the digits of a positive number of three digits are in A.P and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number.....
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Answer:
Explanation:
Solution :-
Let the three digit of three digit number be a - d, a , a + d
Their Sum = 15
⇒ a - d + a + a + d = 15
⇒ 3a = 15
⇒ a = 15/3
⇒ a = 5
Required three digit number = 100(a - d) + 10a + a + d
= 100a + 100d + 10a + a + d
= 111a - 99d
Number obtained by revering digit = 1000(a + d) + 10a + a - d
= 100a + 100d + 10a + a - d
= 111a + 99d
According to the Question,
⇒ 111a + 99d = 999a - 99d - 594
⇒ 594 = 111a - 99d - 111a - 99d
⇒ 594 = - 198d
⇒ d = - 594/198
⇒ d = - 3
Number = 111a - 99d
= 111 × 5 - 99 × (- 31)
= 555 + 297
= 852
Hence, the required number is 852 or 258.
Verified answer
Let the digits be a, a+d and a+2d. Their sum i.e. 3a + 3d = 15 => a + d = 5 => second digit must be 5.You an easily guess the answer: the number has to be of descending order as difference between this and reversed one is positive. So your choices are 654, 753, 852 and 951. Clearly, 852 is the one you're looking for. Let's formally solve it anyway.
100a + 10x5 + (a+2d) = given no.
a + 10x5 + 100(a+2d) = reversed no>
Subtracting,
99a - 99(a+2d) = 594 (gn.)
a - (a+2d) = 6
-2d = 6
d = -3
a + d = 5, so a = 5 + 3 = 8
Now substitute for a, a+d, and a+2d and you'll get the digits and hence the number required
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