The digits of a positive integer having three digits are in A.P. sum of the digits is
15. The number obtained by reversing the digit is 594 less than original number
find the number.
Answers
Answer:
963
Step-by-step explanation:
let x,y and z be the 3 digits of the required number
=> the number is 100x+10y+z
given that the 3 digits are in a.p
hence consider the 3 digits be
(a-d), a, (a+d) they are also equivalent to x,y &z resply -------- (1)
also, given that a-d + a + a+d = 15
=> 3a = 15
=> a = 15/3 = 5
also given that 100x+10y+z - (100z+10y+x ) = 594
=> 100x+10y+z - 100z - 10y - x = 594
=> 99x - 99z = 594
=> 99(x-z) = 99(6)
=> x -z = 6
ie, a-d - (a+d) = 6 ( using (1) )
=> a -d -a -d = 6
=> -2d = 6
=> d =6/(-2) = -3
substitute the value of a & d in (1)
we say the digits are (6- (-3)), 6 , (6+(-3))
=> 6+3 , 6 , 6-3
=> 9,6,3
=> the number is 100(9)+10(6)+3 = 900+60+3 = 963
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verification:
condn: The number obtained by reversing the digit is 594 less than original number.
original number = 963
reversed number = 369
difference = 963-369 = 594
hence verified
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