Clock digits

The best is split in two, thinking only the first 2 digits of the time we have the only numbers (00 through 23) in which both digits are pairs are (00,02,04,06,08,20,22); logo are seven possibilities.

Thinking now in minutes, we have (0, 2, 4, 6, 8, 20, 22, 24, 26, 28, 40, 42, 44, 46, 48) where the number 60 does not enter because I would have passed for 00. So we have 15 possibilities for the minutes. Thus, 7 x 15 = 105

Okay, let's do a tree-and-leaf plot.

hours:

-0| 0, 2, 4, 6, 8

-2| 0, 2

Remember not to include 24, and that 0's are even in any place value.

minutes:

-0| 0, 2, 4, 6, 8

-2| 0, 2, 4, 6, 8

-4| 0, 2, 4, 6, 8

In order to find all the combinations, multiply the total number of minute options by the total number of hour options.

(Note: Just count the digits on the right side of the "|" only for the totals.)

For the HH, the number which satisfy this condition are 00, 02, 04, 06, 08, 20 and 22.

For the tens digit of the minutes, 0, 2, 4 are even. For the ones digit, 0, 2, 4, 6, 8.

We can multiply the possibilities for each section, which gives us 7 * 3 * 5 = 105.

$1 * 2 * 15 + 1 * 5 * 15 = 105$

look at the hours! between 00 and 09 there are 5 cases with a pair of even digits. and between 10 and 19 there are none. 20, and 22 have a pair of even digits.

7 of 24 hours.

15 of the 60 minutes in an hour have 2 even digits.

7*15 = 105 times / day

Quando o primeiro dígito é zero, temos 5 possibilidades para o segundo dígito (0,2,4,6,8). Para o terceiro dígito temos 3 possibilidades (0,2,4) e para o quarto também temos 5 possibilidades. Neste caso, temos (1 x 5 x 3 x 5) 75 possibilidades. Quando o primeiro dígito é 2, há dois possíveis números para o 2.º dígito (0,2). Para os dígitos dos minutos, as possibilidades são 3 e 5. Assim, há (1 x 2 x 3 x 5) 30 possibilidades. Ao todo, todos os dígitos do relógio são pares (75 + 30) 105 vezes.

Since all-even digits can only appear 15 times in MM, i.e 00.02,04,06,08,20,22,24,26,28,40,42,44,46,48, and all-even digits can only appear 7 times, i.e 00.02,04,06,08,20,22, so 15*7=105.

There are 15 times in an hour that starts with 0 or 2 and is divisible by 2 that contains only even numbers. There are 7 of these hours: 00, 02, 04, 06, 08, 20, and 22. 15 x 7=105.

In every 10 numbers 5 are even.Minutes range from 00 to 59 so the ten's digit is even for 0,2,4(3 numbers) and for every even ten's digit the unit's hand is even for 0,2,4,6,8(5 numbers).So multiplying we get 3.5=15.Hour hand ranges from 00 to 23.The ten's hand is even for 0(1.5=5).For 2 we have only 0 and 2 as the unit's hand can't exceed 4 in the case of 2 as max value is 23.So we have only 0 and 2(1.2)=2.Adding these 5+2=7.Multiplying with 15 as these are combinations we get 15.7=105

The possible numbers for HH are $00,02,04,06,08,20,22$ and the possible numbers for MM are $00,02,04,06,08,20,22,24,26,28,40,42,44,46,48$ so the possible permutations are $7 \times 15 = 105$

(1x5x3x5)+(1x3x3x5)=105

At first, divide in 2 cases. The first case, all combinations begin with 0. In the second and fourth space there are 5 possibilities (0,2,4,6,8), already in the third space, can only be (0,2,4) as the maximum for 59 minutes. When arriving in 60 minutes, the time is reset and the hour increases by one unit. For the second case (which starts with 2) the second space admits only 0 and 2. The minutes are equivalent to the first case. To find the amount of times that all the digits are pairs, it is necessary to multiply the quantity of possibilities of each case and sum the result of two cases.