The digits of a (+ve) integer having 3 digits are in A.P and the sum of digits is 15,the number obtained by reversing the digits is 594 less than the original number.find the original no. guys please solve it,i did but stuck with it and my friend as well..
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this is A.P
so, let nos. be (a-d),a and (a+d)
by 1st condition,
(a-d)+a + (a+d)=15
so......a=5
now....bye 2 nd condition,
100(a+d)+10a+(a-d)=100(a-d)+10a+)a+d)
so.....198d=594
so.....d=3
We got a=5 and d=3
since we have asummed numbers as (a-d),a and (a+d),,,,,
we get digits..8,5 and 2
so required number is 852
so, let nos. be (a-d),a and (a+d)
by 1st condition,
(a-d)+a + (a+d)=15
so......a=5
now....bye 2 nd condition,
100(a+d)+10a+(a-d)=100(a-d)+10a+)a+d)
so.....198d=594
so.....d=3
We got a=5 and d=3
since we have asummed numbers as (a-d),a and (a+d),,,,,
we get digits..8,5 and 2
so required number is 852
kvnmurty:
how did you assume that digit a+d is in the 100s place? it could be in the 10s place also or units place also.
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