Question

A number consists of four digit having 8 in its unit place.If the digit in extreme left is shifted to the immidiate right of the unit place, keeping all the other number as they are, the new number formexeeds the original number by 1305.Find the original number​

Answer 1

Let the number be abc8
Its value = 1000a + 100b + 10c + 8

New number formed = bc8a
Its value = 1000b + 100c + 80 + a

Given (1000b + 100c + 80 + a) = (1000a + 100b + 10c + 8) + 1305
=> 1000b + 100c + 80 + a = 1000a + 100b + 10c + 1313
=> 900b + 90c - 999a = 1233
=> 100b + 10c - 111a = 137 ---equation (I)

Last digit of 100b is 0. Similarly last digit of 10c is also 0.
Therefore last digit of 111a must be 3 for the above equation to be true because last digit of the result 137 is 7

only a=3 satisfies this condition. Therefore we get a=3

a=3 put in Equation (I)
100b + 10c - 111×3 = 137
=> 100b + 10c - 333 = 137
=> 100b + 10c = 470
=> 10b + c = 47 .......equation (II)

Last digit of 10b is 0. So, for the above equation to be true, c must be 7 because last digit of 47 is 7

c=7 put in equation (II)
10b + 7 = 47
10b = 40
b = 4

Therefore the original number is 3478

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