im thinking of a number between 3 299 875 and 3 300 015 it is a multiple of 10 but not a multiple of 100 list the 12 possible numbersno

Expert Answers
hustoncmk eNotes educator| Certified Educator

 All of the multiples of 10 will end with a zero.  You want to exclude any that are multiples of 100, or end with 2 zeroes.

You need to begin by finding the first multiple of 10 that comes after 3,299,875.

The first number would be 3,299,880.  It has only 1 zero at the end, so it counts towards the 12 that you need. From there, you need to begin counting by 10s and look at each answer to see if it has 1 or 2 zeroes at the end.

3,299,880      (from above)

3,299,890     1 zero at the end, it counts

3,299,900     2 zeroes at the end, this does not count

3,299,910

3,299,920

3,299,930

3,299,940

3,299,950

3,299,960

3,299,970

3,299,980

3,299,990

3,300,000        2 zeroes at the end, this answer does not count

3,300,010        1 zero, this counts

You would need to stop here because the next number in the sequence 3,300,020 is higher than the maximum number in the range you are looking at.

hala718 eNotes educator| Certified Educator

3,299,875     and  3,300,015

numbers between that are multiple of 10 and not 100:

include all numbers that are multiply of 10 except (3,299,900 and 3,300,00) because they are multiples of 100.

3,299,880

3,299,890

3,299,910

3,299,920

3,299,930

3,299,940

3,299,950

3,299,960

3,299,970

3,299,980

3,299,990

3,300,010

 

krishna-agrawala | Student

Any number that is divisible by 10 but not by 100 will have a 0 in the units place but a number other that 0 in the tens place.

Thus the first 12 numbers between 3,299,875 and 3,300,015 that meet the required condition are:

3,299,880

3,299,800

3,299,910

3,299,920

3,299,930

3,299,940

3,299,950

3,299,960

3,299,970

3,299,980

3,299,990

3,299,880

3,300,110

neela | Student

To find nmbers 12 numbers between 3 299 875 and 3 300 015, which should be a multiple of 10 but not mutiple of 100.

Between 3299875 and 3300015 , there are (3300015 - 3299875) - 1= 140-1 = 139 integers.

The first number divisible by 10 = 3299880 = say T1.

The last number divisible by 10 = 3300010 = Tn say

The common diffrence = 10.

Tn = T1 +(n-1)d.  d= 10. Therefore  n-1 = (Tn -T1)/10 = (32900010-3299880)/10 = 13

Therefore , n-1 = 13. Or n = 14 of which two number 3299900 and 3300000 are not eligible (due to the reason that  the intervening  number 3299900 and 3300000 being a number divisible by 100 is not in the set)

So, 3299880 , 3299890, 3299910 , 3299920 , 3299930, 3299940, 3299950 , 3299960,  3299970, 3299980, 3299990, 3300010 are the 12 numbers required in the set according to the give conditions.