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Permutations And Combinations
The digits of a positive integer having 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.
Let a  d, a, a + d be the three digits of a three digit number respectively at unit place, ten's place, hundreth place.
∴ The number = 100 (a + d) + 10 + a + a  d
Since the sum of digits is 15
∴ a d + a + a + d = 15 or 3a = 15 or a = 5
The number obtained by reversing the digits = 100 (a  d) + 10a + (a + d) ...(ii)
According to the given condition
100 (a  d) + 10 a + a + d = 100 (a + d) + 10a + a  d  594
or 100a  100d + 10a + a + d = 100a + 100d + 10a + a  d  594
or 198d = 594 or d = 3
∴ Original number = 100 (5 + 3) + 10 (5) + 5  3 = 800 + 50 + 2 = 852
Verification: On reversing the digits, the number is 258
852  258 = 594
Sponsor Area
Determine K, so that K + 2, 4K – 6 and 3K – 2 are three consecutive terms of an A.P.
Sponsor Area