Find the quotient when difference of 985 and 958 is divided by 9.

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Solution

The numbers of three digits are 985 and 958 in which tens and ones digits are reversed, then
$¯ ¯¯¯¯¯¯ ¯ a b c - ¯ ¯¯¯¯¯¯ ¯ a c b = 9 (b - c)$
$∴ 985 - 958 = 9 (8 - 5) = 9 \times 3$
i.e, it is divisible by 9, then quotient =b-c=8-5=3

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Find the quotient when difference of 985 and 958 is divided by 9.

The numbers of three digits are 985 and 958 in which tens and ones digits are reversed, then ¯¯¯¯¯¯¯¯abc−¯¯¯¯¯¯¯¯acb=9(b−c) ∴985−958=9(8−5)=9×3 i.e, it is divisible by 9, then quotient =b-c=8-5=3

The numbers of three digits are 985 and 958 in which tens and ones digits are reversed, then
$¯ ¯¯¯¯¯¯ ¯ a b c - ¯ ¯¯¯¯¯¯ ¯ a c b = 9 (b - c)$
$∴ 985 - 958 = 9 (8 - 5) = 9 \times 3$
i.e, it is divisible by 9, then quotient =b-c=8-5=3