Answer:
14
Step-by-step explanation:
Ordinarily, a quick multiplication of 7 by other integers up to 10 indicates that only 7*9 yields 63, i.e ends with 3 as required.
Thus the set of possible multiples of the integer 7 ending with the digit 3 will form the arithmetic series with the first term being Ao = 63 and the common difference being d= 7*10= 70. That is we can see the series in details....
the nth term could be evaluated from the formular
An=Ao+(n-1)d (1)
The series could be explicitly depicted as follows:
9*7=63= 63+70*0
(10+9)*7=133 = 63+70*1
(20+9)*7=203=63+70*2
(30+9)*7=273=63+70*3
.................................
(130+9)*7=973=63+70*13
The last 'n' corresponding to the problem statement could be evaluated from equation (1), assuming An=1000:
1000=63+(n-1)*70
1000-63=70(n-1)
937/70=13.38=n-1
n=14.38
Thus the number of possible multiples of 7 less than 1000 ending with digit 3 will be 14.
Check: 7 times 142 is 994, so there are exactly 142 positive multiples of 7 less than 1000.
One tenth of these, ignoring the decimal fraction, end with a digit of 3.
