6 x 6 x 6 x 3 = 648 6 because the first digit can be any of the numbers 6 again, because the second digit can be any of the numbers 6, again, because the third digit can be any of the numbers 3, because the fourth/last digit can only be 2, 4, or 6
4 digit number having 0 at last position = 9*8*7 =504 4 digit even number having 2 or 4 or 6 or 8 at last position = 8*8*7*4 =1792 (in first position, 0 cannot be fix and in last position one of digit 2,4,6,8 will be fix, so, number of choice left for first position is 8) So, total 4 digit even number possible = 504 + 1792 = 2296
This is a problem involving combinations. We know we need a 4 digit number. So I'll write 4 blank spaces, to symbolize each digit. ___ ___ ___ ___ The number also has to be even. That means that the last digit must be even. (2, 4, 6, 8, 0) Your list of numbers is 1, 2, 3, 5, 6, 8, 0. How many numbers from this list will work in the last digit place? 2, 6, 8, 0. -- 4 numbers. ___ ___ ___ _4_ For the third digit, any number will work. We have 7 choices. ___ ___ _7_ _4_ for the 2nd digit, the same applies. any number from your list will work. ___ _7_ _7_ _4_ For the first digit, only 6 will work. if you put a 0 in the first digit place, it becomes a 3 digit number. _6_ _7_ _7_ _4_ Now, all we do it multiply these numbers together. 6 * 7 * 7 * 4 = 1176. This means, we can create 1176 unique 4 digit numbers that are even with the list of numbers available. 1000 1002 1006 1008 1010 1012 etc.
The 6 is in the ones column.
Look at the first digit after the first non-zero digit;if it is 5 or more, add one to the first digit, otherwise keep the first digit as it is;replace all the digits after the first digit by 0s.6592:First digit is 6, first digit after it is 5is 5 greater than or equal to 5 - yes, so add 1 to the first digit: 6 + 1 = 7replace the 592 by 000, to give 7000
6
The last digit: the 6.The last digit: the 6.The last digit: the 6.The last digit: the 6.
All powers of 6 end in 6, just like all powers of 5 end in 5.
The answer will depend on how many digits there are in each of the 30 numbers. If the 30 numbers are all 6-digit numbers then the answer is NONE! If the 30 numbers are the first 30 counting numbers then there are 126 combinations of five 1-digit numbers, 1764 combinations of three 1-digit numbers and one 2-digit number, and 1710 combinations of one 1-digit number and two 2-digit numbers. That makes a total of 3600 5-digit combinations.
its odd. u can find it out by just seeing that the last digit is not 0, 2, 4, 6, 8. if the last digit is either one of the above mentioned number then its even.
6
In this case with 234.896 the last digit, '6' determines the precision since in is the last non-zero digit.
Yes, the last digit "6" is even.
The answer is 62. 6+2+8. 6 x 2=12. it could be 26 or 62. But becuase it's inbetween 30 and 60, it has to be 62.
If the last two digits are divisible by 4. This is equivalent to: Last digit = 0, 4, 8 and the digit before (in tens place) is even or Last digit = 2 or 6 and the digit before (in tens place) is odd.
1917 Winchester 30-06 has a 6 digit serial number I have a Winchester 30-06 w/a 5 digit serial number 485--
The digit with which a multiple of 4 ends depends on the last digit of the other factor. If the last digit is a zero, the product ends with zero; if the last digit is a 1, the product ends with 4; etc. The only options for the last digit of the product are 0, 2, 4, 6, 8.