Problem: 2/7 (4+c)=6
1. 7/2 x 2/7 x (4+c)=7/2 x 6
2. 4+c=7 x 3
3. 4+c=21
4. c=21-4
Answer: c=17
Answer:
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Step-by-step explanation:
its 1 am for me but I'm just bored and answering peoples questions lol but it's weird that they are doing school work now I was thinking that but was like oh well lol
There is a certain way to find the answer for your question. First, you must be familiar with the divisibility rule for 11: the difference between the sums of even digits and odd digits should either be 0 or 11 only. For example, 1430, here the odd digits are 4 and 0 (which are the 3rd and 1st digits); the even digits on the other hand are 1 and 3, placed as the 4th and 2nd digits respectively.
Adding the sums of odd and even digits will give 4 (4+0) and 4 (3+1). Now, the difference of these sums is obviously equal to 0. Therefore, 1430 is divisible by 11. Same case will hold for numbers whose sums of odd and even digits is 11.
Moving on, now let us try to find the biggest 6-digit number divisible by 11. Intuitively, the last digit should be "9" - the hundred thousands digit - for it is the biggest number among numbers from 0-9. Furthermore, since there is no restriction as to whether the same number could or could not repeat, we can still use "9" all throughout the remaining 5 digits (ten thousands down to ones digit), giving the greatest possible 6-digit number, 999,999.
To check if the number we have obtained, 999,999, is divisible by 11, we need to use the divisibility rule for 11. Sum of even digits is 18 (9+9+9); sum of odd digits is 18 (9+9+9) as well. The difference of the sums is 0. Therefore, 999,999 is divisible by 11, hence the greatest 6-digit number multiple of 11.
As for the 6-digit number whose digits' sum is 40 and multiple of 11, we can assume its first 4 digits to be as 999,9** - four 9s is already equal to 36 - we only need to find two numbers whose sum is 4 to make the sum's total be 40. These remaining numbers should also be divisible by 11 since 9999 is already divisible by 11. The remaining numbers should only be 2 and 2 since the sum of its digit is 4, and that if we put them together, 22, a multiple of 11. This is the only combination of numbers that will give a sum of 4 while maintaining its structure a multiple of 11.
At last, the greatest 6-digit number whose digits' sum equal to 40 and is a multiple of 11 is 999,922. True, for 999,922 / 11 is equal to 90,902.
