the digits of a positive integer having three digits are in AP 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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Let the number be (a-d), a, (a+d)
(a+d)×100+a×10+(a−d)=111a+99d
on reversing
(a-b) 100 + a x 10 + (a+b) = 111a-99b
a+d+a+a-d=15
a=5 and d = 3
hence = 111 x 5 + 99 x 3 = 852
i hope its help you
mark brainliest
(a+d)×100+a×10+(a−d)=111a+99d
on reversing
(a-b) 100 + a x 10 + (a+b) = 111a-99b
a+d+a+a-d=15
a=5 and d = 3
hence = 111 x 5 + 99 x 3 = 852
i hope its help you
mark brainliest
tinu21:
markcbrainliest
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