A 5-digit number overline{abcde},when multipl | vedclass

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(B) Let the number be $N = \overline{abcde} = 10^4a + 10^3b + 10^2c + 10d + e$.Given $9 	imes \overline{abcde} = \overline{edcba}$.Since $9 	imes \o

Vedclass · Accepted Answer

$27$

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(B) Let the number be $N = \overline{abcde} = 10^4a + 10^3b + 10^2c + 10d + e$.Given $9 	imes \overline{abcde} = \overline{edcba}$.Since $9 	imes \overline{abcde}$ is a $5$-digit number,$a$ must be $1$ (if $a \ge 2$,$9 	imes 20000 = 180000$,which is $6$ digits).If $a = 1$,then $9 	imes (1bcde) = edcb1$. This implies $e$ must be $9$ (since $9 	imes 1 = 9$).Now we have $9 	imes (1bcd9) = 9dcb1$.$9 	imes (10000 + 1000b + 100c + 10d + 9) = 90000 + 1000d + 100c + 10b + 1$.$90000 + 9000b + 900c + 90d + 81 = 90000 + 1000d + 100c + 10b + 1$.$9000b + 900c + 90d + 81 = 1000d + 100c + 10b + 1$.Comparing the thousands place: $9b = d$ (with potential carry). Checking $b=0$,$d=9$ (but $e=9$,so $d$ cannot be $9$).Testing $a=1, e=9$: $10989 	imes 9 = 98901$. Here $a=1, b=0, c=9, d=8, e=9$.Sum of digits $= a+b+c+d+e = 1+0+9+8+9 = 27$.

$A$ $5$-digit number $\overline{abcde}$,when multiplied by $9$,gives the $5$-digit number $\overline{edcba}$. The sum of the digits in the number is:

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