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Step-by-step Solution
Step 1: Understand the Divisibility by 55
A number is divisible by 55 if and only if it is divisible by 5 and by 11.
• Divisible by 5: The last digit of the number must be 5 or 0. Here, since the allowed digits are 2, 3, 4, 5, 6 and repetition is not permitted, the last digit must be 5.
• Divisible by 11: For a number \overline{a b c d} , the difference between the sum of the digits in the odd positions and the sum of the digits in the even positions must be 0 or a multiple of 11. In symbolic form,
(a + c) - (b + d) = 0 \text{ or } \pm 11, \pm 22, \dots
Step 2: Determine the Possible Number of Digits
We need the natural numbers lying between 1012 and 23421, formed from digits {2,3,4,5,6} with no repetition.
A 3-digit number is too small (below 1012).
A 5-digit number from the given digits would exceed 23421 or not lie in the given range (for instance, 23456 is already above 23421 or repeats digits not allowed). So, no 5-digit number is possible in the required range.
Hence, only 4-digit numbers are possible candidates.
Step 3: Form 4-Digit Numbers Divisible by 5 and 11
Since the number ends in 5, let our 4-digit number be \overline{a b c 5} . The remaining digits must be chosen from {2,3,4,6} without any repetition.
Next, apply the divisibility rule for 11:
(a + c) - (b + 5) = 0 \text{ or a multiple of 11.}
By testing each permissible combination of a, b, and c from {2,3,4,6}, we find the valid sets (a, b, c) that satisfy the 11-divisibility condition. The valid triplets are:
(6, 4, 3)
(3, 4, 6)
(2, 3, 6)
(6, 3, 2)
(3, 2, 4)
(4, 2, 3)
Each triplet corresponds to a unique 4-digit number \overline{a b c 5} , giving us a total of 6 possible numbers.
Step 4: Verify the Range
All these 4-digit numbers are clearly larger than 1012 and smaller than 23421, so they lie in the desired interval.
Step 5: Conclude the Result
Hence, there are 6 natural numbers between 1012 and 23421 that can be formed using the digits {2,3,4,5,6} without repetition and are divisible by 55.
