Question
Let S be the set of all four-digit positive integers whose digits are 3, 5, 7 and 9, with no digit repeated in the same integer. Calculate the remainder when the sum of all of the integers in S is divided by 9.
Answer: Option C
:
C
There are 24 numbers formed with the four given numbers.
Six of these numbers have a 3 in the 1000s position, six have a 5 in the 1000s position, six have a 7 in the 1000s position and six have a 9 in the 1000s position.
The same can be said about the distribution of numbers in the 100s, 10s and units positions.
Therefore, the sum of the 24 numbers is
6(3 + 5 + 7 + 9)(1000) + 6(3 + 5 + 7 +9)(100) + 6(3 + 5 + 7 + 9)(10) + 6(3 + 5 + 7 + 9) = 159984
The remainder is 0 when 159984 is divided by 9.
Shortcut -
Sum of the digits will be 24 in each case, which will leave a remainder 6 when divided by 9. there are 4! numbers possible. Remainder when 4!×6 is divided by 9 is zero. Hence option (c).
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:
C
There are 24 numbers formed with the four given numbers.
Six of these numbers have a 3 in the 1000s position, six have a 5 in the 1000s position, six have a 7 in the 1000s position and six have a 9 in the 1000s position.
The same can be said about the distribution of numbers in the 100s, 10s and units positions.
Therefore, the sum of the 24 numbers is
6(3 + 5 + 7 + 9)(1000) + 6(3 + 5 + 7 +9)(100) + 6(3 + 5 + 7 + 9)(10) + 6(3 + 5 + 7 + 9) = 159984
The remainder is 0 when 159984 is divided by 9.
Shortcut -
Sum of the digits will be 24 in each case, which will leave a remainder 6 when divided by 9. there are 4! numbers possible. Remainder when 4!×6 is divided by 9 is zero. Hence option (c).
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