Digits of a positive number of three digits number are in A P and thei

1.	Digits of a positive number of three digits number are in A P and their sum is 15. The numberobtain by reversing the digits is 594 less than the original number. Find the number
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

Digits of a positive number of three digits number are in A P and their sum is 15. The numberobtain by reversing the digits is 594 less than the original number. Find the number

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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