Question

Prove that the sum of the given numbers and the numbers obtained by reversing their digits is divisible by 11. (a) 89 (b) ab (c) 69 (d) 54

Answer 1

Answer:

I think The answer is D 54

Answer 2

$\underline{\blue{ Divisibility \:rule \:by \:11 }}$

If the difference between the sum of digits in odd places and even places of a number is a multiple of 11 or equal to zero, then the given number is divisible by 11.

$a) Given \: number \: 89 .$

$Reversing \: the \: digits \:of \: given \:number\\= 98$

$Sum \:of \: the \: numbers\\ = 89 + 98 \\= 117$

$i) Sum\: of \: the \:digits \: at \:odd \:places \\(from \:left) = 1+7 = 8$

$ii) Sum\: of \: the \:digits \: at \:even \:places \\(from \:left) = 8$

$\red{ Difference } \\= 8 -8\\= 0$

$\green {\therefore 117 \: divisible \: by \: 11}$

$b) Given \: number \: ab .$

$Reversing \: the \: digits \:of \: given \:number\\= ba$

$Sum \:of \: the \: numbers\\ = (10a+b)+ (10b+a) \\= 10a+b+10b+a\\= 11a+11b\\= 11(a+b)\:\green { ( Divisible \: by \: 11) }$

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