Question

A number consist of three consecutive digits. If the sequence of the number are reversed then the difference between the new number and old number is equal to 33 times the largest digit. the sum of the digits of the number is​

Answer 1

Step-by-step explanation:

let 3 digits = a-1, a and a+1

number = 100(a-1) + 10a + a+1 = 100a - 100 + 10a + a + 1 = 111a - 99

if digits are reversed then the number = 100(a+1) + 10a + a-1 = 100a + 100 + 10a + a - 1 = 111a + 99

(111a + 99) - (111a - 99) = 33×(a+1)

=> 111a + 99 - 111a + 99 = 33a + 33

=> 198 = 33a + 33

=> 33a = 198 - 33 = 165

=> a = 165/33 = 5

=> a-1 = 4 and a+1 = 6

the digits are = 4, 5 and 6

sum of the digits are = 4+5+6 = 15

Answer 2

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