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If S is the sum of all three-digit numbers that give a remainder of 5 when they are divided by 11, what is the remainder that S gives on division by 7?
Correct answer is '0'. Can you explain this answer?
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Verified Answer
If S is the sum of all three-digit numbers that give a remainder of 5 ...
Solution: Any number that when divided by 11 gives a remainder of 5 will be of the form 11k + 5.
Since we only need three digit numbers, k will range from 9 to 90, where 11(9) + 5 = 104 and 11(90)+ 5 = 995
The sum of these numbers is an AP with first term = 104, common difference = 11 and number of terms = 82.
S = 41(208 + 891)
Dividing each term in S by 7, the remainder is -1 ( - 2 + 2 ) = 0
Answer: 0
Since we only need three digit numbers, k will range from 9 to 90, where 11(9) + 5 = 104 and 11(90)+ 5 = 995
The sum of these numbers is an AP with first term = 104, common difference = 11 and number of terms = 82.
S = 41(208 + 891)
Dividing each term in S by 7, the remainder is -1 ( - 2 + 2 ) = 0
Answer: 0
Most Upvoted Answer
If S is the sum of all three-digit numbers that give a remainder of 5 ...
Solution:
Finding the Numbers:
Let us first list down all the three-digit numbers that give a remainder of 5 when divided by 11. We can start from 105 and keep adding 11 to it until we reach 999.
105, 116, 127, ..., 994
We can see that each number in the list is 11 more than the previous one. We can also find the number of terms in this list by using the formula for the nth term of an arithmetic sequence:
a_n = a_1 + (n - 1)d
where a_1 = 105, d = 11, and a_n = 994. Solving for n, we get:
994 = 105 + (n - 1)11
n = 82
So, there are 82 three-digit numbers that give a remainder of 5 when divided by 11.
Calculating the Sum:
We can find the sum of these numbers by using the formula for the sum of an arithmetic sequence:
S_n = n/2[a_1 + a_n]
where S_n is the sum of the first n terms, a_1 is the first term, and a_n is the nth term. Substituting the values, we get:
S = 82/2[105 + 994]
S = 82/2[1099]
S = 45,079
Finding the Remainder:
To find the remainder that S gives when divided by 7, we can use the fact that if a number is divisible by 7, then its alternating digit sum (the sum of every other digit) must also be divisible by 7. We can apply this rule repeatedly until we get a single-digit number.
45,079 -> 4 + 5 + 0 + 7 + 9 -> 25 -> 2 + 5 -> 7
Since 7 is divisible by 7, we can conclude that S leaves a remainder of 0 when divided by 7.
Finding the Numbers:
Let us first list down all the three-digit numbers that give a remainder of 5 when divided by 11. We can start from 105 and keep adding 11 to it until we reach 999.
105, 116, 127, ..., 994
We can see that each number in the list is 11 more than the previous one. We can also find the number of terms in this list by using the formula for the nth term of an arithmetic sequence:
a_n = a_1 + (n - 1)d
where a_1 = 105, d = 11, and a_n = 994. Solving for n, we get:
994 = 105 + (n - 1)11
n = 82
So, there are 82 three-digit numbers that give a remainder of 5 when divided by 11.
Calculating the Sum:
We can find the sum of these numbers by using the formula for the sum of an arithmetic sequence:
S_n = n/2[a_1 + a_n]
where S_n is the sum of the first n terms, a_1 is the first term, and a_n is the nth term. Substituting the values, we get:
S = 82/2[105 + 994]
S = 82/2[1099]
S = 45,079
Finding the Remainder:
To find the remainder that S gives when divided by 7, we can use the fact that if a number is divisible by 7, then its alternating digit sum (the sum of every other digit) must also be divisible by 7. We can apply this rule repeatedly until we get a single-digit number.
45,079 -> 4 + 5 + 0 + 7 + 9 -> 25 -> 2 + 5 -> 7
Since 7 is divisible by 7, we can conclude that S leaves a remainder of 0 when divided by 7.
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If S is the sum of all three-digit numbers that give a remainder of 5 when they are divided by 11, what is the remainder that S gives on division by 7?Correct answer is '0'. Can you explain this answer?
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