Let x be a four-digit number with exactly three consecutive digits being same and is a multiple of 9. How many such x?S are possible?
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First method :-
Let the four digit number be ′aaab′ or ‘baaa′.
Since, the number has to be a multiple of 9, therefore 3a+b should be either 9, 18 or 27.
Case I: 3a+b=9
Possible cases are: (1116,6111,2223,3222,3330,9000)
Case II: 3a+b=18
Possible cases are: (3339,9333,4446,6444,5553,3555,6660)
Case III: 3a+b=27
Possible cases are: (6669,9666,8883,3888,7776,6777,9990)
Hence, total number of cases is 20.
Hope it helps you
Please make me as brainliest
Let the four digit number be ′aaab′ or ‘baaa′.
Since, the number has to be a multiple of 9, therefore 3a+b should be either 9, 18 or 27.
Case I: 3a+b=9
Possible cases are: (1116,6111,2223,3222,3330,9000)
Case II: 3a+b=18
Possible cases are: (3339,9333,4446,6444,5553,3555,6660)
Case III: 3a+b=27
Possible cases are: (6669,9666,8883,3888,7776,6777,9990)
Hence, total number of cases is 20.
Hope it helps you
Please make me as brainliest
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