When the sum of one set of alternate digits of a 6-digit number is added to the number itself, we get a 7-digit number which is a multiple of 10. What is the difference between the largest and smallest number that can be formed that satisfies the above condition?
Let the 6-digit number be $ABCDEF$. So the $10^5A + 10^4B+10^3C + 10^2D +10E+F + (10C+E)$ must be divisible by $10$. Simplifying gives $100001A + 10001B + 1001D + 101E + 11F = 10(-10C-E)\Longrightarrow 9091A + 910B + 91D + 10E + F = -10C - E$. Note that if this equation holds, then $0\le -10C - E < 1000000$. Thus $-10C - E$ can be between $0$ and $999999$.
If $E=0$, then $0\le -10C < 999999 \Longrightarrow 0 \le C \le 99999$, and $90910A + 910B + 91D + F = -10C \Longrightarrow -9091A - B - 91D - F = C$. Thus we see that $C$ goes through four or five digit numbers. The largest and smallest values of $C$ will come from the largest and smallest possible values of $-9091A - B - 91D - F.$ In order to maximize $C$, we should minimize $-9091A - B - 91D - F.$ We minimize by making each digit as large as possible in the numbers to the left of the digit we are working on. Thus if we start from left to right, we choose $0$ for each of $A,B,$ and $D$ so that each of those variables contributes nothing to the sum. Thus we are trying to minimize $91F$. The smallest digit is $0$ and the largest digit is $9$, so the smallest possible value of $F$ is $0$ and the largest possible value is $9$, giving $0\le C \le 99990$ as an answer for each case.
If $E=1$, then $90001A + 9001B + 901D + 100F = -10C$. We cannot choose values to minimize $C$ here since adding anything to $C$ would simply increase the sum, so to figure out what works or what doesn't we must figure where $-10C\le 999999$. Doing this gives $0 \le C\le 99999$. Thus each value of $C$ between $0$ and $99999$ works.
If $E=2$, then $89001A+8901B + 981D + 12F = -10C$. Minimizing as above gives us $-8910A - 891B - 981D - 2F = C$. maximimzing $F$ means minimizing $C$, so we choose $9$ for each of $A,B$ and $D$, and $0$ for $F$. Thus $0 \le C \le 8901$. Doing this for each digit gives no extra values of $C$ that haven't appeared somewhere else.
Notice that $E$ being $3,4,5,6,7,8,$ or $9$, yields values of $C$ that we can obtain as $0\le C \le 99999$. Since $99990$ is the maximum value of $C$ that we can obtain, and $8901$ is the maximum value of $C$ that we can obtain without using $9$, the largest value of $C$ that satisfies the condition $0 \le -10C - E < 1000000$ is $99990$. The smallest value of $C$ that satisfies the condition $0\le -10C - E < 1000000$ is $0$. Thus the difference between the largest and smallest value of $C$ is $99990 - 0 = \boxed{99990}$. ~sweetMango77
To find the difference between the largest and smallest number that satisfies the given condition, let's break down the problem step by step:
Step 1: Determine the possible values for the alternate digits of the 6-digit number. Since the sum of these digits must result in a number that is a multiple of 10, there are only a few possibilities for the sum:
- If the sum is 0, then all the alternate digits must be 0.
- If the sum is 10, then all the alternate digits must be 1.
- If the sum is 20, then all the alternate digits must be 2.
- If the sum is 30, then all the alternate digits must be 3.
- And so on, until the sum is 90, which means all the alternate digits must be 9.
Step 2: Generate the largest and smallest numbers using the determined values.
- For the largest number, we start with the digit 9, alternating with the digit 0. So the largest number is 9,090,90.
- For the smallest number, we start with the digit 0, alternating with the digit 9. So the smallest number is 0,909,09.
Step 3: Calculate the difference between the largest and smallest numbers.
Difference = Largest number - Smallest number
Difference = 9,090,90 - 0,909,09
Difference = 8,181,81
Therefore, the difference between the largest and smallest number that satisfies the given condition is 8,181,81.