the digits of a positive number of 3 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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Let the number be (a-d), a, (a+d)
(a+d)×100+a×10+(a−d) = 111a + 99d (a+d) × 100 + a × 10 + (a−d) = 111a+99don reversing
(a−d)×100+a×10+(a+d)a+d+a+a−d
⟹=111a−99d=15a=5 and d = 3(a−d)×100+a×10+(a+d)=111a−99da+d+a+a−d=15
⟹a=5 and d = 3Hence the number is 111×5+99×3=852
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(a+d)×100+a×10+(a−d) = 111a + 99d (a+d) × 100 + a × 10 + (a−d) = 111a+99don reversing
(a−d)×100+a×10+(a+d)a+d+a+a−d
⟹=111a−99d=15a=5 and d = 3(a−d)×100+a×10+(a+d)=111a−99da+d+a+a−d=15
⟹a=5 and d = 3Hence the number is 111×5+99×3=852
Hope it helps..
Plz mark it as brainliest if it was helpful.
:)
Takshika:
Plz mark it as brainliest if it was helpful.
ya sure
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thank u .
is ur answer correct
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