Which of the two statements below is/are necessarily false?

Statement A: There were 2 new cases in Tyhrmisto on Day 3.

Statement B: There were no new cases in Pesmisto on Day 2.

Neighbourhoods	Day 1	Day 2	Day 3	Day 4	Day 5	Total
Levmisto						12
Tyhrmisto						12
Pesmisto						5
Kitmisto						14
Total						43

Clue 6 says,

The total numbers of new cases in Levmisto, Tyhrmisto, Pesmisto and Kitmisto over the five-day period were 12, 12, 5 and 14 respectively.

we can directly put this data in the table.

Also the total of all cases will be = $12 + 12+ 5+ 14 = 43$

Neighbourhoods	Day 1	Day 2	Day 3	Day 4	Day 5	Total
Levmisto						12
Tyhrmisto						12
Pesmisto						5
Kitmisto		3				14
Total						43

Clue 5 says,

Kitmisto is the only place to have 3 new cases on Day 2.

=> Other then Kitmisto, all neigbourhoods on day 2 will have 0,1,2 cases.

Now we know the total number of cases in Kitmisto = 14

=> only possibility of cases in 5 days for Kitmisto = $(3, 3, 3, 3, 2)$ in any order.

Neighbourhoods	Day 1	Day 2	Day 3	Day 4	Day 5	Total
Levmisto						12
Tyhrmisto						12
Pesmisto						5
Kitmisto		3				14
Total						43

Clue 4 says,

The maximum number of new cases in a day in Pesmisto was 2, and this happened only once during the five-day period.

=> Number of cases in Pesmisto will be 0,1

we know,

total number of cases in Pesmisto = 5

=> only possibility = $(2, 1, 1, 1, 0)$ in any order.

Neighbourhoods	Day 1	Day 2	Day 3	Day 4	Day 5	Total
Levmisto						12
Tyhrmisto						12
Pesmisto						5
Kitmisto		3				14
Total						43

Statement 3 -

The number of new cases in the city in a day kept increasing during the five-day period.

=> the total number of cases kept increasing everyday.

Now, we know,

Maximum possible new cases in pesmisto = $2$

=> The maximum number of cases in the city on day 5 = $3 + 3 + 3+ 2 = 11$

Neighbourhoods	Day 1	Day 2	Day 3	Day 4	Day 5	Total
Levmisto					3	12
Tyhrmisto					3	12
Pesmisto					2	5
Kitmisto		3			3	14
Total					11	43

Case 1 :

Let us consider the maximum number of cases on Day 5 as 10.

Thus the maximum number of cases possible for the remaining days will be 9, 8, 7, and 6

So, the total number of maximum cases possible for this case will be 40(less than 43)

=> Number of new cases on day 5 = 11

=> All the neighborhoods will have 3 new case except Pesmisto who will have only 2 new cases.

Neighbourhoods	Day 1	Day 2	Day 3	Day 4	Day 5	Total
Levmisto					3	12
Tyhrmisto					3	12
Pesmisto					2	5
Kitmisto		3			3	14
Total				10	11	43

Now, , if the number of cases on Day 4 is 9,

the maximum number of cases possible for the remaining days will be 8, 7, and 6.

Thus, the maximum number of cases, in this case, will be 41(less than 43).

So, the number of cases on day 4 will be 10.

Neighbourhoods	Day 1	Day 2	Day 3	Day 4	Day 5	Total
Levmisto					3	12
Tyhrmisto					3	12
Pesmisto					2	5
Kitmisto		3			3	14
Total	5	8	9	10	11	43

Now, if the number of cases on Day 3 is 8,

the number of cases on day 2 will be 7,

and the maximum possible number of cases on Day 1 will be 6.

Thus, the number of cases, in this case, will be 42(less then 43)

Therefore, the total number of cases on day 3 = 9

But clue 3 says,

The number of new cases on Day 3 was exactly one more than that on Day 2.

Therefore, the total number of cases on day 2 = 8

=> total number of cases on day 1 = $43 - 11- 10- 9- 8 = 5$

Neighbourhoods	Day 1	Day 2	Day 3	Day 4	Day 5	Total
Levmisto	1				3	12
Tyhrmisto	1				3	12
Pesmisto	1				2	5
Kitmisto	2	3	3	3	3	14
Total	5	8	9	10	11	43

Clue 1 says,

There was at least one new case in every neighbourhood on Day 1.

=> number of cases on day 1 = $(1, 1, 1, 2)$ in any order

But, we know the possible cases for Kitmisto = $(2,3, 3, 3, 3)$

Therefore, number of cases on day 1 for Kitmisto = 2

and all other neighbourhoods = 1

Neighbourhoods	Day 1	Day 2	Day 3	Day 4	Day 5	Total
Levmisto	1			3	3	12
Tyhrmisto	1			3	3	12
Pesmisto	1			1	2	5
Kitmisto	2	3	3	3	3	14
Total	5	8	9	10	11	43

Day 4

Total number of days = 10

We know, Pesmisto can have 0,1 cases.

Therefore, to make sum as 10 only possible value = $(3, 3, 1, 3)$

Neighbourhoods	Day 1	Day 2	Day 3	Day 4	Day 5	Total
Levmisto	1	2	3	3	3	12
Tyhrmisto	1	2	3	3	3	12
Pesmisto	1	1	0	1	2	5
Kitmisto	2	3	3	3	3	14
Total	5	8	9	10	11	43

Day 2,

Kitmisto is the only place to have 3 new cases (clue 5)

=>All other neighbourhoods will have 0,1 or 2 cases

Now, to make sum = 8

Also, Pesmisto can't be more then 1

=> only possibility = $(2, 2, 1, 3)$

=> Pesmisto on day 3 will have 0 new cases.

and, Levmisto and Tyhrmisto will have $12 - 3- 3- 2- 1 =3$ new cases on day 3.

Neighbourhoods	Day 1	Day 2	Day 3	Day 4	Day 5	Total
Levmisto	1	2	3	3	3	12
Tyhrmisto	1	2	3	3	3	12
Pesmisto	1	1	0	1	2	5
Kitmisto	2	3	3	3	3	14
Total	5	8	9	10	11	43

Statement A: There were 2 new cases in Tyhrmisto on Day 3
False
Statement B: There were no new cases in Pesmisto on Day 2
False
Hence, both Statement a and Statement b are false