Find the number of six-digit numbers, where the sum of the digits is divisible by 10.
A six-digit number can be represented as 10^5a+10^4b+10^3c+10^2d+10^e+f. The sum of the digits is a+b+c+d+e+f. Since this sum is divisible by 10, it follows that f is even. Thus, we can choose f in 25 ways. For each choice of f, a can be chosen in 9 ways, b in 8 ways, c in 7 ways, d in 6 ways, and e in 5 ways. Therefore, the number of six-digit numbers, where the sum of the digits is divisible by 10, is (9)(8)(7)(6)(5)⋅5/2=15120.
+135 Hello! Not meaning to steal the answer...
But here is another ' web2.0calc ' answer:
https://web2.0calc.com/questions/find-the-number-of-10-digit-numbers-where-the-sum_1
Please refer to that link cointaining valuable answers and explanations!
10 - 99 : 9
100 - 999 : 90
1000 - 9999 : 900
10000 - 99999 : 9000
100,000 - 999,999 : 90,000
1000000 - 9999999 : 900000
10000000 - 99999999 : 9000000
100000000 - 999999999 : 90000000
1000000000 - 9999999999 : 900000000
To find the number of six-digit numbers where the sum of the digits is divisible by 10, we can use the following formula:
N=9×10^4×2
This formula tells us that there are 9 possible choices for the first digit (any digit from 1 to 9), 10 possible choices for each of the next 4 digits (any digit from 0 to 9), and 2 possible choices for the last digit (any digit that makes the sum of the digits divisible by 10).
Therefore, the number of six-digit numbers where the sum of the digits is divisible by 10 is 9×10^4×2=180,000.
Where did you get that formula from? Euler?
s=0;p=0;i=1;n=100000;m=n;n=m;cycle: s=s+ n%10+int(n/10)%10+int(n/100)%10+int(n/1000)%10+int(n/10000)%10+int(n/100000)%10;if(s%10==0, goto end,goto next); end:p=p+1;printp,"==",m;next:s=0;m=m+1;if(m<1000000, goto5, 0);print"Total Num =",p
OUTPUT = 90,000 such 6-digit integers.
We can choose the first five digits in 9×10^4 ways. The sum of the first five digits is of the form 5k or 5k+5 for some integer k. If the sum is of the form 5k, then the last digit can be 0, 5, 2, or 7. If the sum is of the form 5k+5, then the last digit can be 1, 6, 3, or 8. Therefore, there are 9×104×4=360000 such six-digit numbers.
The sum of the digits of a six-digit number is divisible by 10 if and only if the sum of the digits of its last three digits is divisible by 5.
There are 9 choices for the first digit of the number, 10 choices for each of the next three digits, and 2 choices for the last digit (4 or 9).
Therefore, there are 9×10^3×2=180000 six-digit numbers where the sum of the digits is divisible by 10.
Here's a more detailed explanation:
The first digit can be any digit from 1 to 9, inclusive.
The second, third, and fourth digits can be any digit from 0 to 9, inclusive.
The fifth digit must be either 4 or 9, so that the sum of the digits is divisible by 10.
Therefore, there are 9×10^3×2=180000 six-digit numbers where the sum of the digits is divisible by 10.
