How many positive multiples of 7 that are less than 1000 end with the digit 3?

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.

you probably already have the answer but anyways it's b. 15.0

step-by-step explanation:

6 equilateral triangles