Question 1
There is a 6-digit number in which the first and the fourth digit from the first are the same, the second and the fifth digit from the first are the same and the third and the sixth digit from the first are the same. Then the number is always
Solution
Let the number be $$abcabc$$. Its value is $$100000a+10000b+1000c+100a+10b+c=1001(100a+10b+c)$$. Since $$1001=7\times 11\times 13$$, the number is always divisible by $$11$$.
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