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The number of 7-digit numbers which are multiples of 11 and are formed using all the digits 1, 2, 3, 4, 5, 7 and 9 is _____. (In integers)
Correct answer is '576'. Can you explain this answer?
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The number of 7-digit numbers which are multiples of 11 and are formed...
Solution:
To form a number which is a multiple of 11, we add the alternate digits and find their difference. If the difference is 0 or a multiple of 11, then the number is a multiple of 11.
• Let's find the number of ways of selecting 7 digits from the given 7 digits.
7 digits can be selected in ^7C_7 ways = 1 way.
• The 7 digits can be arranged in 7! ways.
• The digits at the odd places (1st, 3rd, 5th and 7th places) can be selected in 4! ways.
• The digits at the even places (2nd, 4th and 6th places) can be selected in 3! ways.
• Now, let's fix the first digit. We can select any one of the 7 digits.
• Let's form the number by adding the alternate digits and find their difference. If the difference is 0 or a multiple of 11, then the number is a multiple of 11.
• If the first digit is 1, then the difference of the sum of alternate digits is (2+4+9) - (1+3+5+7) = 0. Hence, we need to select the even places (2nd, 4th and 6th places) in such a way that the sum of digits at these places is equal to the sum of digits at the odd places.
• If the first digit is 2, then the difference of the sum of alternate digits is (1+3+5+7) - (2+4+9) = -9, which is not a multiple of 11.
• If the first digit is 3, then the difference of the sum of alternate digits is (1+5+9) - (3+4+7) = -9, which is not a multiple of 11.
• If the first digit is 4, then the difference of the sum of alternate digits is (1+3+7) - (4+5+9) = -9, which is not a multiple of 11.
• If the first digit is 5, then the difference of the sum of alternate digits is (1+4+9) - (5+3+7) = 0. Hence, we need to select the even places (2nd, 4th and 6th places) in such a way that the sum of digits at these places is equal to the sum of digits at the odd places.
• If the first digit is 7, then the difference of the sum of alternate digits is (1+5+9) - (7+3+4) = 0. Hence, we need to select the even places (2nd, 4th and 6th places) in such a way that the sum of digits at these places is equal to the sum of digits at the odd places.
• If the first digit is 9, then the difference of the sum of alternate digits is (1+3+7) - (9+4+5) = -9, which is not a multiple of 11.
• Hence, the number of 7-digit numbers which are multiples of 11 and are formed using all the digits 1, 2, 3, 4,
To form a number which is a multiple of 11, we add the alternate digits and find their difference. If the difference is 0 or a multiple of 11, then the number is a multiple of 11.
• Let's find the number of ways of selecting 7 digits from the given 7 digits.
7 digits can be selected in ^7C_7 ways = 1 way.
• The 7 digits can be arranged in 7! ways.
• The digits at the odd places (1st, 3rd, 5th and 7th places) can be selected in 4! ways.
• The digits at the even places (2nd, 4th and 6th places) can be selected in 3! ways.
• Now, let's fix the first digit. We can select any one of the 7 digits.
• Let's form the number by adding the alternate digits and find their difference. If the difference is 0 or a multiple of 11, then the number is a multiple of 11.
• If the first digit is 1, then the difference of the sum of alternate digits is (2+4+9) - (1+3+5+7) = 0. Hence, we need to select the even places (2nd, 4th and 6th places) in such a way that the sum of digits at these places is equal to the sum of digits at the odd places.
• If the first digit is 2, then the difference of the sum of alternate digits is (1+3+5+7) - (2+4+9) = -9, which is not a multiple of 11.
• If the first digit is 3, then the difference of the sum of alternate digits is (1+5+9) - (3+4+7) = -9, which is not a multiple of 11.
• If the first digit is 4, then the difference of the sum of alternate digits is (1+3+7) - (4+5+9) = -9, which is not a multiple of 11.
• If the first digit is 5, then the difference of the sum of alternate digits is (1+4+9) - (5+3+7) = 0. Hence, we need to select the even places (2nd, 4th and 6th places) in such a way that the sum of digits at these places is equal to the sum of digits at the odd places.
• If the first digit is 7, then the difference of the sum of alternate digits is (1+5+9) - (7+3+4) = 0. Hence, we need to select the even places (2nd, 4th and 6th places) in such a way that the sum of digits at these places is equal to the sum of digits at the odd places.
• If the first digit is 9, then the difference of the sum of alternate digits is (1+3+7) - (9+4+5) = -9, which is not a multiple of 11.
• Hence, the number of 7-digit numbers which are multiples of 11 and are formed using all the digits 1, 2, 3, 4,
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Community Answer
The number of 7-digit numbers which are multiples of 11 and are formed...
Sum of all given numbers = 31
The difference between odd and even positions must be 0, 11, or 22, but 0 and 22 are not possible.
∴ Only difference 11 is possible.
This is possible only when 1, 2, 3, or 4 is filled at odd positions in some order and the remaining in other order.
There are similar arrangements of 2, 3, 5 or 7, 2, 1 or 4, 5, 1 at even positions.
∴ Total possible arrangements = (4! × 3!) × 4 = 576
The difference between odd and even positions must be 0, 11, or 22, but 0 and 22 are not possible.
∴ Only difference 11 is possible.
This is possible only when 1, 2, 3, or 4 is filled at odd positions in some order and the remaining in other order.
There are similar arrangements of 2, 3, 5 or 7, 2, 1 or 4, 5, 1 at even positions.
∴ Total possible arrangements = (4! × 3!) × 4 = 576
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The number of 7-digit numbers which are multiples of 11 and are formed using all the digits 1, 2, 3, 4, 5, 7 and 9 is _____. (In integers)Correct answer is '576'. Can you explain this answer?
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