If in ‘4798267583’ each digit at the odd place from the left end remain unchanged while each digit at the even place from the left end is increased by one, then how many such pairs of the digits are there in the number thus formed, which have as many digits in the number both forward and backward direction as they have between them in the number system?
Given number: 4798267583 After rearrangement we get: 4899277593 So, there are 8 such pairs i.e., 89, 25, 79, 35, 37, 57, 59 and 79.
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