The digts of a positive no. of three digits are
in AP and their sum
is 15. The number
Obtained by reversing the digits is 594 less than the
original number. Find the reimber.
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The digts of a positive no. of three digits are
in AP and their sum
is 15. The number
Obtained by reversing the digits is 594 less than the
original number. Find the reimber.
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Answer:
Original number - 852
Step-by-step explanation:
Let the digits of the number be a-d, a, a+d
The sum of the digits are - 15
a-d+a+a+d = 15
3a = 15 => a = 5
Original number - 100(a+d) + 10a + a-d
New number obtained by reversing the digits - 100(a-d)+ 10 a + a + d
100(a+d) + 10a + a-d - 594 = 100(a-d)+ 10 a + a + d
100a + 100 d + 10 a + a-d - 594 = 100 a - 100 d +10 a+ a+d
200d-2d = 594
198d = 594
d = 3
The digits are - 5-3, 5, 5+3 = 2, 5, 8
original number = 100(5+3) + 10(5) + (5-3) = 800 + 50 + 2 = 852.
New number obtained by reversing the digits - 852 - 594 = 258
Kyu....................