ 3.75

The closest multiple of two is a number 4

| Informatics and information and communication technologies | Preparation for the exam | Control training tasks

Control training tasks

Part 1

Assignment 9

Answers to tasks 1 - 23 are a number, a sequence of letters or numbers that should be written in ANSWER FORM No. 1 to the right of the number of the corresponding task, starting from the first cell, without spaces, commas and other additional characters. Write each character in a separate box in accordance with the samples given in the form.

Example 1

What is the minimum amount of memory (in KB) that must be reserved to store any 64 × 64 pixel bitmap image, provided the image can use 256 different colors? In the answer, write down only an integer, you do not need to write a unit of measurement.

Example 2

To store an arbitrary raster image with a size of 1024 × 1024 pixels, 512 KB of memory are allocated, while a binary number is stored for each pixel - the color code of this pixel. For each pixel, the same number of bits are allocated to store the code. Data is not compressed. What is the maximum number of colors that can be used in an image?

Answer: ___________________________.

Example 3

An automatic camera produces bitmap images with a size of 640 × 480 pixels. At the same time, the size of the file with the image cannot exceed 320 Kbytes, the data is not packed. What is the maximum number of colors that can be used in the palette?

The author of the assignments is a leading specialist who is directly involved in the development of control measuring materials for the exam.
Typical test tasks in computer science contain 14 options for sets of tasks, compiled taking into account all the features and requirements of the Unified State Exam in 2019. The purpose of the manual is to provide readers with information about the structure and content of the KIM 2019 in informatics, the degree of difficulty of tasks.
The collection provides answers to all test options, provides solutions to all tasks of one of the options, as well as solutions to the problems of part 2.
The manual is intended for teachers to prepare students for the computer science exam, as well as high school students for self-study and self-control.

Examples.
An automatic camera produces raster images of 640 x 480 pixels. At the same time, the size of the file with the image cannot exceed 170 Kbytes, the data is not packed. What is the maximum number of colors that can be used in the palette?

Artist A16 converts the number written on the screen.
The performer has three teams, which are assigned numbers:
1. Add 1
2. Add 2
3. Multiply by 2
The first of them increases the number on the screen by 1, the second increases it by 2, the third multiplies by 2.
The program for the performer A16 is a sequence of commands.
How many programs are there that convert the original number 3 to the number 12 and the path of the program's computation contains the number 10?
The program computation path is a sequence of results of execution of all program commands. For example, for program 132 with an initial number of 7, the trajectory will consist of numbers 8, 16, 18.

Free download the e-book in a convenient format, watch and read:
Download the book Unified State Exam 2019, Informatics, 14 options, Typical test tasks, V.R. Leshchiner - fileskachat.com, fast and free download.

• I will pass the exam, Informatics, Typical assignments, Leshchiner V.R., Krylov S.S, Ushakov D.M., 2019
• Unified State Exam, Informatics, grade 11, Preparing for the final certification, Leshchiner V.R., Krylov S.S., Yakushkin A.P., 2020
• Unified State Exam 2020, Informatics, 16 options, Typical options for exam tasks from the developers of the Unified State Exam, Leshchiner V.R.
• Informatics, Unified State Exam, Preparing for the final certification, Leshchiner V.R., Krylov S.S., Yakushkin A.P., 2020

The following tutorials and books.

Analysis of the 9th task of the exam 2018 in computer science and ICT from the demo version. This is a basic level of difficulty task. Estimated time to complete the task is 5 minutes.

Checked content items:
- Ability to determine the speed of information transfer at a given channel bandwidth, the amount of memory required to store sound and graphic information.

Assignment 9

An automatic camera produces bitmap images with a size of 640 × 480 pixels. At the same time, the size of the file with the image cannot exceed 320 Kbytes, the data is not packed. What is the maximum number of colors that can be used in the palette?

Answer: ________

Analysis of the 9 tasks of the exam 2018 in computer science

The volume of the bitmap is found as the product of the number of pixels in the image by the amount of memory i needed to store the color of one pixel

In our case

640 480 i ≤ 320 1024 8 bit

i ≤ 32010248 / (640480),

since a bit can only be an integer, we get

i \u003d 8 (in our problem it is said that the size of the file with the image cannot exceed 320 KB, therefore, we cannot round up! The resulting value i round down to 8 bits!)

For this we use the formula N \u003d 2 i

N \u003d 2 8 \u003d 256

This means that no more than 256 flowers.

. Information coding, volume and transmission of information: demo version of exam informatics 2019; state final exam 2019; training options for the exam in computer science, thematic test tasks and tasks from the simulator in computer science 2019

✍ Solution:

• Let's recall the basic formula for the size of an image file:

I \u003d m * n * i

m, n - resolution (width and height in pixels)
i - color coding depth

We express all values \u200b\u200bin powers of two:

200|2 100|2 50 |2 25 | 25 * 2 3

Resolution:

25 * 2 3 * 2 8 = 25 * 2 11

Substitute all the values:

i * 25 * 2 11

Insofar as i must be less than the obtained value, then i \u003d 10.

Let's count the number of colors:

colors \u003d 2 i \u003d 2 10 \u003d 1024

Result:1024

9 tasks of the exam 2019 for training

Option No. 3, 2019 (10 options), S.S. Krylov, T.E. Churkin:

800 × 2800 700 kB

What

Answer: 4

Option number 4, Typical exam options 2019 (10 options), S.S. Krylov, T.E. Churkin:

An automatic camera produces bitmaps in size 1000 × 1600 pixels. The same number of bits is used to encode the color of each pixel, and the pixel codes are written to the file one by one without gaps. The size of the image file cannot exceed 2100 Kbytes without considering the size of the file header.

What maximum number of colors can be used in palette?

Answer: 1024

Option number 8, Typical exam options 2019 (10 options), S.S. Krylov, T.E. Churkin:

What minimum memory size (in KB) needs to be reserved so that you can save any bitmap size 1280 x 80 pixels provided that the image can use 64 different colors? In the answer, write down only an integer, you do not need to write a unit of measurement.