Question

18) When a two digit number is reversed and added to the original number, the result is divisible by 13. How
many such numbers are possible?
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Answer 1

Answer:

this is nice question

Step-by-step explanation:

there is no any no. that is divide by 13 in your given situation

Answer 2

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$\large\mathcal\red{solution}$

let the unit place digit is X and

2nd place digit is y .......

therefore the number is =10y+X

now the reverse number is =10x+y

therefore.....

10y+X+10x+y=11x+11y=11(X+y)

now this is divisible by 13 .....

so ...

X+y should be a minimum value of 13 to be divisible by 13....

so ....the possible combination are ...

9,4;,8,5;7,6;6,7;5,8;4,9

therefore..there are 6 such numbers are possible...

$\underline{\large\mathcal\red{hope\: this \: helps \:you......}}$