the four digit no. 2652 is such that any two consecutive digits from it make a multipleof 13. another no. n has this same property, is 100 digits long and begins in a 9. the last digit of n is ? pls explain how this type of ans are solved ?

Dear jayati

any two consecutive digits from it make a multipleof 13

so next digist after 9 should be 1

coz..91 is multiple of 13

then 3i..e 913

next agin 9i..e 9139

then 9139139...

it has 100 digists = 3*33+1

so 913 will be 33 times then 9

answer is 9

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