Math Problem (Level 3) : Two machines displaying numbers on the rule (III2)

There is the machine A which displays the number of double-digit figures, and when a switch is turned on, a number will be displayed as the following rule for every second.

[Rule]
As for tens digit, it changes every second in order of 1, 2, 3, 4, 5, 6, 7, 8 and after 8 it return to 1 again. It changes in a same cycle.

As for ones digit, it changes every second in order of 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 and after 0 it return to 1 again. It changes in a same cycle.

For example, when a switch is turned on, one second after 11 will be displayed, two seconds after 22 will be displayed, three seconds after 33 will be displayed, nine seconds after 20 will be displayed and eleven seconds after 31 will be displayed.

Answer the following questions.

(1) After turning on a switch, in how many seconds is it that 73 will be displayed for the first time ?

(2) If each number of double-digit figures which can be displayed by this machine A is added altogether, how much is it ?

(3) There is the machine B which displays the number of triple-digit figures according to the rule as machine A. 


It displays both tens digit and ones digit as same as the machine A. 

As for hundreds digit, it changes every second in order of 7, 6, 5, 4, 3, 2, 1 and after 1 it return to 7 again. 

It changes in a same cycle.

For example, when a switch is turned on, one second after 711 will be displayed, two seconds after 622 will be displayed and eight seconds after 788 will be displayed.

(a) After turning on a switch, in how many seconds is it that 711 is displayed on the 2nd times?

(b) After turning on a switch, in how many seconds is it that 773 is displayed for the first time?



Answer
(1) 23 seconds after
(2) 1980
(3) 
(a) 281 seconds after
(b) 183 seconds after

Solution
 (1) The time when 7 comes to tens digit is in 7 seconds, 15 seconds, and 23 seconds that is the number added the multiple of eight to seven. 

The time when 3 comes to ones digit is in 3 seconds, 13 seconds, and 23 seconds that is the number added the multiple of ten to three. 

Therefore, as you can find, it turns out that it is set to 73 in 23 seconds. 

(2) The number of tens digit is the repetitions from 1 to 8 and that of ones digit is the repetitions from 1 to 0. 

The number of double digit figures displayed by this machine is 40 which is the least common multiple of 8 and 10. 

There are five cycles from 1 to 8 and four cycles from 1 to 0 in 40. 

The sums from 1 to 8 is 36 and that from 1 to 9 is 45. 

With careful attention to the numbers from 1 to 8 being allocated to tens digit, calculate the summation. 

Then it is 36 × 5 × 10 + 45 × 4 × 1 = 1800 + 180 = 1980. 

(3)(a) It turns out that 40 numbers are displayed and the machine A returns first in 40 seconds. 

Since seven numbers are added to hundreds digit as for the machine B, the same number is displayed every 280 seconds of the least common multiple of 7, 8, and 10. 

Therefore, 711 displayed in 1 second is displayed on the 2nd time after 1+280 = 281 seconds.

(b) Since it turned out that 73 is displayed for the first time in 23 seconds according to (1), it is 23+40 =63 seconds after that 73 is displayed on the 2nd time. 

The 3rd time is 63+40=103 seconds after. 

7 of hundreds digit being displayed on the 1st time is in 1 second, and the 2nd time is 1+7= 8 seconds after. 

In order to find the first time for 773 to be displayed, the time when 73 is displayed should just be same as the time when 7 of hundreds digit is displayed. 

That is, the number which subtracted 1 from the time when 73 is displayed should be multiples of 7. 

Then the time when 7 is displayed is investigated. 

Then numbers of 23,63,103,143 are not applied and 183 is applied for the first time and the 183 seconds after is to be an answer.